Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises: 12

Answer

$\frac{T^5}{5} - 4T^2 + 7T $

Work Step by Step

$\int^T_{0}(x^4-8x+7)dx$ $=[\frac{x^5}{5} - 4x^2 + 7x] ^T_{0}$ Sub in the bounds and subtract the lower bound (0) from the upper bound (T): $=[\frac{(T)^5}{5} - 4(T)^2 + 7(T)] - [\frac{(0)^5}{5} - 4(0)^2 + 7(0)] $ $=\frac{T^5}{5} - 4T^2 + 7T - 0$ $=\frac{T^5}{5} - 4T^2 + 7T $.
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