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

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

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

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

1 - \frac{1}{e}

Decimal Notation

0.63212 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

= [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 1 - \frac{1}{e}

\frac{d}{d x} (4 x^{4} + 2 x^{3} - x + 1)

\int sin^{2} (n x) d x

y = \frac{( x + b ) ( x - a )}{( x - b )}

\int e^{(x - y)} d y

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