## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 7

#### Answer

$$\frac{e-1}{2} \mathbf{i}+\frac{e-1}{e} \mathbf{i}+\mathbf{k}$$

#### Work Step by Step

We evaluate the integral of the vector function as follows: \begin{align*} \int_{0}^{1}\left(t e^{t^{2}} \mathbf{i}+e^{-t} \mathbf{j}+\mathbf{k}\right) d t&=\left(\frac{1}{2} e^{t^{2}}\right)\bigg|_{0}^{1} \mathbf{i}-\left(e^{-t}\right)\bigg|_{0}^{1} \mathbf{j}+t\bigg|_{0}^{1} \mathbf{k}\\ &=\frac{e-1}{2} \mathbf{i}+\frac{e-1}{e} \mathbf{i}+\mathbf{k} \end{align*}

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