## Calculus with Applications (10th Edition)

$${\text{A}} = 5 - {e^{ - 4}}$$
\eqalign{ & f\left( x \right) = 1 + {e^{ - x}}{\text{ on the interval }}\left[ {0,4} \right] \cr & {\text{The area between two curves is defined by }}\int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & {\text{where }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right]{\text{for this exercise }} \cr & g\left( x \right) = 0{\text{ }}\left( {x{\text{ - axis}}} \right){\text{ and the interval }}\left[ {0,4} \right]{\text{ implies that }}a = 0{\text{ and }}b = 4 \cr & {\text{then}} \cr & {\text{A}} = \int_0^4 {\left[ {1 + {e^{ - x}} - 0} \right]} dx \cr & {\text{A}} = \int_0^4 {\left( {1 + {e^{ - x}}} \right)} dx \cr & {\text{integrate by using }}\int_a^b {{e^{kt}}} dt = \left( {\frac{{{e^{kt}}}}{k}} \right)_a^b{\text{ and }}\int_a^b {dx} = \left[ x \right]_a^b \cr & {\text{Then}} \cr & {\text{A}} = \left( {x - {e^{ - x}}} \right)_0^4 \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & {\text{A}} = \left( {4 - {e^{ - 4}}} \right) - \left( {0 - {e^{ - 0}}} \right) \cr & {\text{simplifying}} \cr & {\text{A}} = \left( {4 - {e^{ - 4}}} \right) - \left( { - 1} \right) \cr & {\text{A}} = 4 - {e^{ - 4}} + 1 \cr & {\text{A}} = 5 - {e^{ - 4}} \cr}