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 - Page 327: 28



Work Step by Step

Evaluate the integral: $\int^{4}_{0}(4-t)\sqrt t dt$ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(t)$: $F(t)=\int(4-t)\sqrt t dt$ $F(t)=\int(4t^{1/2} - t^{3/2}) $ $F(t)=\frac{2}{3}4t^{3/2} -\frac{2}{5}t^{5/2}$ $F(t)=\frac{8}{3}t^{3/2} -\frac{2}{5}t^{5/2}$ Now evaluate $F(b)−F(a)$: $F(4)−F(0)$ $(\frac{8}{3}(4)^{3/2} -\frac{5}{2}(4)^{5/2}) -(\frac{8}{3}(0)^{3/2} -\frac{5}{2}(0)^{5/2})$ $(\frac{8}{3}(\sqrt 4)^{3} -\frac{2}{5}(\sqrt 4)^{5}) -0$ $\frac{8(8)}{3} -\frac{2(32)}{5}$ $\frac{64}{3} - \frac{64}{5}$ $\frac{320}{15}-\frac{192}{15} = \frac{128}{15}$
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