Q.Evaluate $\displaystyle\int_{0}^{1} 3x^{2}\,e^{x^{3}}\,dx$ using substitution.

The factor $3x^2$ is the derivative of $x^3$, so substitute $u=x^3$ and change the limits.

Follow the substitution method of §4.

Choose the substitution. Let $u = x^3$. Then $du = 3x^2\,dx$ — exactly the factor $3x^2\,dx$ present in the integrand.

Change the limits. When $x = 0$: $u = 0^3 = 0$. When $x = 1$: $u = 1^3 = 1$. So the limits stay $0 \to 1$ (in $u$).

Rewrite and evaluate in $u$.

$\int_{0}^{1} 3x^2 e^{x^3}\,dx = \int_{0}^{1} e^{u}\,du = \big[e^{u}\big]_{0}^{1} = e^{1} - e^{0} = e - 1.$ …

Direct antiderivative: $\displaystyle\int 3x^2 e^{x^3}\,dx = e^{x^3}$, so the integra …

Although the numerical limits happen to remain $0$ and $1$ here (because $g(0)=0$, $g(1)=1$), this is a coincidence of $u = x^3$; always recompute $g(a)$ and $g(b)$ rather than …