Answer
$\displaystyle \frac{1}{\ln 2}$
Work Step by Step
FTC: Let $f$ be a continuous function defined on the interval $[a, b]$,
and let $F$ be any antiderivative of f defined on $[a, b]$. Then
$\displaystyle \int_{a}^{b}f(x)dx=\left[F(x)\right]_{a}^{b}=F(b)-F(a)$.
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$f(x)=2^{x}$, which is continuous on [$-1,1$].
First, find an antiderivative $F(x)$:
...From sec.13-1, $\displaystyle \int b^{x}dx=\frac{b^{x}}{\ln b}+C$ ...
$\displaystyle \int f(x)dx= \displaystyle \frac{2^{x}}{\ln 2}+C$
(we take C=0, since by FTC we need ANY antiderivative)
$F(x)=\displaystyle \frac{2^{x}}{\ln 2}$
Next, evaluate :
$F(1)=\displaystyle \frac{2}{\ln 2}$
$F(0)=\displaystyle \frac{1}{\ln 2}$
Finally, apply the theorem:
$\displaystyle \int_{0}^{1}f(x)dx=F(1)-F(0)$
$=\displaystyle \frac{2}{\ln 2}-\frac{1}{\ln 2}=\frac{1}{\ln 2}$
