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Question Number 75809 by aliesam last updated on 17/Dec/19

Commented by kaivan.ahmadi last updated on 19/Dec/19

$${\int }_0^3{\int }_0^{\frac{x}{3}}{e}^{x^2}dydx={\int }_0^3{e}^{x^2}y{\mid }_0^{\frac{x}{3}}dx=\frac{1}{3}{\int }_0^3{xe}^{x^2}dx=$$$$\frac{1}{6}\left(e^{x^2}{\mid }_0^3\right)=\frac{1}{6}(e^9-1)$$$$$$

Commented by kaivan.ahmadi last updated on 19/Dec/19

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