Question: Evaluate $\int_{0}^{2}\int_{0}^{x}\int_{0}^{2x+2y}e^{x+y+z}dz\,dy\,dx$ .
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Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : Medium

am2(82) • 286 views
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modified 10 months ago by gravatar for Juilee Juilee2.1k written 12 months ago by gravatar for smitapn612 smitapn6120
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$I = \int_{x=0}^2\int^x_{y=0}\int^{2x+2y}_{z=0}e^{x+y+z}dz\hspace{0.1cm}dy\hspace{0.1cm}dx\\ \hspace{0.1cm}= \int_0^2 \int_0^x e^{x+y}[e^z]_0^{2x+2y} dx \hspace{0.1cm}dy\\ \hspace{0.1cm}= \int_0^2\int_0^x e^{x+y}[e^{2x+2y}-e^0]dx \hspace{0.1cm}dy\\ \hspace{0.1cm}= \int_0^2\int_0^x(e^{3x+3y}-e^{x+y}) dx\hspace{0.1cm}dy\\ \hspace{0.1cm}=\int_0^2\Big[e^{3x}.\frac{e^{3y}}{3}-e^xe^y\Big]_0^x dx\\ \hspace{0.1cm}=\int_0^2\big[\frac{e^{3x}}{3}(e^{3x}-e^0)-e^x(e^x - e^0)\big]dx\\ \hspace{0.1cm}=\int_0^2\Big[\frac{e^{6x}-e^{3x}}{3}-e^{2x}+e^x\Big]dx\\ \hspace{0.1cm}= \frac{1}{3}\Big[\big[\frac{e^{6x}}{6}-\frac{e^{3x}}{3}\big]\Big]_0^2 - \big[\frac{e^{2x}}{2}\big]_0^2 +[e^x]_0^2\\ \hspace{0.1cm}= \frac{1}{3}\Big[\frac{e^{12}}{6}-\frac{e^6}{3} - \frac{1}{6} +\frac{1}{3} \Big] - \frac{1}{2} [e^4-1] +[e^2-1]\\ \hspace{0.1cm} = \frac{e^{12}}{18}-\frac{e^6}{9}-\frac{e^4}{2}+e^2-\frac{4}{9} $

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written 10 months ago by gravatar for Juilee Juilee2.1k
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