Question

Prove that:

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

Prove that:

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

Hence show that:

`int_0^pi sin x  dx = 2 int_0^(pi/2) sin x  dx`

Answer 1

Since ‘a’ lies between 0 and 2a,
we have

`int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx,  .......(byint_a^bf(x)dx=int_a^cf(x)dx+int_c^bf(x)dx)`

`=I_1+I_2`     ........................(say)

`I_2 = int_a^(2a)f(x)dx`

Put x = 2a − t

Therefore, dx = −dt

When x = a, 2a − t = a

t = a

When x = 2a, 2a − t = 2a

t = 0

`I_2 = int_0^(2a) f(x) dx = int_a^0 f(2a - t) (-dt)`

`= -int_a^0 f(2a - t)dt = int_0^a f(2a - t)dt      ...................... (By int_a^b f(x)dx = -int_b^a f(x)dx)`

`=int_0^a f(2a - x)dx    ..............(By int_a^b f(X)dx = int_a^b f(t)dt)`

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

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

To show that:

`int_0^pi sin x  dx = 2 int_0^(pi/2) sin x  dx`

We use the proven property by setting f(x) = sin x and 2a = π, which means a = `pi/2`.

The property tells us that:

`int_0^pi sin x  dx = int_0^(pi/2) sin  x  dx + int_0^(pi/2) sin (pi - x)  dx`

Knowing the trigonometric identity sin (π - x) = sin x, the equation simplifies to:

`int_0^pi sin x  dx = 2 int_0^(pi/2) sin x  dx`

This directly applies the property to the integral of sin x over [0, π] to show it equals twice the integral of sin x over `[0, pi/2]`, demonstrating the utility of this property in simplifying integrals with symmetric functions over specific intervals​.