Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.2 Regions Between Curves - 6.2 Exercises - Page 417: 7

Answer

$\dfrac{5}{2}-\dfrac{1}{\ln 2}$

Work Step by Step

Let us consider that two continuous functions $f(x)$ and $g(x)$ with $f(x)\geq g(x)$ on the interval $[a,b]$ . The area (A) of the region bounded by the graph of $f(x)$ and $g(x)$ on the interval $[a,b]$ can be calculated as: $Area(A)=\int_a^b [f(x)-g(x)] \ dx$ The given curves are $y=2^x$ and $y=3-x$.The intersection points of these curves satisfy the given equations are: $2^x=3-x \implies x=0 \ \text{and}\ x=1$ . These points will be the lower and upper limit of the integration. Thus, the area of the region is: $A=\int_a^b [f(x)-g(x)] \ dx= \int_{0}^{1} [3-x-2^x] \ dx\\ =[3x-\dfrac{x^2}{2}-\dfrac{2^x}{\ln 2}]_{0}^1\\ = (3-\dfrac{1^2}{2}-\dfrac{2^1}{\ln 2})-(3(0)-\dfrac{0^2}{2}-\dfrac{2^0}{\ln 2})\\=\dfrac{5}{2}-\dfrac{1}{\ln 2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.