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

The integral from 1 to 2 of 8xe^{x^2+2} is 4(e^6-e^3)

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

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

4 (e^{6} - e^{3})

Decimal Notation

1533.37302 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{6} \frac{e ^{u}}{2} d u

= 8 \cdot \int_{3}^{6} \frac{e ^{u}}{2} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int_{3}^{6} e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} [e^{u}]_{3}^{6}

Simplify

8 \cdot \frac{1}{2} [e^{u}]_{3}^{6} : 4 [e^{u}]_{3}^{6}

= 4 [e^{u}]_{3}^{6}

Compute the boundaries:

e^{6} - e^{3}

= 4 (e^{6} - e^{3})

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What is the integral from 1 to 2 of 8xe^{x^2+2} ?
The integral from 1 to 2 of 8xe^{x^2+2} is 4(e^6-e^3)