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Can anybody do the following integral?

\[\int\limits_1^2\int\limits_{x^3}^{x}e^{y/x}dydx\]

Note by Esraa Ibrahim 3 years, 6 months ago

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You start by evaluating the inner integral

\( \int_1^2 \! \int_{x^3}^{x} \! e^{\frac{y}{x}} \, dy \, dx = \int_1^2 \! [xe^{\frac{y}{x}}]_{x^3}^{x} \, dx = \int_1^2 \! (ex-xe^{x^2}) \, dx \)

You then follow by evaluating the outer integral

\( \int_1^2 \! (ex-xe^{x^2}) \, dx = [ \frac{e}{2} x^2 - \frac{1}{2} e^{x^2} ]_1^2 = ( 2e - \frac{1}{2} e^{4} ) - ( \frac{1}{2} e - \frac{1}{2} e ) = \boxed{\frac{1}{2} e (4 - e^{3})} \)

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thanks alot ^^

Aw... i know the answer but i dont know how to explain it in this website....

thanks...no problem^^

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## Comments

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TopNewestYou start by evaluating the inner integral

\( \int_1^2 \! \int_{x^3}^{x} \! e^{\frac{y}{x}} \, dy \, dx = \int_1^2 \! [xe^{\frac{y}{x}}]_{x^3}^{x} \, dx = \int_1^2 \! (ex-xe^{x^2}) \, dx \)

You then follow by evaluating the outer integral

\( \int_1^2 \! (ex-xe^{x^2}) \, dx = [ \frac{e}{2} x^2 - \frac{1}{2} e^{x^2} ]_1^2 = ( 2e - \frac{1}{2} e^{4} ) - ( \frac{1}{2} e - \frac{1}{2} e ) = \boxed{\frac{1}{2} e (4 - e^{3})} \)

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thanks alot ^^

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Aw... i know the answer but i dont know how to explain it in this website....

Log in to reply

thanks...no problem^^

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