Answer
$\frac{7}{2}$
Work Step by Step
$\int^{0}_{1}(3x^{2}+x-5)dx=3\int^{0}_{1}x^{2}dx+\int^{0}_{1}xdx-5\int^{0}_{1}dx $
$=3(\frac{0^{3}}{3}-\frac{1^{3}}{3})+(\frac{0^{2}}{2}-\frac{1^{2}}{2})-5(0-1)$
$=-1+(-\frac{1}{2})-(-5)=\frac{7}{2}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 309: 50
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