Answer
$\displaystyle \frac{3}{2\ln 2}$
Work Step by Step
The method of example 4, as suggested by the hint, involves substitution.
Remember to express the LIMITS of integration accordingly.
$\left[\begin{array}{ll}
u=-x+1 & du=-dx\\
dx=-du & \\
& \\
x=0\Rightarrow & u=1\\
x=2\Rightarrow & u=-1
\end{array}\right]$
$\displaystyle \int_{0}^{2}2^{-x+1}dx=\int_{1}^{-1}2^{u}(-du)=-\int_{1}^{-1}2^{u}du$
... Apply the FTC,
... def.integral = [antiderivative$]_{a}^{b}$
$=-\left[\dfrac{2^{u}}{\ln 2}\right]_{1}^{-1}$
$=-(\displaystyle \frac{2^{-1}}{\ln 2}-\frac{2}{\ln 2})$
$=\displaystyle \frac{1}{\ln 2}(2-\frac{1}{2})$
$=\displaystyle \frac{3}{2\ln 2}$
