Question

If `f(a + b - x) = f(x)`, then `int_0^b x f(x)  dx` is equal to

Options

• `(a + b)/2 int_0^b f(b - x)  dx`

• `(a + b)/2 int_0^b f(b + x)  dx`

• `(b - a)/2 int_0^b f(x)  dx`

• `(a + b)/2 int_0^b f(x)  dx`

Answer 1

`(a + b)/2 int_0^b f(x)  dx`

Explanation:

Let I = `int_0^b x f(x)  dx`

= `int_0^b (a + b - x) f(a + b - x)  dx`  ......`[because int_0^b f(x)  dx = int_a^b (a + b - x)  dx]`

I = `int_a^b [(a + b) f(x) - xf(x)]  dx`

= `(a + b) int_a^b f(x)  dx - int_0^b  xf(x)  dx`

= `(a + b) int_a^b f(x)  dx - I`

∴ 2I = `(a + b) int_a^b f(x)  dx`

∴ I = `((a + b))/2 int_a^b f(x)`