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Q. 15

Expert-verified

Calculus

Found in: Page 510

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

For each pair of definite integrals in exercise 13-18 decide which if either is larger without computing the integrals
$π \int_{0}^{2} x^{4} d x a n d π \int_{0}^{2} (x^{4} - 16) d x$

The integral $π \int_{0}^{2} x^{4} d x$ is bigger

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given information

We are given two integrals

$π \int_{0}^{2} x^{4} d x a n d π \int_{0}^{2} (x^{4} - 16) d x$

Step 2: Explaination

The term $x^{4} i s b i g g e r t h a n t e r m x^{4} - 16$

Hence on integrating we will get the term 16x in second integral And after substituting the limits the second term will be 16 less than that of first term. Hence the first term is bigger

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In the process of solving the differential equation $\frac{d y}{d x} = 1 - y$ by separation of variables, we obtain the equation $- \ln | 1 - y | = x + C$ . After solving for $y$ , this equation becomes $y = 1 - A e^{- x}$ . Given that $y > 1$ , how is A related to $C$ ?

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