What is the integral from 0 to 1 of 2pix(4e^{-x^2}) ?

The integral from 0 to 1 of 2pix(4e^{-x^2}) is 4pi(1-1/e)

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integral from 0 to 1 of 2pix(4e^{-x^2})

\int_{0}^{1} 2 π x (4 e^{- x^{2}}) d x

4 π (1 - \frac{1}{e})

Decimal Notation

7.94346 \dots

Solution steps

\int_{0}^{1} 2 π x \cdot 4 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 4 \cdot \int_{0}^{1} x e^{- x^{2}} d x

Simplify

= 8 π \cdot \int_{0}^{1} x e^{- x^{2}} d x

Apply u-substitution

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

= 8 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

= 8 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

= 8 π (- (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u))

Use the common integral:

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

= 8 π (- (- \frac{1}{2} [e^{u}]_{- 1}^{0}))

Simplify

8 π (- (- \frac{1}{2} [e^{u}]_{- 1}^{0})) : 4 π [e^{u}]_{- 1}^{0}

= 4 π [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= 4 π (1 - \frac{1}{e})

y = ln (3^{3 x})

{\frac{s}{( s + 2 ) ( s ^{2} + 4 )}}

\frac{d}{d x} (2 e^{3 x^{2} + 6 x + 7})

\int cos (3 x - 2) d x

f (x) = \frac{1}{x + 1}, at x = 1

What is the integral from 0 to 1 of 2pix(4e^{-x^2}) ?
The integral from 0 to 1 of 2pix(4e^{-x^2}) is 4pi(1-1/e)