What is the integral from 0 to 1 of 7xe^{-x^2} ?

The integral from 0 to 1 of 7xe^{-x^2} is (7(e-1))/(2e)

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integral from 0 to 1 of 7xe^{-x^2}

\int_{0}^{1} 7 x e^{- x^{2}} d x

\frac{7 ( e - 1 )}{2 e}

Decimal Notation

2.21242 \dots

Solution steps

\int_{0}^{1} 7 x e^{- x^{2}} d x

Take the constant out:

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

= 7 \cdot \int_{0}^{1} x e^{- x^{2}} d x

Apply u-substitution

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

= 7 \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

= 7 (- \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

= 7 (- (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u))

Use the common integral:

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

= 7 (- (- \frac{1}{2} [e^{u}]_{- 1}^{0}))

Simplify

7 (- (- \frac{1}{2} [e^{u}]_{- 1}^{0})) : \frac{7}{2} [e^{u}]_{- 1}^{0}

= \frac{7}{2} [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= \frac{7}{2} (1 - \frac{1}{e})

Simplify

= \frac{7 ( e - 1 )}{2 e}

\frac{\partial}{\partial x} (x y^{2} + 7)

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f (x) = cos (2 \cdot x) \cdot sin (5 \cdot x)

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

\frac{d}{d x} ((x^{4} - 3 x^{2} + 5)^{3})

What is the integral from 0 to 1 of 7xe^{-x^2} ?
The integral from 0 to 1 of 7xe^{-x^2} is (7(e-1))/(2e)