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Essential Calculus: Early Transcendentals
Found in: Page 707
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

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

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 \( = \frac{1}{{16}}\left( {3{e^4} + 1} \right)\)

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