What is the integral from 0 to infinity of 2xe^{-2x} ?

The integral from 0 to infinity of 2xe^{-2x} is 1/2

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integral from 0 to infinity of 2xe^{-2x}

\int_{0}^{\infty} 2 x e^{- 2 x} d x

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{\infty} 2 x e^{- 2 x} d x

Compute the indefinite integral:

\int 2 x e^{- 2 x} d x = - e^{- 2 x} x - \frac{1}{2} e^{- 2 x} + C

Compute the boundaries:

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

= 0 - (- \frac{1}{2})

Simplify

= \frac{1}{2}

\frac{d}{d x} (x x + 1)

\int (x^{2} + 5) cos (3 x) d x

f (x) = x^{3} cos (2 x)

x \frac{d y}{d x} = x^{3} + 9 x^{3} y

y^{^{''}} = 6 x, y (0) = 1, y^{^{'}} (0) = - 1

What is the integral from 0 to infinity of 2xe^{-2x} ?
The integral from 0 to infinity of 2xe^{-2x} is 1/2