What is the integral from 0 to 2pi of integral from 0 to 1 of re^{r^2} ?

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integral from 0 to 2pi of integral from 0 to 1 of re^{r^2}

\int_{0}^{2 π} \int_{0}^{1} r e^{r^{2}} d r d θ

π (e - 1)

Decimal Notation

5.39814 \dots

Solution steps

\int_{0}^{2 π} \int_{0}^{1} r e^{r^{2}} d r d θ

\int_{0}^{1} r e^{r^{2}} d r = \frac{e - 1}{2}

= \int_{0}^{2 π} (\frac{e - 1}{2}) d θ

\int_{0}^{2 π} (\frac{e - 1}{2}) d θ = π (e - 1)

= π (e - 1)

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\int_{0}^{\frac{π}{2}} \int_{0}^{2} x^{2} y cos (x y^{2}) d y d x

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What is the integral from 0 to 2pi of integral from 0 to 1 of re^{r^2} ?
The integral from 0 to 2pi of integral from 0 to 1 of re^{r^2} is pi(e-1)