Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 55

Answer

\[\left| {Area} \right| = \frac{{94}}{3}\]

Work Step by Step

\[\begin{gathered} f\left( x \right) = {x^2} - 25{\text{ on }}\left[ {2,4} \right] \hfill \\ \hfill \\ Area = \int_a^b {f\left( x \right)dx} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ Area = \int_2^4 {\left( {{x^2} - 25} \right)} dx \hfill \\ \hfill \\ {\text{integrate}} \hfill \\ \hfill \\ Area = \left[ {\frac{{{x^3}}}{3} - 25x} \right]_2^4 \hfill \\ \hfill \\ evaluate{\text{ the limits}} \hfill \\ \hfill \\ Area = \left[ {\frac{{{{\left( 4 \right)}^3}}}{3} - 25\left( 4 \right)} \right] - \left[ {\frac{{{{\left( 2 \right)}^3}}}{3} - 25\left( 2 \right)} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ Area = - \frac{{236}}{3} + \frac{{142}}{3} \hfill \\ \hfill \\ Area = - \frac{{94}}{3} \hfill \\ \hfill \\ \left| {Area} \right| = \frac{{94}}{3} \hfill \\ \hfill \\ \end{gathered} \]
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.