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

Page / Exercise

Chapter 8 - Applications of Integrals 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

Chapter 8 - Applications of Integrals Ex. 8.2

Solution 1

Solution 2

Solution 3
Solution 4
Solution 5
Solution 6
Solution 7

Chapter 8 - Applications of Integrals 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 4units.

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 3units.

Solution 18

Solution 19