What is the integral from 0 to 1 of x^2e^x ?

The integral from 0 to 1 of x^2e^x is e-2

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integral from 0 to 1 of x^2e^x

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

e - 2

Decimal Notation

0.71828 \dots

Solution steps

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

Apply Integration By Parts

: [x^{2} e^{x} - \int 2 x e^{x} d x]_{0}^{1}

= [x^{2} e^{x} - \int 2 x e^{x} d x]_{0}^{1}

\int 2 x e^{x} d x = 2 (e^{x} x - e^{x})

= [x^{2} e^{x} - 2 (e^{x} x - e^{x})]_{0}^{1}

Compute the boundaries:

e - 2

= e - 2

y = \frac{16}{x ^{2} - 6 x - 55}, y = 0, x = - 3, x = 3

\frac{d}{d x} (\frac{x ^{3}}{( 1 - x ) ^{2}})

\frac{d}{d x} (x^{2} + y^{2} - z)

\int x^{r} d x

What is the integral from 0 to 1 of x^2e^x ?
The integral from 0 to 1 of x^2e^x is e-2