What is the integral from 0 to 1 of integral from 4y to 4 of e^{x^2} ?

The integral from 0 to 1 of integral from 4y to 4 of e^{x^2} is (e^{16}-1)/8

Popular >

integral from 0 to 1 of integral from 4y to 4 of e^{x^2}

\int_{0}^{1} \int_{4 y}^{4} e^{x^{2}} d x d y

\frac{e ^{16} - 1}{8}

Solution steps

\int_{0}^{1} \int_{4 y}^{4} e^{x^{2}} d x d y

Change order of integration

= \int_{0}^{4} \int_{0}^{\frac{x}{4}} e^{x^{2}} d y d x

\int_{0}^{\frac{x}{4}} e^{x^{2}} d y = \frac{x e ^{x^{2}}}{4}

= \int_{0}^{4} (\frac{x e ^{x^{2}}}{4}) d x

\int_{0}^{4} (\frac{x e ^{x^{2}}}{4}) d x = \frac{e ^{16} - 1}{8}

= \frac{e ^{16} - 1}{8}

\int \frac{x ^{2}}{( x + 3 ) ^{3}} d x

\frac{d v}{d t} + \frac{( v ^{^{'}} \cdot v )}{t} = v^{^{'}} \cdot (\frac{t}{2} - \frac{1}{2})

\frac{d}{d x} (2 x ln (x))

x \to 0 lim (3 x^{2} + 2 x h + h^{2})

What is the integral from 0 to 1 of integral from 4y to 4 of e^{x^2} ?
The integral from 0 to 1 of integral from 4y to 4 of e^{x^2} is (e^{16}-1)/8