Class 12-science NCERT Solutions Maths Chapter 8 - Applications of Integrals

Ex. 8.1

Ex. 8.2

Misc. Ex.

Applications of Integrals Exercise Ex. 8.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Applications of Integrals Exercise Ex. 8.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Applications of Integrals Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Required area = A(DCOD) + A(OAB)

$A left parenthesis D C O D right parenthesis equals integral subscript negative 2 end subscript superscript 0 x cubed space d x equals right enclose x to the power of 4 over 4 end enclose subscript negative 2 end subscript superscript 0 equals negative open parentheses negative 2 close parentheses to the power of 4 over 4 equals negative 16 over 4$

Since the area is positive

$therefore A left parenthesis D C O D right parenthesis equals 16 over 4$

$N o w comma space A left parenthesis O A B right parenthesis equals integral subscript 0 superscript 1 x cubed space d x space space space space space space space space space equals right enclose x to the power of 4 over 4 end enclose subscript 0 superscript 1 equals 1 fourth therefore A left parenthesis O A B right parenthesis equals 1 fourth$

Thus, the total area = $16 over 4 plus 1 fourth equals 17 over 4$ units

Hence, the required area of the region is $17 over 4$ units.

Solution 17

Required area = A(DCOD) + A(OAB)

$A left parenthesis D C O D right parenthesis equals integral subscript negative 1 end subscript superscript 0 x vertical line x vertical line space d x equals negative integral subscript negative 1 end subscript superscript 0 space x squared space d x equals right enclose negative open parentheses x cubed over 3 close parentheses end enclose subscript negative 1 end subscript superscript 0 equals open parentheses negative 1 close parentheses cubed over 3 equals negative 1 third$

Since the area is positive

$therefore A left parenthesis D C O D right parenthesis equals 1 third$

$N o w comma space A left parenthesis O A B right parenthesis equals integral subscript 0 superscript 1 x vertical line x vertical line space d x equals integral subscript 0 superscript 1 x squared space d x equals right enclose x cubed over 3 end enclose subscript 0 superscript 1 equals 1 third therefore A left parenthesis O A B right parenthesis equals 1 third$

Thus, the total area = $1 third plus 1 third equals 2 over 3$ units

Hence, the required area of the region is $2 over 3$ units.

Class 12-science NCERT Solutions Maths Chapter 8 - Applications of Integrals

Applications of Integrals Exercise Ex. 8.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Applications of Integrals Exercise Ex. 8.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Applications of Integrals Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

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