What is the integral from-infinity to 1 of xe^{2x} ?

The integral from-infinity to 1 of xe^{2x} is (e^2)/4

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integral from-infinity to 1 of xe^{2x}

\int_{- \infty}^{1} x e^{2 x} d x

\frac{e ^{2}}{4}

Decimal Notation

1.84726 \dots

Solution steps

\int_{- \infty}^{1} x e^{2 x} d x

Compute the indefinite integral:

\int x e^{2 x} d x = \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) + C

Compute the boundaries:

\int_{- \infty}^{1} x e^{2 x} d x = \frac{e ^{2}}{4} - 0

= \frac{e ^{2}}{4} - 0

Simplify

= \frac{e ^{2}}{4}

\int x^{2} - a^{2} d x

- \frac{1}{4 s}

f (x) = x^{π^{2}} + (π^{2})^{x}

n = 1 \sum \infty ln (\frac{1}{1 + 1})

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

What is the integral from-infinity to 1 of xe^{2x} ?
The integral from-infinity to 1 of xe^{2x} is (e^2)/4