Prove that : `int_-a^af(x)dx=2int_0^af(x)dx` , if f (x) is an even function.
= 0, if f (x) is an odd function.
Show that: `int_-a^af(x)dx=2int_0^af(x)dx` , if f (x) is an even function.
= 0, if f (x) is an odd function.
Solution 1
`int_-a^af(x)dx=int_-a^0f(x)dx+int_0^af(x)dx`
`int_a^af(x)dx=I+int_0^af(x)dx`
`Now I=int_-a^0f(x)dx`
Put x=-t
dx = - dt
When x = -a, t = a and when x = 0, t = 0
`I=int_a^0f(-t)(-dt)`
`=-int_a^0f(-t)dt`
`=int_0^af(-t)dt ...........[because int_a^bf(x)dx=-int_b^af(x)dx]`
`=int_0^af(-x)dt ...........[because int_a^bf(x)dx=-int_b^af(t)dx]`
Equation (i) becomes
`int_-a^af(x)dx=int_0^af(-x)dx+int_0^af(x)dx`
`=int_0^a[f(-x)+f(x)]dx.......(ii)`
case 1: If f(x) is an even function, then f(-x) = f(x).
Thus, equation (ii) becomes
`int_-a^af(x)dx=int_0^a[f(x)+f(x)]dx=2int_0^af(x)dx`
Case 2: If f(x) is an odd function, then f(-x) = -f(x).
Thus, equation (ii) becomes
`int_-a^af(x)dx=int_0^a[-f(x)+f(x)]dx=0`
Solution 2
We shall use the following results :
