Q.Evaluate the integral using substitution $\int_{0}^{2}x\sqrt{x+2}\,dx$ (Put $x+2=t^2$)

Concept: U Substitution (with the given substitution $x+2 = t^2$).

Step 1 – Change of variable

Let $x+2 = t^2 \implies x = t^2 - 2$, and $dx = 2t\,dt$.

Step 2 – Update limits

When $x = 0$, $t^2 = 2 \implies t = \sqrt{2}$ (positive root, since $t \ge 0$).

When $x = 2$, $t^2 = 4 \implies t = 2$.

Step 3 – Rewrite and integrate

The integral becomes:

Step 4 – Evaluate

At $t=2$: $\frac{32}{5} - \frac{16}{3} = \frac{96 - 80}{15} = \frac{16}{15}$. …

Step-by-step solution

1. Substitute.

Let $x+2=t^2$, so $x=t^2-2$ and $dx=2t\,dt$. Also $\sqrt{x+2}=t$.

Limits: $x=0\Rightarrow t=\sqrt{2}$; $x=2\Rightarrow t=2$.

2. Rewrite the integral.

3. Integrate.

4. Evaluate at the limits.

$F(2)=\frac{2(32)}{5}-\frac{4(8)}{3}=\frac{64}{5}-\frac{32}{3}=\frac{32}{15}.$ …

Method: Rationalising substitution for an integrand with a square root

A term $\sqrt{\text{linear}}$ is removed by setting that linear expression equal to $t^2$; the radical becomes $t$ and the whole integrand turns into a polynomial.

Steps

Step 1: Substitute the radicand to a perfect square.

Put $x+c=t^2$, so $\sqrt{x+c}=t$ and $x=t^2-c$. Differentiate: $dx=2t\,dt$.

Step 2: Rewrite every $x$ in terms of $t$.

Replace $x$, $\sqrt{x+c}$ and $dx$; the integrand becomes a polynomial in $t$. …

Common Mistakes

Mistake 1: Taking the wrong sign of the square root at a limit.

Why it's wrong: $x=0$ gives $t^2=2$; since $t=\sqrt{x+2}\ge 0$, the correct lower limit is $t=+\sqrt2$, not $-\sqrt2$. Correct approach: always keep $t\ge0$ when solving $t=\sqrt{x+2}$.

Mistake 2: Forgetting $x=t^2-2$ when replacing the $x$ outside the root. …