Substitution for Definite Integrals

So far we have been finding antiderivatives using substitution.  What if we have a definite integral?  Consider

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

1. 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;


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


3. 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


4. 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:

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


2. 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                                                                                                                
   
3. Evaluate the integral in terms of $u$, so if $F$ is an antiderivative of $f$, getting $\displaystyle\int_a^bf(g(x))g'(x)\,dx=\int_c^df(u)\,du=F(u)\left|\begin{array}{c} ^d \\ _c \end{array}\right .=F(d)-F(c).$