What is the integral from 1 to 2 of 8xe^{2x^2} ?

The integral from 1 to 2 of 8xe^{2x^2} is 2(e^8-e^2)

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integral from 1 to 2 of 8xe^{2x^2}

\int_{1}^{2} 8 x e^{2 x^{2}} d x

2 (e^{8} - e^{2})

Decimal

5947.13786 \dots

Solution steps

\int_{1}^{2} 8 x e^{2 x^{2}} d x

Take the constant out:

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

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

Apply u-substitution

: \int_{2}^{8} \frac{e ^{u}}{4} d u

= 8 \cdot \int_{2}^{8} \frac{e ^{u}}{4} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int_{2}^{8} e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{4} [e^{u}]_{2}^{8}

Simplify

8 \cdot \frac{1}{4} [e^{u}]_{2}^{8} : 2 [e^{u}]_{2}^{8}

= 2 [e^{u}]_{2}^{8}

Compute the boundaries:

e^{8} - e^{2}

= 2 (e^{8} - e^{2})

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What is the integral from 1 to 2 of 8xe^{2x^2} ?
The integral from 1 to 2 of 8xe^{2x^2} is 2(e^8-e^2)