Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{2}x\sqrt{2-x}\,dx$

Concept: Definite Integral Symmetry — use the substitution $t = 2 - x$ to transform the integral into a standard form.

Step 1: Let $t = 2 - x$. Then $x = 2 - t$, $dx = -dt$, and when $x = 0$, $t = 2$; when $x = 2$, $t = 0$.

Step 2: Substitute and reverse the limits:

Step 3: Split and integrate:

Step 4: Evaluate at $t = 2$: …

The key insight here is that the integrand $x\sqrt{2-x}$ is not symmetric in any obvious way over $[0,2]$, but the factor $\sqrt{2-x}$ suggests a natural substitution: let $t = 2 - x$. This flips the limits and often simplifies the square root into a power of $t$, while the $x$ becomes $2 - t$. The result is a sum of two simple power integrals — no tricks, just clean algebra.

Let’s work through it step by step.

1. Set up the substitution. Let $t = 2 - x$. Then $x = 2 - t$, and $dx = -dt$. When $x = 0$, $t = 2$; when $x = 2$, $t = 0$. The integral becomes:

2. Simplify the limits. The negative sign in $dx$ and the reversed limits cancel:

Notice the limits are now $0$ to $2$ again, but the integrand is in terms of $t$.

3. Expand the integrand. Write $\sqrt{t} = t^{1/2}$, so:

4. Integrate term by term. Using $\int t^{n}\,dt = \frac{t^{n+1}}{n+1}$:

So:

5. Evaluate at the limits. At $t = 2$:

At $t = 0$, both terms are $0$. …

Method: The substitution $t=a-x$ to rationalise $x\sqrt{a-x}$

For $\displaystyle\int_0^a x\sqrt{a-x}\,dx$, put $t=a-x$: the square root becomes $\sqrt t$ and $x$ becomes $a-t$, leaving a sum of simple fractional-power terms.

Steps

Step 1: Substitute $t=a-x$.

Then $x=a-t$, $\sqrt{a-x}=\sqrt t$, $dx=-dt$; swapping the limits cancels the minus sign.

Step 2: Expand the integrand in $t$.

$(a-t)\sqrt t = a\,t^{1/2}-t^{3/2}$.

Step 3: Integrate with the power rule. …

Common Mistakes

Mistake 1: Reaching for integration by parts on $x\sqrt{2-x}$.

Why it's wrong: by-parts works but is longer and more error-prone than the substitution $t=2-x$, which rationalises the root immediately. Correct approach: put $t=2-x$ so the integrand becomes $2t^{1/2}-t^{3/2}$.

Mistake 2: Not converting the $x$ outside the root to $2-t$. …