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# Use integration by parts to find the integrals. $\int_{0}^{3} \frac{3-x}{3 e^{x}} d x$

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As indicated in the problem, we will solve this using integration by parts.

\begin{align*} \int_0 ^3 \frac{3-x}{3e^x}\,dx &= \frac{1}{3}\underbrace{\int_0 ^3 (3-x)e^{-x}\,dx}_{\boxed{\begin{smallmatrix} \text{Integration by parts}\\ u=3-x\quad dv = e^{-x}\,dx\\ du = -\,dx\quad v = -e^{-x} \end{smallmatrix}}} && \text{Factor out the constant }\frac{1}{3}. \\ &=\frac{1}{3}\left(-(3-x)e^{-x}\right)\bigg|_0 ^3 -\frac{1}{3}\int_0 ^3 e^{-x}\,dx && \text{You can evaluate }uv \text{ part here.}\\ &=\frac{1}{3}\cdot 3 - \frac{1}{3}\int_0 ^3 e^{-x}\,dx \\ &=1-\frac{1}{3}\left( -e^{-x}\right)\bigg|_0 ^3 && \int e^{-x}\,dx = -e^{-x}+C\\ &=1-\frac{1}{3}\left(-\frac{1}{e^3}+1\right) \\ &=1+\frac{1}{3e^3}-\frac{1}{3} \\ &=\boxed{\color{#4257b2}\frac{2}{3}+\frac{1}{3e^3}} \\ &\approx \boxed{\color{#4257b2}0.6833} \end{align*}

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