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Q15E

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Essential Calculus: Early Transcendentals
Found in: Page 699
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Calculate the integrated integral \(\int\limits_{ - 3}^3 {\int\limits_0^{\pi /2} {(y + {y^2}\cos x)dxdy} } \)

\(\int\limits_{ - 3}^3 {\int\limits_0^{\pi /2} {(y + {y^2}\cos x)dxdy} } = 18\)

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1:- Given to find

\(\int\limits_{ - 3}^3 {\int\limits_0^{\pi /2} {(y + {y^2}\cos x)dxdy} } \)

The value of integral (to find)

Step 2: - Find the value of integral with respect to x

\(\begin{aligned}\int\limits_{ - 3}^3 {\int\limits_0^{\pi /2} {(y + {y^2}\cos x)dxdy} } &= \int\limits_{ - 3}^3 {(y\int\limits_0^{\pi /2} {dx} + {y^2}\int\limits_0^{\pi /2} {(\cos x)dx} )dy} \\ &= \int\limits_{ - 3}^3 {(y(x)_0^{\pi /2} + {y^2}(\sin x)_0^{\pi /2})dy} \\ &= \int\limits_{ - 3}^3 {(y(\frac{\pi }{2} - 0) + {y^2}(\sin \frac{\pi }{2} - \sin 0))dy} \\ &= \int\limits_{ - 3}^3 {(y(\frac{\pi }{2}) + {y^2}(1 - 0))dy} \\ &= \int\limits_{ - 3}^3 {\frac{{\pi y}}{2} + {y^2}dy} \\ &= \frac{\pi }{2}\int\limits_{ - 3}^3 {(y)dy + \int\limits_{ - 3}^3 {({y^2})dy} } \end{aligned}\)

Step: - 3 Find the solution of final integral

\(\begin{aligned}\int\limits_{ - 3}^3 {\int\limits_0^{\pi /2} {(y + {y^2}\cos x)dxdy} } &= \frac{\pi }{2}\int\limits_{ - 3}^3 {(y)dy + \int\limits_{ - 3}^3 {({y^2})dy} } \\ &= \frac{\pi }{2}\left( {\frac{{{y^2}}}{2}} \right)_{ - 3}^3 + \left( {\frac{{{y^3}}}{3}} \right)_{ - 3}^3\\ &= \frac{\pi }{2}\left( {\frac{{{3^2}}}{2} - \frac{{{{( - 3)}^{^2}}}}{2}} \right) + \left( {\frac{{{3^3}}}{3} - \frac{{{{( - 3)}^3}}}{3}} \right)\\ &= \frac{\pi }{2}\left( {\frac{{9 - 9}}{2}} \right) + \left( {\frac{{27 - ( - 27)}}{3}} \right)\\ &= \frac{\pi }{2}\left( {\frac{0}{2}} \right) + \left( {\frac{{54}}{3}} \right)\\ &= 0 + 18\\ &= 18\end{aligned}\)

Therefore, \(\int\limits_{ - 3}^3 {\int\limits_0^{\pi /2} {(y + {y^2}\cos x)dxdy} } = 18\)

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