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: 14

Answer

$$A= \frac{8}{3} Area\thinspace Units$$
1517858793

Work Step by Step

Find the intersections by setting the functions equal to each other. $$4x-x^{2} = x^{2}$$ $$0=2x^{2} - 4x$$ $$0 = 2x(x-2)$$ $$x = 0,\space 2$$ These will be the integral limits. The top function is $y = 4x-x^{2}$; the bottom is $y = x^{2}$. Build the integral: $$A=\displaystyle\int_{0}^{2}((4x-x^{2})-x^{2})dx$$ $$=\displaystyle\int_{0}^{2}(4x-2x^{2})dx$$ $$=\left[2x^{2}-2\times\frac{1}{3}x^{3}\right]^{2}_0$$ $$2\times2^{2}-\frac{2}{3}\times2^{3}-(0-0)$$ $$= 8 -\frac{16}{3} = \frac{24-16}{3}=\frac{8}{3}$$
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.