Substitution in a definite integral

For a definite integral, with bounds, substitution has a useful trap in store: the bounds refer to $x$, but after substitution we integrate in $u$. So the bounds must be changed too.

The rule: the new bounds are the values of $u$ at the old bounds.

$$\int_a^b f(u(x))\,u'(x)\,\mathrm{d}x = \int_{u(a)}^{u(b)} f(u)\,\mathrm{d}u$$

Let's compute $\displaystyle\int_0^1 2x\,e^{x^2}\,\mathrm{d}x$. We set $u = x^2$, so:

Old bound	New bound
$x = 0$	$u = 0^2 = 0$
$x = 1$	$u = 1^2 = 1$

The integral becomes, entirely in $u$:

$$\int_0^1 e^u\,\mathrm{d}u = \left[e^u\right]_0^1 = e^1 - e^0 = e - 1 \approx 1.718$$

The advantage is clear: by changing the bounds, we no longer need to go back to $x$ at the end. We compute directly with the bounds in $u$. It is faster and avoids the common mistake of forgetting to return to the original variable.

Type of integral	What we do
definite	transform the bounds, never go back to $x$
indefinite	same steps, but go back to $x$ at the end

Remember this reflex: as soon as there are bounds, carry them along with the substitution. It is the most classic mistake, and the easiest one to avoid.