Question

Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`

Answer 1

`LHS=int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx.........(1)`

Substitute x = a + t in the second integral

dx=dt

When x = a, t = 0.

When x = 2a, t = a.

`thereforeint_a^(2a)f(x)dx=int_0^af(a+t)dt`

`=int_0^af(a+(a-t))dt (therefore int_0^af(x)dx=int_0^a f(a-x)dx)`

 

`=int_0^af(2a-t)dt`

 

`int_a^(2a)f(x)dx=int_0^af(2a-x)dx (therefore int_0^af(t)dt=int_0^af(x)dx)`

Using the above in (1), we get

`int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx`

`=int_0^af(x)dx+int_0^af(2a-x)dx=RHS ("Proved")`