Answer
$\int_{0}^{5}f(x)~dx = 17$
Work Step by Step
We can evaluate the integral:
$\int_{0}^{5}f(x)~dx$
$=\int_{0}^{3}f(x)~dx+\int_{3}^{5}f(x)~dx$
$=\int_{0}^{3}3~dx+\int_{3}^{5}x~dx$
$=3(3-0)+(\frac{5^2-3^2}{2})$
$= 9+8$
$= 17$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 390: 50
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