Question

Prove that:

`{:(int_(-a)^a f(x) dx  = 2 int_0^a f(x) dx",", "If"  f  "is an even function"),(                                       = 0",", "if"  f  "is an odd function"):}`

Hence find the value of `int_-1^1tan^-1x  dx`.

Answer 1

We shall use the following results:

`int_a^b f(x) dx = -int_b^a f(x) dx`  ...(1)

`int_a^b f(x) dx = int_a^b f(t) dt`  ...(2)

If c is between a and b, then

`int_a^b f(x)  dx = int_a^c f(x) dx + int_c^b f(x) dx`  ...(3)

Since 0 lies between −a and a, by (3), we have

`int_-a^af(x)dx = int_(-a)^0 f(x) dx + int_0^a f(x) = I_1 + I_2`  ...(Say)

In I₁, put x = −t. Then dx = −dt

When x = −a, −t = −a ∴ t = a

When x = 0, −t = 0 ∴ t = 0

∴ `int_(-a)^0 f(x) dx = int_a^0 f(-t)(-dt) = -int_a^0 f(-t) dt`

= `int_0^a f(-t) dt`   ...[By (1)]

= `int_0^a f(-x) dx`  ...[By (2)]

∴ `int_(-a)^a f(x) dx = int_0^a f(-x) dx + int_0^a f(x) dx`

(i) If f is an even function, then

f(−x) = f(x)

∴ In this case,

`int_(-a)^a f(x) dx = int_0^a f(x) dx + int_0^a f(x) dx = 2 int_0^a f(x) dx`

(ii) If f is an odd function, then

f (−x) = −f(x)

∴ In this case,

`int_(-a)^a f(x) dx = int_0^a -f(x) dx + int_0^a f(x) dx`

= `-int_0^a f(x) dx + int_0^a f(x) dx = 0`

To find: `int_-1^1tan^-1x  dx`

Let f(x) = tan⁻¹x

Then f(−x) = tan⁻¹(−x) = −tan⁻¹x = −f(x)

f(x) is an odd function.

∴ `int_(-a)^a f(x) dx = 0`

∴ `int_(-1)^1 tan^-1x  dx = 0`