Question

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

Hence evaluate: `int_0^3 sqrtx/(sqrtx + sqrt(3 - x)) dx`

Answer 1

Proving the Property:

We need to prove,

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

Using substitution t = a + b − x, so dt = −dx

Changing the limits, 

When x = a, then t = b,

When x = b, then t = a

Thus, the integral transforms as,

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

Reversing the limits removes the negative sign,

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

Property is proved.

Evaluating the given integral:

We need to evaluate,

`I = int_0^3 sqrtx/(sqrtx + sqrt(3 - x)) dx`

Using the proved property, let:

`f(x) = sqrtx/(sqrtx + sqrt(3 - x))`

Applying the transformation,

`I = int_0^3 sqrt((3 - x))/(sqrt((3 - x)) + sqrtx) dx`

Now adding both integrals,

`I + I = int_0^3 [sqrtx/(sqrtx + sqrt(3 - x)) + sqrt(3 - x)/(sqrt(3 - x) + sqrtx)]dx`

Since,

`sqrtx/(sqrtx + sqrt(3 - x)) + sqrt(3 - x)/(sqrt(3 - x) + sqrtx) = 1`

We get,

`2 I = int_0^3 1dx = 3`

`I = 3/2`