What is the integral from 0 to 1 of 2pixe^{-x^2} ?

The integral from 0 to 1 of 2pixe^{-x^2} is pi(1-1/e)

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integral from 0 to 1 of 2pixe^{-x^2}

\int_{0}^{1} 2 π x e^{- x^{2}} d x

π (1 - \frac{1}{e})

Decimal Notation

1.98586 \dots

Solution steps

\int_{0}^{1} 2 π x e^{- x^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π \cdot \int_{0}^{1} x e^{- x^{2}} d x

Apply u-substitution

: \int_{0}^{- 1} - \frac{e ^{u}}{2} d u

= 2 π \cdot \int_{0}^{- 1} - \frac{e ^{u}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 2 π (- \int_{- 1}^{0} - \frac{e ^{u}}{2} d u)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 π (- (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u))

Use the common integral:

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

= 2 π (- (- \frac{1}{2} [e^{u}]_{- 1}^{0}))

Simplify

2 π (- (- \frac{1}{2} [e^{u}]_{- 1}^{0})) : π [e^{u}]_{- 1}^{0}

= π [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= π (1 - \frac{1}{e})

\frac{d}{d x} (x y^{^{'}} - y)

\int \frac{1}{x ^{2} + x - 6} d x

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\frac{d}{d x} (ln (x^{2} - 5 x + 2))

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What is the integral from 0 to 1 of 2pixe^{-x^2} ?
The integral from 0 to 1 of 2pixe^{-x^2} is pi(1-1/e)