## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 396: 79

#### Answer

$$2\pi (e^2+1)$$

#### Work Step by Step

Since \begin{align*} V&=\int_{a}^{b}(2 \pi r) h d x\\ &=2 \pi \int_{0}^{2} x e^{x} d x \end{align*} Let \begin{align*} u&=x\ \ \ \ \ \ \ \ \ \ \ \ dv= e^xdx\\ du&=dx\ \ \ \ \ \ \ \ \ \ \ v=e^x \end{align*} Then \begin{align*} V&=\int_{a}^{b}(2 \pi r) h d x\\ &=2 \pi \int_{0}^{2} x e^{x} d x\\ &=2\pi \left( xe^x \bigg|_{0}^{2} - \int_{0}^{2} e^{x} d x\right)\\ &= 2\pi \left( xe^x -e^x\bigg|_{0}^{2} \right)\\ &= 2\pi (e^2+1) \end{align*}

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