Question

Prove that: `int_a^b f(x)dx = int_a^b f(a + b - x) * dx`. Hence, find `int_(π/6)^(π/3) sin^2x * dx`

Answer 1

Part 1: Proof

We prove:

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

Proof:

Let I = `int_a^b f(x) dx`

Make substitution:

x = a + b – t

Then, dx = – dt

When x = a,

t = b

when x = b,

t = a

So, I = `int_b^a f(a + b - t)(-dt)`

Reversing limits:

I = `int_a^b f(a + b - t) dt`

Since variable is dummy:

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

Proved.

Part 2: Evaluate

`int_(π/6)^(π/3) sin^2 x  dx`

Using the property:

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

Here, `a = π/6, b = π/3`

`a + b = π/6 + π/3`

= `π/6 + (2π)/6`

= `(3π)/6`

= `π/2`

So, `int_(π/6)^(π/3) sin^2 x  dx = int_(π/6)^(π/3) sin^2 (π/2 - x) dx`

But `sin(π/2 - x) = cos x`

So, `int_(π/6)^(π/3) sin^2 x  dx = int_(π/6)^(π/3) cos^2 x  dx`

Now add both integrals:

2I = `int_(π/6)^(π/3) (sin^2x + cos^2x) dx`

2I = `int_(π/6)^(π/3) 1  dx`

2I = `[x]_(π//6)^(π//3)`

2I = `π/3 - π/6`

2I = `π/6`

I = `π/12`