Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises: 2

Answer

$$\frac{e-1}{2}$$

Work Step by Step

Area = $\int^b_a f(x) - g(x) dx$ Where f(x) is above g(x) on the coordinate system. In this question $f(x) = e^x$ and $g(x) = xe^{x^2}$. The limits of integration will be $a$ = line (x = 0) and $b$ = the x-coord of the right interception (where $f(x)$ = $g(x)$). $$\int^b_a f(x) - g(x) dx$$ $$= \int^1_0 e^x - xe^{x^2}$$ Next step is to solve the definite integral. Split the integral into two and use the substitution rule for the integral $xe^{x^2}$. $$\int^1_0e^x dx - \int^1_0xe^{x^2}$$ Substitution: $u = x^2$ $du = 2x dx$ $\frac{1}{2}du = x dx$ $a = 0 \rightarrow u = 0$ ||| $b = 1 \rightarrow u = 1$ $$= \int^1_0 e^x dx - \frac{1}{2}\int^1_0 e^u du$$ $$= (e^x)|^1_0 - \frac{1}{2}(e^u)|^1_0$$ $$= (e^1 - e^0)-\frac{1}{2}(e^1 - e^0)$$ $$= e - 1 -\frac{e-1}{2}$$ $$= \frac{e-1}{2}$$
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