## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 7 - Section 7.7 - Approximate Integration - 7.7 Exercises - Page 524: 9

#### Answer

$$\int_{0}^{2}\frac{e^{x}}{1+x^{2}}dx, \quad n=10$$ (a) The approximation of the given integral by using the Trapezoidal Rule is $\approx 2.660833$ (b) The approximation of the given integral by using the midpoints Rule is $\approx 2.664377$ (c) The approximation of the given integral by using Simpson’s Rule is $\approx 2.663244$

#### Work Step by Step

$$\int_{0}^{2}\frac{e^{x}}{1+x^{2}}dx, \quad n=10$$ (a) Use the Trapezoidal Rule to approximate the given integral with the specified value of n. With $n =10, a = 0$, and $b = 2$ we have $$\Delta x=\frac{b-a}{n}=\frac{2-0}{10}=\frac{1}{5}$$ and so the Trapezoidal Rule gives: \begin{aligned} \int_{0}^{2}\frac{e^{x}}{1+x^{2}}dx &= \frac{\Delta x}{2}\left[f\left(x_{0}\right)+2 f\left(x_{1}\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right]\\ & \approx T_{10} \\ & =\frac{1}{5 \cdot 2}[f(0)+2 f(0.2)+2 f(0.4)+2 f(0.6)+\\ & \quad\quad+2 f(0.8)+2 f(1) +2 f(1.2)+2 f(1.4)+\\ & \quad\quad+2 f(1.6)+2 f(1.8)+f(2)] \\ & \approx 2.660833 \end{aligned} (b) The midpoints Rule to approximate the given integral with the specified value of $n=10$ gives : \begin{aligned}\int_{0}^{2}\frac{e^{x}}{1+x^{2}}dx &=\Delta x\left[f\left(\overline{x}_{1}\right)+f\left(\overline{x}_{2}\right)+\cdots+f\left(\overline{x}_{n}\right)\right] \\ & = M_{10} \\ & \approx \frac{1}{5}[f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9)+\\ &\quad\quad +f(1.1)+f(1.3)+f(1.5)+f(1.7)+f(1.9)] \\ & \approx 2.664377 \end{aligned} (c) Simpson’s Rule to approximate the given integral with the specified value of $n=10$ gives : \begin{aligned} \int_{0}^{2}\frac{e^{x}}{1+x^{2}}dx & = \frac{\Delta x}{3}\left[f\left(x_{0}\right)\right. +4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)+\cdots \\ & \quad \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \\ &= S_{10}\\ & \approx \frac{1}{5 \cdot 3}[f(0)+4 f(0.2)+2 f(0.4)+4 f(0.6)+\\ &\quad\quad + 2 f(0.8) +4 f(1)+2 f(1.2)+4 f(1.4)+\\ & \quad\quad+2 f(1.6)+4 f(1.8)+f(2)] \\ & \approx 2.663244 \end{aligned}

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