Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises: 21

Answer

2

Work Step by Step

Evaluate the integral: $\int^2_0 (\frac{4}{5}t^3-\frac{3}{4}t^2+\frac{2}{5}t) dx$ Recall the 2nd part of the Fundamental Theorem of Calculus: $\int_a^bf(x)dx = F(b) - F(a)$ Find $F(x)$: $F(x) = \int^2_0 (\frac{4}{5}t^3-\frac{3}{4}t^2+\frac{2}{5}t) dx$ $F(x) =\frac{1}{4}(\frac{4}{5}t^4) - \frac{1}{3}(\frac{3}{4}t^3) + \frac{1}{2}(\frac{2}{5}t^2) $ $F(x) = \frac{t^4}{5} - \frac{t^3}{4} +\frac{t^2}{5}$ Now evaluate $F(b) - F(a)$: $F(2) - F(0)$ $( \frac{2^4}{5} - \frac{2^3}{4} +\frac{2^2}{5}) - ( \frac{0^4}{5} - \frac{0^3}{4} +\frac{0^2}{5})$ $\frac{16}{5} - \frac{8}{4} + \frac{4}{5} - 0$ $ 4 - 2 = 2$
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