## Calculus: Early Transcendentals 8th Edition

$\displaystyle\int\limits_0^3(2\sin x-e^{x})dx=3-2\cos3-e^{3}$
$\displaystyle\int\limits_0^3(2\sin x-e^{x})dx$ Integrate each term separately and then apply the second part of the fundamental theorem of calculus to get the answer: $\displaystyle\int\limits_0^3(2\sin x-e^{x})dx=2\int\limits_0^3\sin xdx-\int\limits_0^3e^{x}dx=-2\cos x-e^{x}\Big|_0^3=$ $...=-2\cos3-e^{3}+2\cos0+e^{0}=-2\cos3-e^{3}+2+1=...$ $...=3-2\cos3-e^{3}$