RD SHARMA Solutions for Class 12-science Maths Chapter 20 - Definite Integrals

Chapter 20 - Definite Integrals Exercise Ex. 20.1

Question 1
Solution 1
Question 2

Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5

Solution 5

T h e r e f o r e comma integral subscript 2 superscript 3 fraction numerator x over denominator x squared plus 1 end fraction equals 1 half log 2

Question 6

Solution 6

Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16

Question 17
Solution 17
Question 18

Solution 18
Question 19
Solution 19

Question 20

Solution 20

Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24

Solution 24

Question 25
Solution 25
Question 26
Solution 26
Question 27
Solution 27
Question 28
Solution 28
Question 29

Solution 29

Question 30
Solution 30
Question 31
Solution 31
Question 32
Solution 32
Question 33

Solution 33

Question 34
Solution 34
Question 35

Solution 35

Question 36
Solution 36
Question 37
Solution 37
Question 38
Solution 38
Question 39
Solution 39
Question 40
Solution 40
Question 41
Solution 41
Question 42

Solution 42
Question 43
Solution 43
Question 44

Evaluate the Integral in using substitution.

begin mathsize 12px style integral subscript negative 1 end subscript superscript 1 fraction numerator dx over denominator straight x squared space plus space 2 straight x space plus space 5 end fraction end style

Solution 44

begin mathsize 12px style integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator x squared plus 2 x plus 5 end fraction equals integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator open parentheses x squared plus 2 x plus 1 close parentheses plus 4 end fraction equals integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator open parentheses x plus 1 close parentheses squared plus open parentheses 2 close parentheses squared end fraction
Let space x space plus space 1 space equals space t space rightwards double arrow space d x space equals space d t
When space straight x space equals space minus 1 comma space t space equals space 0 space and space when space x space equals space 1 comma space t space equals space 2
therefore integral subscript negative 1 end subscript superscript 1 fraction numerator d x over denominator open parentheses x plus 1 close parentheses squared plus open parentheses 2 close parentheses squared end fraction equals integral subscript 0 superscript 2 fraction numerator d t over denominator t squared plus 2 squared end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open square brackets 1 half tan to the power of negative 1 end exponent t over 2 close square brackets subscript 0 superscript 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half tan to the power of negative 1 end exponent space 1 minus 1 half tan to the power of negative 1 end exponent space 0
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half open parentheses straight pi over 4 close parentheses equals straight pi over 8 end style

Question 45
Solution 45

Question 46
Solution 46
Question 47
Solution 47
Question 48

Solution 48

Question 49
Solution 49
Question 50
Solution 50
Question 51

Solution 51

Question 52
Solution 52
Question 54

Solution 54

T h e r e f o r e comma space I equals 2 to the power of begin display style 5 over 2 end style end exponent over 3

Question 55

Solution 55

Question 56

  

Solution 56

 

  

  

 

Let cosx =u , Then

  

  

Hence

  

  

  

Question 57

Solution 57

Question 58

Solution 58

Question 59

  

Solution 59

  

Question 60

  

Solution 60

  

  

  

  

  

Given :

  

  

  

  

Question 61

Solution 61

Question 62

  

Solution 62

  

  

  

  

 

Question 63

  

Solution 63

Question 64

  

Solution 64

 

  

  

  

  

  

We know , By reduction formula 

  

For n=2

  

  

For n=4

  

  

Hence

  

  

 

Note: Answer given at back is incorrect.

Question 65

  

Solution 65

Using Integration By parts

  

  

 

 

  

 

Question 66

  

Solution 66

  

  

  

  

  

  

  

  

Question 67

  

Solution 67

 

Note: Answer given in the book is incorrect. 

Question 68

  

Solution 68

 =(1/4)log(2e)

 

Chapter 20 - Definite Integrals Exercise Ex. 20.2

Question 1

Solution 1

Question 2

Solution 2
Question 3
Solution 3
Question 4
Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17

Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20

Question 21
Solution 21
Question 22
Solution 22

Question 23
Solution 23
Question 24

  

Solution 24

Using Integration By parts

  

  

  

  

  

Hence

  

  

  

 

Question 25

  

Solution 25

  

Question 26
Solution 26
Question 27

Evaluate begin mathsize 11px style integral subscript 0 superscript straight pi fraction numerator 1 over denominator 5 plus 3 space cos space straight x end fraction dx end style

Solution 27

Question 28
Solution 28
Question 29

Solution 29

Question 30

?

Solution 30

Question 31

  

Solution 31

  

Question 32
Solution 32

Question 33
Solution 33

Question 34
Solution 34
Question 35
Solution 35
Question 36
Solution 36
Question 37
Solution 37

Question 38
Solution 38

Question 39

begin mathsize 12px style Evaluate space integral subscript negative 1 end subscript superscript 1 space 5 straight x to the power of 4 space square root of straight x to the power of 5 plus 1 end root dx. end style

Solution 39

Question 40

  

Solution 40

  

Question 41

Solution 41

Question 42
Solution 42
Question 43
Solution 43
Question 44

Solution 44
Question 45
Solution 45
Question 46
Solution 46

Question 47
Solution 47
Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

 

Solution 55

Question 56

Solution 56

Question 57

  

Solution 57

  

Question 58

  

Solution 58

 

  

 

Question 59

  

Solution 59

  

Question 60

  

Solution 60

  

  

  

 

Question 61

  

Solution 61

  

 

Question 62

  

Solution 62

  

Chapter 20 - Definite Integrals Exercise Ex. 20.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

  

Solution 5

2x+3 is positive for x>-1.5 . Hence

  

  

  

  

  

 

Question 6

  

Solution 6

  

  

  

  

  

  

  

 

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Evaluate the integral integral subscript 1 superscript 4 open curly brackets open vertical bar x minus 1 close vertical bar plus open vertical bar x minus 2 close vertical bar plus open vertical bar x minus 4 close vertical bar close curly brackets d x

Solution 17

L e t space I equals integral subscript 1 superscript 4 open curly brackets open vertical bar x minus 1 close vertical bar plus open vertical bar x minus 2 close vertical bar plus open vertical bar x minus 4 close vertical bar close curly brackets d x
equals integral subscript 1 superscript 2 open curly brackets open parentheses x minus 1 close parentheses minus open parentheses x minus 2 close parentheses minus open parentheses x minus 4 close parentheses close curly brackets d x plus integral subscript 2 superscript 4 open curly brackets open parentheses x minus 1 close parentheses plus open parentheses x minus 2 close parentheses minus open parentheses x minus 4 close parentheses close curly brackets d x
equals integral subscript 1 superscript 2 open curly brackets open parentheses x minus 1 minus x plus 2 minus x plus 4 close parentheses close curly brackets d x plus integral subscript 2 superscript 4 open curly brackets open parentheses x minus 1 plus x minus 2 minus x plus 4 close parentheses close curly brackets d x
equals integral subscript 1 superscript 2 open parentheses 5 minus x close parentheses d x plus integral subscript 2 superscript 4 open parentheses x plus 1 close parentheses d x
equals open square brackets 5 x minus x squared over 2 close square brackets table row 2 row 1 end table plus open square brackets x squared over 2 plus x close square brackets table row 4 row 2 end table
equals open square brackets 10 minus 2 minus 5 plus 1 half close square brackets plus open square brackets 8 plus 4 minus 2 minus 2 close square brackets
equals 7 over 2 plus 8
I equals 23 over 2

Question 18

Solution 18

Question 19

Solution 19

Question 20

  

Solution 20

  

  

  

  

  

Question 21

  

Solution 21

  

For

  

Using Integration By parts

  

  

  

  

  

For

  

Using Integration By parts

  

  

  

  

Question 22

  

Solution 22

  

  

 

Question 23

  

Solution 23

  

Question 24

  

Solution 24

Question 25

  

Solution 25

  

  

Question 27

  

Solution 27

[x]=0 for 0 < x

and [x]=1 for 1< x < 2

Hence

Question 28

  

Solution 28

  

Question 26

Evaluate the following integrals:

begin mathsize 12px style integral from negative straight pi divided by 2 to straight pi divided by 2 of fraction numerator negative straight pi divided by 2 over denominator square root of cosx space sin squared straight x end root end fraction dx end style


Solution 26

NOTE: Answer not matching with back answer.

Chapter 20 - Definite Integrals Exercise Ex. 20.4A

Question 1

  

Solution 1

We know

  

Hence

  

We know

  

  

If

  

Then also

  

Hence

  

Question 2

  

Solution 2

We know

  

Hence

  

If

  

Then

  

Question 3

  

Solution 3

We know

  

Hence

  

If

  

Then

  

So

  

 

Question 4

  

Solution 4

We know

  

Hence

  

If

  

Then

  

Hence

Question 5

  

Solution 5

We know

  

Hence

  

If

  

Then

  

So

  

We know

 

If f(x) is even

  

If f(x) is odd

  

Here

  

f(x) is even, hence

  

 

Note: Answer given in the book is incorrect.

Question 6

  

Solution 6

We know

  

Hence

  

If

  

Then

  

So

Question 7

  

Solution 7

We know

  

Hence

  

If

  

Then

  

So

Question 8

  

Solution 8

We know

  

Hence

  

If

  

Then

  

 

So

 

Note: Answer given in the book is incorrect. 

Question 9

  

Solution 9

  

If f(x) is even

  

If f(x) is odd

  

Here

  is odd and

  is even. Hence

  

Question 10

  

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

  

Solution 16

  

 

Chapter 20 - Definite Integrals Exercise Ex. 20.4B

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

B

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Error: the service is unavailable.

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

  

Solution 18

  

Question 19

  

Solution 19

  

  

  

Hence

  

Question 20

  

Solution 20

  

Question 21

  

Solution 21

  

Now

  

Let cosx=t

  

  

  

  

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25 (i)

Solution 25 (i)

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

  

Solution 29

  

Question 30

  

Solution 30

  

Question 31

  

Solution 31

  

Question 32 (i)

  

Solution 32 (i)

  

Question 33

Solution 33

Question 34

Evaluate the integral integral subscript 0 superscript 1 log open parentheses 1 over x minus 1 close parentheses d x

Solution 34

L e t space I equals integral subscript 0 superscript 1 log open parentheses 1 over x minus 1 close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses fraction numerator 1 minus x over denominator x end fraction close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses 1 minus x close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
A p p l y i n g space t h e space p r o p e r t y comma space integral subscript 0 superscript a f open parentheses x close parentheses d x equals integral subscript 0 superscript a f open parentheses a minus x close parentheses d x
T h u s comma space I equals integral subscript 0 superscript 1 log open parentheses 1 minus open parentheses 1 minus x close parentheses close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses 1 minus 1 plus x close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
equals integral subscript 0 superscript 1 log open parentheses x close parentheses d x minus integral subscript 0 superscript 1 log open parentheses x close parentheses d x
equals 0

Question 35

  

Solution 35

  

Question 36

  

Solution 36

  

  

  

  

Question 37

  

Solution 37

  

Question 38

  

Solution 38

We know

  

Also here

  

So

  

  

Hence

Question 39

  

Solution 39

  

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45


Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

  

Solution 49

  

Chapter 20 - Definite Integrals Exercise Ex. 20RE

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Evaluate the following integrals

begin mathsize 12px style integral subscript 0 superscript 1 cos to the power of negative t end exponent x d x end style

Solution 4

begin mathsize 12px style integral subscript 0 superscript 1 cos to the power of negative 1 end exponent xdx equals integral subscript 0 superscript 1 cos to the power of negative 1 end exponent straight x times 1 dx
space space space space space space space space space space space space equals cos to the power of negative 1 end exponent straight x integral subscript 0 superscript 1 dx minus integral subscript 0 superscript 1 open curly brackets straight d over dx cos to the power of negative 1 end exponent straight x integral dx close curly brackets dx space space space space space space space open square brackets Using space Partial space Fraction close square brackets
space space space space space space space space space space space space equals xcos to the power of negative 1 end exponent straight x vertical line subscript 0 superscript 1 minus integral subscript 0 superscript 1 open curly brackets negative fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction close curly brackets dx
space space space space space space space space space space space space equals integral subscript 0 superscript 1 fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction dx
space space space space space space space space space space space space equals integral subscript 0 superscript 1 tdt over straight t dx space space space space space space space space space space space space space space space space space space space space open square brackets 1 minus straight x squared equals straight t squared close square brackets
space space space space space space space space space space space space equals 1 end style

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

begin mathsize 12px style We space have comma
integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx over denominator square root of 1 plus cos end root end fraction dx
we space konw space that
space sin space straight x space equals space 2 sin straight x over 2 cos straight x over 2 space and space 1 plus cos space straight x equals 2 cos squared straight x over 2
therefore integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx over denominator square root of 1 space plus space cos end root end fraction
equals integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 2 sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style over denominator square root of 2 cos squared begin display style straight x over 2 end style end root end fraction dx
equals fraction numerator 2 over denominator square root of 2 end fraction integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style over denominator cos begin display style straight x over 2 end style end fraction dx
equals square root of 2 integral subscript 0 superscript straight pi over 2 end superscript sin space straight x over 2 space dx
equals square root of 2 space open square brackets negative 2 cos straight x over 2 close square brackets subscript 0 superscript straight pi over 2 end superscript
equals square root of 2 open square brackets 1 minus fraction numerator 1 over denominator square root of 2 end fraction close square brackets
equals 2 open parentheses square root of 2 minus 1 close parentheses
therefore integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator square root of 1 plus cosx end root end fraction dx equals 2 open parentheses square root of 2 minus 1 close parentheses end style

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

begin mathsize 12px style Evaluate space integral subscript 1 superscript 2 1 over straight x squared straight e to the power of fraction numerator negative 1 over denominator straight x end fraction end exponent dx end style

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Evaluate the integral integral subscript 0 superscript pi over 2 end superscript x squared cos 2 x d x

Solution 21

begin mathsize 12px style integral subscript 0 superscript straight pi over 2 end superscript straight x squared cos space 2 xdx equals straight x squared integral subscript 0 superscript straight pi over 2 end superscript cos space 2 xdx minus integral subscript 0 superscript straight pi over 2 end superscript open curly brackets straight d over dx straight x squared integral cos space 2 xdx close curly brackets dx open square brackets Using space by space Parts close square brackets
space space space space space space equals straight x squared fraction numerator sin 2 straight x over denominator 2 end fraction vertical line subscript 0 superscript straight pi over 2 end superscript minus integral subscript 0 superscript straight pi over 2 end superscript open curly brackets 2 straight x fraction numerator sin space 2 straight x over denominator 2 end fraction close curly brackets dx
space space space space space space equals integral subscript 0 superscript straight pi over 2 end superscript open curly brackets straight x space sin space 2 straight x close curly brackets dx
space space space space space space equals straight x fraction numerator cos 2 straight x over denominator 2 end fraction vertical line subscript 0 superscript straight pi over 2 end superscript plus 1 half fraction numerator sin 2 straight x over denominator 2 end fraction vertical line subscript 0 superscript straight pi over 2 end superscript space space space space space space space space space space space space space space space space space space space space space space space space open square brackets Using space by space Parts space again close square brackets
space space space space space space equals negative straight pi over 4 end style

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Evaluate the following integrals

begin mathsize 12px style integral subscript 0 superscript straight pi over 4 end superscript tan to the power of 4 xdx end style

Solution 28

begin mathsize 12px style integral subscript 0 superscript straight pi over 4 end superscript tan to the power of 4 xdx equals integral subscript 0 superscript straight pi over 4 end superscript tan squared straight x space tan squared xdx
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript tan squared straight x open parentheses sec squared straight x minus 1 close parentheses dx open square brackets tan squared straight x equals sec squared space straight x minus 1 close square brackets
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript open parentheses tan squared xsec squared straight x minus tan squared straight x close parentheses dx
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript open parentheses tan squared xsec squared straight x minus sec squared straight x plus 1 close parentheses dx open square brackets tan squared straight x equals sec squared straight x minus 1 close square brackets
space space space space space space space space space space space space equals integral subscript 0 superscript straight pi over 4 end superscript tan squared xsec squared xdx minus integral subscript 0 superscript straight pi over 4 end superscript sec squared xdx plus integral subscript 0 superscript straight pi over 4 end superscript dx
space space space space space space space space space space space space equals 1 third tan cubed straight x minus tanx plus straight x vertical line subscript 0 superscript straight pi over 4 end superscript
space space space space space space space space space space space space equals straight pi over 4 minus 2 over 3 end style

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Evaluate the following integrals

begin mathsize 12px style integral subscript 0 superscript 2 straight pi end superscript cos to the power of 7 xdx end style

Solution 38

begin mathsize 12px style Let space straight f left parenthesis straight x right parenthesis equals cos to the power of 7 straight x. space Then space straight f left parenthesis 2 straight pi minus straight x right parenthesis equals open curly brackets cos left parenthesis 2 straight pi minus straight x right parenthesis close curly brackets to the power of 7 equals cos to the power of 7 straight x
space space space space space space space space space integral subscript 0 superscript 2 straight pi end superscript cos to the power of 7 xdx equals 2 integral subscript 0 superscript straight pi cos to the power of 7 xdx
Now
space space space space space space space space space straight f left parenthesis straight pi minus straight x right parenthesis equals open curly brackets cos open parentheses straight pi minus straight x close parentheses close curly brackets to the power of 7 equals negative cos to the power of 7 straight x equals negative straight f left parenthesis straight x right parenthesis
Therefore
space space space space space space space space integral subscript 0 superscript straight pi cos to the power of 7 xdx equals 0
Hence
space space space space space space space space integral subscript 0 superscript 2 straight pi end superscript cos to the power of 7 xdx equals 2 integral subscript 0 superscript straight pi cos to the power of 7 xdx equals 0 end style

Question 39

Solution 39

begin mathsize 12px style Let comma
straight I equals integral subscript 0 superscript straight a fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx space space space space space space space space space space space space space space space space space minus negative negative left parenthesis straight i right parenthesis
therefore straight l minus integral subscript 0 superscript straight a fraction numerator square root of straight a minus straight x end root over denominator square root of straight a minus straight x end root plus square root of straight x end fraction dx space space space space space space space space space space space space space minus negative negative left parenthesis ii right parenthesis open square brackets therefore integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis dx minus integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis dx close square brackets
Add space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
2 straight l equals integral subscript 0 superscript straight a fraction numerator square root of straight x plus square root of straight a minus straight x end root over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx
therefore 2 straight l equals integral subscript 0 superscript straight a dx space space space space space space space equals open square brackets straight x close square brackets subscript 0 superscript straight a equals straight a
rightwards double arrow straight l equals straight a over 2
therefore integral subscript 0 superscript straight a fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx equals straight a over 2 end style

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

begin mathsize 12px style integral from fraction numerator negative straight pi over denominator 4 end fraction to straight pi over 4 of open vertical bar tan space straight x close vertical bar dx equals negative integral from fraction numerator negative straight pi over denominator 4 end fraction to 0 of tan space straight x space dx space plus integral from 0 to straight pi over 4 of tan space straight x space dx space space space space space space open square brackets table row cell because tan space straight x greater or equal than 0 space space space end cell cell when space 0 less than straight x less than straight pi over 4 end cell row cell tan space straight x space less or equal than 0 end cell cell when space straight pi over 4 less than straight x less than 0 end cell end table close square brackets
equals open square brackets log space seg space straight x close square brackets subscript fraction numerator negative straight pi over denominator 4 end fraction end subscript superscript 0 minus open square brackets log space sec space straight x close square brackets subscript 0 superscript fraction numerator negative straight pi over denominator 4 end fraction end superscript
equals open square brackets 0 minus log fraction numerator 1 over denominator square root of 2 end fraction close square brackets minus open square brackets log fraction numerator 1 over denominator square root of 2 end fraction minus 0 close square brackets
equals negative 2 log space fraction numerator 1 over denominator square root of 2 end fraction
equals 2 space straight x space 1 half log 2
equals space log 2
therefore integral from fraction numerator negative straight pi over denominator 4 end fraction to straight pi over 4 of open vertical bar tan space straight x close vertical bar dx space equals space log 2 end style

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

Question 65

Solution 65

Question 66

Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Chapter 20 - Definite Integrals Exercise Ex. 20.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

begin mathsize 12px style Evaluate integral subscript 1 superscript 3 left parenthesis 3 straight x minus 2 right parenthesis dx end style

Solution 3

begin mathsize 12px style We space have comma
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus negative negative negative negative plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
where space straight h space equals fraction numerator straight b minus straight a over denominator straight n end fraction
Hear space straight a equals 1 comma space straight b equals 3 space and space straight f space left parenthesis straight x right parenthesis equals 3 straight x space minus 2
straight h equals 2 over straight n rightwards double arrow nh equals 2
Thus comma space we space have comma
straight l equals integral subscript 1 superscript 3 left parenthesis 3 straight x minus space 2 right parenthesis dx end style

begin mathsize 12px style rightwards double arrow straight l equals limit as straight h rightwards arrow 0 of open square brackets straight f left parenthesis 1 right parenthesis plus straight f left parenthesis 1 plus straight h right parenthesis plus straight f left parenthesis 1 plus 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis 1 plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
equals limit as straight h rightwards arrow 0 of straight h open square brackets 1 plus open curly brackets 3 left parenthesis 1 plus straight h right parenthesis minus 2 close curly brackets plus open curly brackets 3 left parenthesis 1 plus 2 straight h right parenthesis minus 2 close curly brackets plus negative negative negative negative plus open curly brackets 3 left parenthesis 1 plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis minus 2 close curly brackets close square brackets
equals limit as straight h rightwards arrow 0 of straight h open square brackets straight n plus 3 straight h left parenthesis 1 plus 2 plus 3 plus negative negative negative negative left parenthesis straight n minus 1 right parenthesis right parenthesis close square brackets
equals limit as straight h rightwards arrow 0 of straight h open square brackets straight n plus 3 straight h fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
because straight h equals 2 over straight n space space space space space space space space space therefore if space straight h rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
therefore limit as straight n rightwards arrow 0 of 2 over straight n open square brackets straight n plus 6 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
equals limit as straight n rightwards arrow 0 of space 2 plus 6 over straight n squared straight n squared open parentheses 1 minus 1 over straight n right parenthesis close parentheses
equals limit as straight n rightwards arrow 0 of space 2 space plus space 6 space equals space 8
therefore integral subscript 1 superscript 3 left parenthesis 3 straight x minus 2 right parenthesis dx equals space 8 end style

Question 4

Solution 4

Question 5

Solution 5

begin mathsize 12px style we space have comma
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f open parentheses straight a close parentheses plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
where space straight h space equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Hear space straight a space equals space 0 comma space straight b space equals space 5 space and space straight f left parenthesis straight x right parenthesis equals left parenthesis straight x plus 1 right parenthesis
therefore straight h equals 5 over straight n rightwards double arrow nh equals 5
Thus comma space we space have comma
straight l equals integral subscript 0 superscript 5 left parenthesis straight x plus 1 right parenthesis dx
straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis straight h right parenthesis plus straight f left parenthesis 2 straight h right parenthesis plus negative negative negative negative negative straight f open curly brackets open parentheses straight n minus 1 close parentheses straight h close curly brackets close square brackets
straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets 1 plus left parenthesis straight h plus 1 right parenthesis plus left parenthesis 2 straight h plus 1 right parenthesis plus negative negative negative negative negative left parenthesis left parenthesis straight n minus 1 right parenthesis straight h plus 1 right parenthesis close square brackets
straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight n plus straight h left parenthesis 1 plus 2 plus 3 plus negative negative negative negative negative left parenthesis straight n minus 1 right parenthesis right parenthesis close square brackets
because straight h equals 5 over straight n space and space if space straight h space rightwards arrow 0 comma space straight n rightwards arrow infinity
straight l equals limit as straight h rightwards arrow 0 of space 5 over straight n open square brackets straight n plus 5 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
straight l equals limit as straight h rightwards arrow 0 of space 5 plus fraction numerator 25 over denominator 2 straight n squared end fraction straight n squared open parentheses 1 minus 1 over straight n close parentheses
equals 5 plus 25 over 2
therefore integral subscript 0 superscript 5 left parenthesis straight x plus 1 right parenthesis dx equals 35 over 2 end style

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

begin mathsize 12px style We space have
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a space plus space straight h right parenthesis plus straight f left parenthesis straight a space plus space 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis straight n minus straight a right parenthesis straight h close square brackets
where space straight h space equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Hear space straight a space equals 0 comma space straight b equals space 2 space and space straight f left parenthesis straight x right parenthesis space equals straight x squared plus space 4
therefore straight h equals 2 over straight n rightwards double arrow nh equals 2
Thus comma space we space have comma
straight l equals integral subscript 0 superscript 2 open parentheses straight x squared plus space 4 close parentheses dx
equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis straight h right parenthesis plus straight f left parenthesis 2 straight h right parenthesis plus negative negative negative negative negative straight f left parenthesis 0 plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
equals limit as straight h rightwards arrow 0 of space straight h open square brackets 4 left parenthesis straight h squared plus 4 right parenthesis plus open curly brackets left parenthesis 2 straight h right parenthesis squared plus close curly brackets plus negative negative negative negative negative open curly brackets left parenthesis straight n minus 1 right parenthesis straight h squared plus 4 close curly brackets close square brackets
because straight h equals 2 over straight n space & space if space straight h space rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
equals limit as straight n rightwards arrow infinity of space 2 over straight n open square brackets 4 straight n plus 4 over straight n squared fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction close square brackets
equals limit as straight n rightwards arrow infinity of space 8 space plus space fraction numerator 4 over denominator 3 straight n squared end fraction straight n cubed open parentheses 1 minus 1 over straight n close parentheses open parentheses 2 minus 1 over straight n close parentheses
equals 8 space plus space fraction numerator 4 space straight x space 2 over denominator 3 end fraction equals 32 over 3
therefore integral subscript 0 superscript 2 open parentheses straight x squared space plus space 4 close parentheses dx equals 32 over 3 space end style

Question 13

Solution 13

Question 14

Solution 14

Question 15

begin mathsize 12px style Evahuate space the space following space in space tegrals space as space limit space of space sums
integral subscript 0 superscript 2 straight e to the power of straight x dx end style

Solution 15

Question 16

Solution 16

Question 17

Solution 17

begin mathsize 12px style We space have comma
space space space space space space space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a space plus space straight h right parenthesis plus straight f left parenthesis straight a space plus space 2 straight h right parenthesis space plus....... plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets comma space where space straight h space equals fraction numerator straight b minus straight a over denominator straight n end fraction.
Since space we space have space to space find space integral subscript straight a superscript straight b cosxdx
we space have comma space space space space space space space straight f left parenthesis straight x right parenthesis equals cos space straight x
therefore space space space space space space space straight l equals integral subscript straight a superscript straight b cosxdx
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets cos space straight a space plus cos left parenthesis straight a space plus space straight h right parenthesis space plus cos left parenthesis straight a plus 2 straight h right parenthesis plus..... plus cos left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis space straight h right parenthesis close square brackets
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets fraction numerator cos left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis begin display style straight h over 2 end style right parenthesis sin begin display style nh over 2 end style over denominator sin begin display style straight h over 2 end style end fraction close square brackets equals limit as straight h rightwards arrow 0 of space straight h open square brackets fraction numerator cos open parentheses straight a plus begin display style nh over 2 end style minus begin display style straight h over 2 end style space sin begin display style nh over 2 end style close parentheses over denominator sin begin display style straight h over 2 end style end fraction close square brackets
rightwards double arrow space space space space space straight l equals limit as straight h rightwards arrow 0 of space straight h open square brackets fraction numerator cos open parentheses straight a plus begin display style fraction numerator straight b minus straight a over denominator 2 end fraction end style minus begin display style straight h over 2 end style close parentheses sin open parentheses begin display style fraction numerator straight b minus straight a over denominator 2 end fraction end style close parentheses over denominator sin begin display style straight h over 2 end style end fraction close square brackets space space space space space space space space space space space space space space space space space space space open square brackets because nh space equals space straight b minus straight a close square brackets
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space open square brackets fraction numerator begin display style straight h over 2 end style over denominator sin begin display style straight h over 2 end style end fraction space straight x space 2 cos open parentheses fraction numerator straight a plus straight b over denominator 2 end fraction minus straight h over 2 close parentheses sin open parentheses fraction numerator straight b minus straight a over denominator 2 end fraction close parentheses close square brackets
rightwards double arrow space space space space space space straight l equals limit as straight h rightwards arrow 0 of space open square brackets fraction numerator begin display style straight h over 2 end style over denominator sin begin display style straight h over 2 end style end fraction close square brackets straight x limit as straight h rightwards arrow 0 of space 2 cos open parentheses fraction numerator straight a plus straight b over denominator 2 end fraction minus straight h over 2 close parentheses sin open parentheses fraction numerator straight b minus straight a over denominator 2 end fraction close parentheses equals 2 cos open parentheses fraction numerator straight a plus straight b over denominator 2 end fraction close parentheses sin open parentheses fraction numerator straight b minus straight a over denominator 2 end fraction close parentheses
rightwards double arrow space space space space space straight l equals sin space straight b space equals sin space straight alpha space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because 2 space cos space straight A space sin space straight B equals sin left parenthesis straight A minus straight B right parenthesis minus space sin space left parenthesis straight A space plus space straight B right parenthesis close square brackets end style

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

begin mathsize 12px style We space have comma
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis dx equals limit as straight h rightwards arrow 0 of open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus negative negative negative negative straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h close square brackets
space space space space space space space space space space space where space straight h space equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Thus comma space we space have comma
straight l equals integral subscript 0 superscript 2 open parentheses straight x squared plus space straight x close parentheses dx
equals limit as straight h rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis 0 plus straight h right parenthesis plus straight f left parenthesis 0 plus 2 straight h right parenthesis plus negative negative negative negative negative straight f open curly brackets open parentheses straight n minus 1 close parentheses straight h close curly brackets close square brackets
equals limit as straight h rightwards arrow 0 of space straight h open square brackets 0 space plus space left parenthesis straight h squared plus space straight h right parenthesis plus open curly brackets left parenthesis 2 straight h right parenthesis squared plus 2 straight h close curly brackets plus negative negative negative negative negative close square brackets
equals limit as straight h rightwards arrow 0 of space straight h open square brackets open curly brackets straight h squared open parentheses 1 plus 2 squared plus 3 squared plus negative negative negative negative left parenthesis straight n minus 1 right parenthesis squared close parentheses close curly brackets plus straight h open curly brackets 1 plus 2 plus 3 minus negative negative negative left parenthesis straight n minus 1 right parenthesis close curly brackets close square brackets
because straight h equals 2 over straight n space & space if space straight h rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
equals limit as straight n rightwards arrow infinity of space 2 over straight n open square brackets 4 over straight n squared fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction plus 2 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
equals limit as straight h rightwards arrow infinity of space fraction numerator 4 over denominator 3 straight n cubed end fraction straight n cubed space open parentheses 1 minus 1 over straight n close parentheses open parentheses 2 minus 1 over straight n close parentheses plus 2 over straight n squared straight n squared open parentheses 1 minus 1 over straight n close parentheses
equals 8 over 3 plus 2 equals 14 over 3
therefore integral subscript 0 superscript 2 open parentheses straight x squared plus straight x close parentheses dx equals 14 over 3 end style

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Evaluate the following in tegrals as limit of sums

begin mathsize 12px style integral subscript 0 superscript 2 left parenthesis straight x squared minus straight x right parenthesis dx end style

Solution 31

begin mathsize 12px style We space have
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis equals dx limit as straight k rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus.... plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
where space straight h equals space fraction numerator straight b minus straight a over denominator straight n end fraction
Here
space space space straight a equals 0 comma space straight b equals 2 space and space straight f left parenthesis straight x right parenthesis equals straight x squared minus straight x
Now
space space space straight h equals 2 over straight n
space space space nh space equals space 2
Thus comma space we space have
straight l equals integral subscript 0 superscript 2 left parenthesis straight x squared minus straight x right parenthesis dx
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 0 right parenthesis plus straight f left parenthesis straight h right parenthesis plus straight f left parenthesis 2 straight h right parenthesis plus.... plus straight f left parenthesis left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open curly brackets left parenthesis 0 right parenthesis squared minus left parenthesis 0 right parenthesis close curly brackets plus open curly brackets left parenthesis straight h right parenthesis squared minus left parenthesis straight h right parenthesis close curly brackets plus open curly brackets left parenthesis 2 straight h right parenthesis squared minus left parenthesis 2 straight h right parenthesis close curly brackets plus...... close square brackets
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open parentheses open parentheses straight h close parentheses squared plus left parenthesis 2 straight h right parenthesis squared plus.... close parentheses minus open curly brackets left parenthesis straight h right parenthesis plus left parenthesis 2 straight h right parenthesis plus.... close curly brackets close square brackets
space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets straight h squared open parentheses 1 plus 2 squared plus 3 cubed plus..... plus left parenthesis straight n minus 1 right parenthesis squared close parentheses minus straight h open curly brackets 1 plus 2 plus 3 plus.... plus left parenthesis straight n minus 1 right parenthesis close curly brackets close square brackets
space space space because straight h equals 2 over straight n & if space straight h rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
space space space equals limit as straight k rightwards arrow infinity of 2 over straight n open square brackets 9 over straight n squared fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction minus 9 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction close square brackets
space space space equals 2 over 3 end style

Question 32

Evaluate the following in tegrals as limit of sums

begin mathsize 12px style integral subscript t superscript 3 left parenthesis 2 straight x squared plus 5 straight x right parenthesis dx end style

Solution 32

begin mathsize 12px style We space have
integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis equals dx limit as straight k rightwards arrow 0 of space straight h open square brackets straight f left parenthesis straight a right parenthesis space plus space straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus.... plus straight f left parenthesis straight a plus left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets
where space straight h space equals fraction numerator straight b minus straight a over denominator straight n end fraction
Here
space space space space straight a equals 1 comma space straight b equals 3 space and space straight f left parenthesis straight x right parenthesis equals 2 straight x squared space plus space 5 straight x
Now
space space space space straight h space equals space 2 over straight n
space space space space nh equals 2
Thus comma space we space have
straight I equals integral subscript straight t superscript 3 left parenthesis 2 straight x squared space plus space 5 straight x right parenthesis dx
space space space equals limit as straight k rightwards arrow 0 of space straight h open square brackets straight f left parenthesis 1 right parenthesis plus straight f left parenthesis 1 plus straight h right parenthesis plus straight f left parenthesis 1 space plus space 2 straight h right parenthesis plus...... plus straight f left parenthesis plus 1 left parenthesis straight n minus 1 right parenthesis straight h right parenthesis close square brackets end style

begin mathsize 12px style space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open parentheses 2 plus 5 close parentheses plus open curly brackets 2 left parenthesis 1 plus straight h right parenthesis squared space plus space 5 left parenthesis 1 plus straight h right parenthesis close curly brackets plus open curly brackets 2 left parenthesis 1 plus 2 straight h right parenthesis squared plus 5 left parenthesis 1 plus 2 straight h right parenthesis close curly brackets plus.... close square brackets
space equals limit as straight k rightwards arrow 0 of space straight h open square brackets open parentheses 7 straight n plus 9 straight h left parenthesis 1 plus 2 plus 3 plus..... close parentheses plus 2 straight h squared left parenthesis 1 plus 2 squared plus 3 cubed plus..... right parenthesis right parenthesis close square brackets
space because straight h equals 2 over straight n & if space straight h rightwards arrow 0 rightwards double arrow straight n rightwards arrow infinity
space space equals limit as straight k rightwards arrow 0 of space 2 over straight n open square brackets 7 straight n plus 18 over straight n fraction numerator straight n left parenthesis straight n minus 1 right parenthesis over denominator 2 end fraction plus 8 over straight n squared fraction numerator straight n left parenthesis straight n minus 1 right parenthesis left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction close square brackets
space space equals 112 over 3 end style

Question 33

Solution 33

Chapter 20 - Definite Integrals Exercise MCQ

Question 1

Mark the correct alternative in each of the following:

(a) p/2

(b) p/4

(c) p/6

(d) p/8

Solution 1

Correct option: (d)

  

Question 2

(a) 0

(b) 1/2

(c) 2

(d) 3/2

Solution 2

Correct option: (c)

  

Question 3

 

Solution 3

Correct option: (a)

  

Question 4

(a) 0

(b) 2

(c) 8

(d) 4

 

Solution 4

Correct option: (c)

  

Note: Answer not matching with back answer.

Question 5

(a) 0

(b) p/2

(c) p/4

(d) None of these

 

Solution 5

Correct option:(c)

 

Question 6

(a) Log 2-1

(b) Log 2

(c) Log 4-1

(d) -log 2

 

Solution 6

Correct option: (b)

  

Question 7

(a) 2

(b) 1

(c) p/4

(d) p2/8

 

Solution 7

Correct option: (a)

  

Question 8

 

Solution 8

Correct option: (d)

  

 

Question 9

 

Solution 9

Correct option: (b)

  

Question 10

 

Solution 10

Note: Answer not matching with back answer.

Question 11

 

 

Solution 11

Correct option: (a)

  

Question 12

(a) p/3

(b) p/6

(c) p/12

(d) p/2

 

Solution 12

Correct option:(c)

  

 

Question 13

 

Solution 13

Correct option: (a)

  

Question 14

(a) 1

(b) e-1

(c) e+1

(d) 0

 

Solution 14

Correct option: (a)

  

 

Question 15

 

Solution 15

Correct option:(a)

  

Question 16

 

Solution 16

Correct option:(a)

  

 

Question 17

 

Solution 17

Correct option:(b)

  

Question 18

(a) 1

(b) 2

(c) -1

(d) -2

 

Solution 18

Correct option: (b)

  

Question 19

 

Solution 19

Correct option: (a)

  

Question 20

(a) 1

(b) e-1

(c) 0

(d) -1

 

Solution 20

Correct option: (b)

  

 

Question 21

 

Solution 21

Correct option:(b)

  

 

Question 22

begin mathsize 12px style If integral from 0 to 1 of straight f open parentheses straight x close parentheses dx space equals space 1 comma integral from 0 to 1 of cross times straight f open parentheses straight x close parentheses dx equals straight a comma integral from 0 to 1 of straight x squared straight f open parentheses straight x close parentheses dx space equals straight a squared
then space integral from 0 to 1 of open parentheses straight a minus straight x close parentheses squared straight f open parentheses straight x close parentheses dx space space equals end style

 

(a) 4a2 

(b) 0

(c) 2a2 

(d) None of these

Solution 22

Correct option: (b)

  

 

Question 23

 

Solution 23

Correct option: (c)

  

Question 24

 

Solution 24

Correct option: (b)

  

 

Question 25

(a) -2

(b) 2

(c) 0

(d) 4

 

Solution 25

Correct option: (b)

  

Question 26

 

Solution 26

Correct option:(c)

  

Question 27

 

Solution 27

Correct option: (b)

  

Question 28

 

Solution 28

Correct option: (d)

 

Note: Question is modified.

 

Question 29

Solution 29

Correct option: (c)

  

Question 30

(a) 4

(b) 2

(c) -2

(d) 0

 

Solution 30

Correct option:(a)

  

Question 31

(a) 0

(b) 1

(c) begin mathsize 12px style straight pi end style/2

(d) begin mathsize 12px style straight pi end style/4

 

Solution 31

Correct option:(d)

  

 

Question 32

(a) begin mathsize 12px style straight pi end style

(b) begin mathsize 12px style straight pi end style/2

(c) begin mathsize 12px style straight pi end style/3

(d) begin mathsize 12px style straight pi end style/4

 

Solution 32

Correct option: (d)

  

 

Question 33

(a) 0

(b) begin mathsize 12px style straight pi end style

(c) begin mathsize 12px style straight pi end style/2

(d) begin mathsize 12px style straight pi end style/4

 

Solution 33

Correct option:(c)

  

Note: Answer not matching with back answer.

Question 34

(a) begin mathsize 12px style straight pi end style/4

(b) begin mathsize 12px style straight pi end style/2

(c) begin mathsize 12px style straight pi end style

(d) 1

 

Solution 34

Correct option:(d)

  

Note: Answer not matching with back answer.

Question 35

(a) begin mathsize 12px style straight pi end style 

(b) begin mathsize 12px style straight pi end style/2

(c) 0

(d) 2begin mathsize 12px style straight pi end style 

 

Solution 35

Correct option: (c)

  

 

Question 36

(a) begin mathsize 12px style straight pi end style/4

(b) begin mathsize 12px style straight pi end style/8

(c) begin mathsize 12px style straight pi end style/2

(d) 0

 

Solution 36

Correct option: (a)

  

Question 37

(a) begin mathsize 12px style straight pi space in space 2 end style

(b) begin mathsize 12px style negative straight pi space in space 2 end style

(c) 0

(d) begin mathsize 12px style negative straight pi over 2 in space 2 end style

 

Solution 37

Correct option:(d)

  

NOTE: Answer is not matching with back answer. 

 

Question 38

 

Solution 38

Correct option: (c)

  

Question 39

 

Solution 39

Correct option: (d)

Question 40

(a) 1

(b) 0

(c) -1

(d) p/4

 

Solution 40

Correct option: (b)

  

Question 41

(a) 2

(b) 

(c) 0

(d) -2

Solution 41

Correct option: (c)

  

 

Question 42

 

(a) 0

(b) 2

(c) begin mathsize 12px style straight pi end style

(d) 1

Solution 42

Correct option: (c)

  

Chapter 20 - Definite Integrals Exercise Ex. 20VSAQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Note: The lower limit is incorrect in textbook. Consider the lower limit as '0'.

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

G i v e n space t h a t space integral subscript 0 superscript a 3 x squared d x equals 8
rightwards double arrow open square brackets fraction numerator 3 x cubed over denominator 3 end fraction close square brackets subscript 0 superscript a equals 8
rightwards double arrow a cubed minus 0 equals 8
rightwards double arrow a cubed equals 8
rightwards double arrow a equals 2

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45