What is the integral from 1 to 2 of x^5e^{x^2} ?

The integral from 1 to 2 of x^5e^{x^2} is (10e^4-e)/2

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integral from 1 to 2 of x^5e^{x^2}

\int_{1}^{2} x^{5} e^{x^{2}} d x

\frac{10 e ^{4} - e}{2}

Solution steps

\int_{1}^{2} x^{5} e^{x^{2}} d x

Apply u-substitution

: \int_{1}^{32} \frac{e ^{u^{\frac{2}{5}}} u ^{\frac{1}{5}}}{5} d u

= \int_{1}^{32} \frac{e ^{u^{\frac{2}{5}}} u ^{\frac{1}{5}}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int_{1}^{32} e^{u^{\frac{2}{5}}} u^{\frac{1}{5}} d u

Apply u-substitution

: \int_{1}^{4} \frac{5 e ^{v} v ^{2}}{2} d v

= \frac{1}{5} \cdot \int_{1}^{4} \frac{5 e ^{v} v ^{2}}{2} d v

Take the constant out:

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

= \frac{1}{5} \cdot \frac{5}{2} \cdot \int_{1}^{4} e^{v} v^{2} d v

Apply Integration By Parts

: [v^{2} e^{v} - \int 2 v e^{v} d v]_{1}^{4}

= \frac{1}{5} \cdot \frac{5}{2} [v^{2} e^{v} - \int 2 v e^{v} d v]_{1}^{4}

\int 2 v e^{v} d v = 2 (e^{v} v - e^{v})

= \frac{1}{5} \cdot \frac{5}{2} [v^{2} e^{v} - 2 (e^{v} v - e^{v})]_{1}^{4}

Simplify

\frac{1}{5} \cdot \frac{5}{2} [v^{2} e^{v} - 2 (e^{v} v - e^{v})]_{1}^{4} : \frac{1}{2} [e^{v} v^{2} - 2 (e^{v} v - e^{v})]_{1}^{4}

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

Compute the boundaries:

10 e^{4} - e

= \frac{1}{2} (10 e^{4} - e)

Simplify

= \frac{10 e ^{4} - e}{2}

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What is the integral from 1 to 2 of x^5e^{x^2} ?
The integral from 1 to 2 of x^5e^{x^2} is (10e^4-e)/2