Integrals · Question 4

The value of 2 0 2f(x)g(x) dx- 0 2g(x) dx is .

Answer

0

Solution

g ( 2-x )= ( 2-x )x=g(x), so with I= 0 /2 2x g(x) dx, King's rule gives I= 0 /2 2x g(x) dx.

Adding: 2I= 0 /2g(x) dx.

Required value =2I- 0 /2g(x) dx=0.

Answer: 0