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Q10E

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Essential Calculus: Early Transcendentals

Found in: Page 707

Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Evaluate the double integral: \begin{gathered}\iint\limits_D {{x^3}dA,} \hfill \\D = \{ x,y)/1 \leqslant x \leqslant e,0 \leqslant y \leqslant lnx\} \hfill \\\end{gathered}

Value of integral is \(\frac{1}{{16}}\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:- Integration on y.

\(\begin{array}{l}I = \int\limits_1^e {{x^3}y\left| {_0^{ln(x)}dx.} \right.} \\ = \int\limits_1^e {{x^3}y\left| {_0^{ln(x)}dx. = \int\limits_1^e {{x^3}ln(x)dx.} } \right.} \end{array}\)

Step 2:- Using integration by parts

\(\begin{array}{l}\int {udv = uv - \int {vdu} } \\I = \frac{1}{4}{x^4}ln(x)\left| {_1^e - \int\limits_1^e {\frac{1}{4}{x^4}\left( {\frac{1}{x}} \right)dx} } \right.\\ = \frac{{{e^4}}}{4} - \left( {\frac{{1{e^4}}}{{16}} - \frac{1}{{16}}} \right)\\ = \frac{1}{{16}}\left( {3{e^4} + 1} \right)\end{array}\)

Hence, value of integral is \( = \frac{1}{{16}}\left( {3{e^4} + 1} \right)\)

Most popular questions for Math Textbooks

The average value of a function\({\bf{f}}\left( {{\bf{x,y}}} \right)\)over a rectangle\({\bf{R}}\)is defined to be,

\({{\bf{f}}_{{\bf{ave}}}}{\bf{ = }}\frac{{\bf{1}}}{{{\bf{A}}\left( {\bf{R}} \right)}}\int {\int\limits_{\bf{R}} {{\bf{f}}\left( {{\bf{x,y}}} \right)} } {\bf{dA}}\)

Find the average value of\({\bf{f}}\)over the given rectangle.

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}{{\bf{x}}^{\bf{2}}}{\bf{y}}\)

\({\bf{R}}\)has vertices\(\left( {{\bf{ - 1,0}}} \right){\bf{,}}\left( {{\bf{ - 1,5}}} \right){\bf{,}}\left( {{\bf{1,5}}} \right){\bf{,}}\left( {{\bf{1,0}}} \right)\).

Calculate the iterated integral.

\(\int {_1^4\int {_0^2\left( {6{x^2}y - 2x} \right)dydx} } \)

Plot the point whose cylindrical coordinates are given. Then find the rectangular coordinates of the point.

a.\(\left( {\sqrt {\rm{2}} {\rm{,3\pi /4,2}}} \right)\)

b.\(\left( {{\rm{1,1,1}}} \right)\)

Evaluate \(\iiint_{\text{E}}{\text{z}}\text{dV}\), where \({\rm{E}}\) is enclosed by the paraboloid \({\rm{z = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}\) and the plane \({\rm{z = 4}}\). Use cylindrical coordinates.

Use a graphing device to draw the solid enclosed by the paraboloids \({\rm{z = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}\) and \({\rm{z = 5 - }}{{\rm{x}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}.\)

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