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: 18


$y' = -\sqrt{1+sin^2(x)}\times cos(x)$

Work Step by Step

We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $ y =\int_{sin(x)} ^{1} \sqrt{1+t^2}dt$ Swap upper and lower bounds: (this makes the expression negative) $ y = -\int_{1} ^{sinx} \sqrt{1+t^2}dt$ Since the upper bound is a function we need to use the chain rule: $u = sin(x)$ $u' = cos(x)$ $ y =-\int^{u} _{1} \sqrt{1+t^2}dt$ Using the FTC $F(u) =\int_0^uf(θ)dθ$ $F'(u) = f(u)$ Thus, $y'= -\sqrt{1+u^2}\times u'$ $y' = -\sqrt{1+sin^2(x)}\times cos(x)$
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