Q.Evaluate the definite integral: $\int_{0}^{1} x e^{x^2} \, dx$

The key idea is U Substitution: we reverse the chain rule by choosing $u = x^2$, so that $du = 2x \, dx$.

Step 1: Let $u = x^2$. Then $du = 2x \, dx$, so $x \, dx = \frac{du}{2}$.

Step 2: Change the limits: when $x = 0$, $u = 0$; when $x = 1$, $u = 1$.

Step 3: Substitute and integrate: …

Why U-Substitution Works Here

The integrand $x e^{x^2}$ has a special structure: the derivative of $x^2$ is $2x$, and we have an $x$ sitting right next to $e^{x^2}$. This is the classic signal for substitution — the chain rule in reverse. When you see a function multiplied by its own derivative (or a constant multiple of it), u-substitution is the natural tool.

Think of it this way: if $F'(u) = e^u$, then by the chain rule, the derivative of $F(x^2)$ is $2x e^{x^2}$. Our integrand is exactly half of that. So we're essentially undoing the chain rule.

Step-by-Step Solution

1. Choose the substitution.

   Let $u = x^2$. Why? Because $du = 2x \, dx$, and we have an $x \, dx$ in the integrand. This choice will absorb the $x$ and turn the exponential into something simple.

2. Rewrite the differential.

   From $du = 2x \, dx$, we get $x \, dx = \frac{du}{2}$. The integral becomes:

3. Change the limits of integration.

   Since the original integral runs from $x=0$ to $x=1$, we need the corresponding $u$ values:

   • When $x = 0$, $u = 0^2 = 0$
   • When $x = 1$, $u = 1^2 = 1$

   So the definite integral transforms to:

4. Evaluate the simpler integral. The antiderivative of $e^u$ is $e^u$ itself. So: …

Method: Reverse chain rule — substitute the "inside" of a composite function

When the integrand is a composite $F'(g(x))$ multiplied by (a constant times) $g'(x)$, put $u=g(x)$ and the integral unwinds into a standard form.

Steps

Step 1: Identify the inner function.

Look for a function tucked inside another (here the exponent of $e^{\square}$) whose derivative also appears as a factor. Set $u$ equal to that inner function.

Step 2: Replace the differential.

From $du = g'(x)\,dx$, express the factor sitting next to the composite in terms of $du$ (e.g. $x\,dx = \tfrac12\,du$). …

Common Mistakes

Mistake 1: Integrating $e^{x^2}$ as if it were $\tfrac{1}{2x}e^{x^2}$ or some elementary form without substituting.

Why it's wrong: $e^{x^2}$ has no elementary antiderivative on its own; only the presence of the $x$ factor makes it integrable. Correct approach: use $u=x^2$ so the $x\,dx$ supplies $\tfrac12\,du$.

Mistake 2: Not changing the limits (or changing them wrongly). …