Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx$

The key idea is to use the symmetry property of definite integrals: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.

Let $I = \int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx$.

Replace $x$ by $a-x$. Since $dx$ becomes $-dx$ but the limits swap, we have:

Now add the two expressions for $I$: …

The key insight here is that the integrand has a beautiful symmetry when you replace $x$ with $a-x$. The denominator $\sqrt{x} + \sqrt{a-x}$ simply swaps its two terms, while the numerator $\sqrt{x}$ becomes $\sqrt{a-x}$. This kind of structure is tailor-made for the property:

This holds for any function $f$ that is integrable on $[0, a]$. It works because as $x$ runs from $0$ to $a$, the quantity $a-x$ runs from $a$ down to $0$ — a perfect reversal. The area under the curve doesn't change.

Let's apply it.

1. Define the integral Let

2. Apply the substitution $x \to a-x$ Using the property above, replace $x$ by $a-x$ everywhere in the integrand. The limits stay the same (0 and $a$), and $dx$ remains $dx$:

Notice the denominator is the same as before — just the two terms swapped — so it's unchanged. The numerator is now $\sqrt{a-x}$.

3. Add the two expressions for $I$ We now have two different-looking integrals that are actually equal. Add them:

Since the denominators are identical, we can combine the integrands over a single integral:

4. Simplify …

Method: The general $\dfrac{f}{f+f(a-x)}$ complementary trick $\Rightarrow \dfrac{a}{2}$

For $\displaystyle\int_0^a \frac{f(x)}{f(x)+f(a-x)}\,dx$, reflection produces the complementary fraction; the two always add to $1$, giving half the interval length regardless of $f$.

Steps

Step 1: Reflect with $x\to a-x$.

Step 2: Add — the denominators are identical. …

Common Mistakes

Mistake 1: Trying to integrate $\frac{\sqrt x}{\sqrt x+\sqrt{a-x}}$ by rationalising directly.

Why it's wrong: direct manipulation is messy and unnecessary — the $\frac{f}{f+f(a-x)}$ pattern gives $\tfrac a2$ by reflection. Correct approach: reflect $x\to a-x$ and add.

Mistake 2: Doubting the answer because it has no $x$ in it. …