What is the integral from 0 to 1 of x*e^{x^2} ?

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

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integral from 0 to 1 of x*e^{x^2}

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

\frac{e - 1}{2}

Decimal Notation

0.85914 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e - 1

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

Simplify

= \frac{e - 1}{2}

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

(sin (x) - y sin (x) - 2 cos (x)) + cos (x) y^{^{'}} = 0

\frac{d}{d x} (\frac{e ^{4 x}}{ln ( 6 x )})

\int \frac{2 x}{e ^{x}} d x

\frac{d}{d x} (36 x^{2})

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