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

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

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

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

z (1 - \frac{1}{e})

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

1 - \frac{1}{e}

= z (1 - \frac{1}{e})

\int_{0}^{zy} z d z

\int_{0}^{\frac{π}{2}} (x - sin (x)) d x

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

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\int_{0}^{y} x (y^{2} - x^{2})^{\frac{3}{2}} d x

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