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# Iterated Integral

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, 2 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

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

thanks alot ^^ · 3 years, 1 month ago

Aw... i know the answer but i dont know how to explain it in this website.... · 3 years, 2 months ago