What is the integral from 0 to 4 of xe^x ?

The integral from 0 to 4 of xe^x is 3e^4+1

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integral from 0 to 4 of xe^x

\int_{0}^{4} x e^{x} d x

3 e^{4} + 1

Decimal Notation

164.79445 \dots

Solution steps

\int_{0}^{4} x e^{x} d x

Apply Integration By Parts

: [e^{x} x - \int e^{x} d x]_{0}^{4}

= [e^{x} x - \int e^{x} d x]_{0}^{4}

\int e^{x} d x = e^{x}

= [e^{x} x - e^{x}]_{0}^{4}

Compute the boundaries:

3 e^{4} + 1

= 3 e^{4} + 1

(ln (x^{2} + 1))^{^{'}}

\frac{d x}{d t} = k x (k - x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (- 1) = 0, y^{^{'}} (- 1) = 1

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

\int e^{3 x} sin (5 x) d x

What is the integral from 0 to 4 of xe^x ?
The integral from 0 to 4 of xe^x is 3e^4+1