So far we have been finding antiderivatives using
substitution. What if we have a definite integral?
Consider $$\int_a^bf(g(x))g'(x)\,dx.$$ It is important to realize
that $a$ and $b$ are $x$-values,
not $u$-values. We cannot plug $a$ and $b$ in for $u$.
We have to do more work. There are two approaches to
computing a definite integral by substitution.

Method 1 - Finding the antiderivative, then evaluating the
integral using FTC II:

Use substitution on the indefinite integral (without the
limits of integration included) and write $\int
f(g(x))g'(x)\,dx=\int f(u)\, du$, as we have been doing;

Find an antiderivative $F(u)$, as we've been doing, writing
$\displaystyle\int f(g(x))g'(x)\,dx=\int f(u)\, du=F(u)+c$;

Rewrite $F(u)$ in terms of $x$ to get $F(g(x))$. Now you
will have written $\displaystyle\int f(g(x))g'(x)\,dx=\int
f(u)\, du=F(u)+c=F(g(x))+c$, as we have been doing; and

Go back to the definite integral, with the limits of
integration for $x$, and write $\displaystyle\int_a^b
f(g(x))g'(x)\,dx=F(g(x))\left|\begin{array}{c} ^b \\ _a
\end{array}\right .=F(g(b))-F(g(a)).$

(Notice: this definite integral is not equal to the indefinite
integrals in the steps above.)

Method 2 - change completely from the $x$ world to the $u$
world and evaluate in the $u$ world:

Convert the integrand to $f(u)\, du$ using substitution (as we
have been doing);

Convert the limits of integration to the corresponding values
of $u$, and forget about $x$, treating $u$ as its own variable,
getting $\displaystyle\int_c^df(u)\,du$, where $c$ is $u$
evaluated at $x=a$ and $d$ is $u$ evaluated at $x=b$; then

Evaluate the integral in terms of $u$, so if $F$ is an
antiderivative of $f$, getting
$\displaystyle\int_a^bf(g(x))g'(x)\,dx=\int_c^df(u)\,du=F(u)\left|\begin{array}{c}
^d \\ _c \end{array}\right .=F(d)-F(c).$

These methods are illustrated in the following video.