What is the integral from 2 to 8 of xe^{x^2} ?

The integral from 2 to 8 of xe^{x^2} is (e^{64}-e^4)/2

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integral from 2 to 8 of xe^{x^2}

\int_{2}^{8} x e^{x^{2}} d x

\frac{e ^{64} - e ^{4}}{2}

Decimal Notation

3.11757 E 27

Solution steps

\int_{2}^{8} x e^{x^{2}} d x

Apply u-substitution

: \int_{4}^{64} \frac{e ^{u}}{2} d u

= \int_{4}^{64} \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_{4}^{64} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{4}^{64}

Compute the boundaries:

e^{64} - e^{4}

= \frac{1}{2} (e^{64} - e^{4})

Simplify

= \frac{e ^{64} - e ^{4}}{2}

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

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\int_{x^{2}}^{x^{4}} (t + t^{2})^{2} d t

What is the integral from 2 to 8 of xe^{x^2} ?
The integral from 2 to 8 of xe^{x^2} is (e^{64}-e^4)/2