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

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

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

\int_{0}^{1} \frac{4 x}{e ^{x^{2}}} d x

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

Decimal Notation

1.26424 \dots

Solution steps

\int_{0}^{1} \frac{4 x}{e ^{x^{2}}} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} \frac{x}{e ^{x^{2}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x}{e ^{x^{2}}} = (x) e^{- x^{2}}

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

Apply u-substitution

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

= 4 \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

= 4 (- \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

= 4 (- (- \frac{1}{2} \cdot \int_{- 1}^{0} e^{u} d u))

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

1 - \frac{1}{e}

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

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{0.5})

\int \frac{4 - x}{x ^{4} - x ^{2}} d x

x \to 2 lim (2 x + c)

f (x) = \frac{1 - ln ( x )}{2 x ^{2}}

\frac{d y}{d x} = ∣ x - 2 ∣

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