Prove that: \(\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx\)Hence evaluate: \(\displaystyle\int_0^3 \dfrac{\sqrt x}{\sqrt x + \sqrt{3 - x}}\,dx\)

Answer: Proof below; the integral equals \(\dfrac32\).

Proof: In the RHS put \(a + b - x = t\), so \(dx = -dt\). When \(x = a\), \(t = b\); when \(x = b\), \(t = a\).

\(\displaystyle\int_a^b f(a + b - x)\,dx = \int_b^a f(t)(-dt) = \int_a^b f(t)\,dt = \int_a^b f(x)\,dx\) (a definite integral does not depend on the variable). Hence proved.

Evaluation: Let \(I = \displaystyle\int_0^3 \dfrac{\sqrt x}{\sqrt x + \sqrt{3 - x}}\,dx\) …(1)

Using the property with \(a = 0\), \(b = 3\) (replace \(x\) by \(3 - x\)):

\(I = \displaystyle\int_0^3 \dfrac{\sqrt{3 - x}}{\sqrt{3 - x} + \sqrt x}\,dx\) …(2)

Adding (1) and (2): \(2I = \displaystyle\int_0^3 \dfrac{\sqrt x + \sqrt{3 - x}}{\sqrt x + \sqrt{3 - x}}\,dx = \int_0^3 1\,dx = 3\).

\(\therefore I = \dfrac32\).