Answer
$\frac{e^4-1}{e^2}$
Work Step by Step
$$\eqalign{
&\text{Let }I= \int_{ - 2}^2 {\left( {\sinh x + \cosh x} \right)} dx \cr
& {\text{Integrate using the formulas }} \cr
& \int {\sinh x} dx = \cosh x + C{\text{ and }}\int {\cosh x} dx = \sinh x + C \cr
& I = \left[ {\cosh x + \sinh x} \right]_{ - 2}^2 \cr
& {\text{Using the fundamental theorem of calculus}}{\text{, part 2}} \cr
& I = \left[ {\cosh \left( 2 \right) + \sinh \left( 2 \right)} \right] - \left[ {\cosh \left( { - 2} \right) + \sinh \left( { - 2} \right)} \right] \cr
& {\text{where cosh}}\left( { - x} \right) = \cosh x{\text{ and }}\sinh \left( { - x} \right) = - \sinh x \cr
& = \left[ {\cosh \left( 2 \right) + \sinh \left( 2 \right)} \right] - \left[ {\cosh \left( 2 \right) - \sinh \left( 2 \right)} \right] \cr
& I = \cosh \left( 2 \right) + \sinh \left( 2 \right) - \cosh \left( 2 \right) + \sinh \left( 2 \right) \cr
& = 2\sinh \left( 2 \right) \cr
& = 2\cdot\frac{e^2-e^{-2}}{2}\cr
&=e^2-\frac{1}{e^2}\cr
&=\frac{e^4-1}{e^2}} $$
