What is the integral from-2 to 0 of 2x^2e^{3x^3+7} ?

The integral from-2 to 0 of 2x^2e^{3x^3+7} is (2(e^{24}-1))/(9e^{17)}

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integral from-2 to 0 of 2x^2e^{3x^3+7}

\int_{- 2}^{0} 2 x^{2} e^{3 x^{3} + 7} d x

\frac{2 ( e ^{24} - 1 )}{9 e ^{17}}

Decimal Notation

243.69625 \dots

Solution steps

\int_{- 2}^{0} 2 x^{2} e^{3 x^{3} + 7} d x

Take the constant out:

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

= 2 \cdot \int_{- 2}^{0} x^{2} e^{3 x^{3} + 7} d x

Apply u-substitution

: \int_{- 17}^{7} \frac{e ^{u}}{9} d u

= 2 \cdot \int_{- 17}^{7} \frac{e ^{u}}{9} d u

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int_{- 17}^{7} e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{9} [e^{u}]_{- 17}^{7}

Simplify

2 \cdot \frac{1}{9} [e^{u}]_{- 17}^{7} : \frac{2}{9} [e^{u}]_{- 17}^{7}

= \frac{2}{9} [e^{u}]_{- 17}^{7}

Compute the boundaries:

e^{7} - \frac{1}{e ^{17}}

= \frac{2}{9} (e^{7} - \frac{1}{e ^{17}})

Simplify

= \frac{2 ( e ^{24} - 1 )}{9 e ^{17}}

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What is the integral from-2 to 0 of 2x^2e^{3x^3+7} ?
The integral from-2 to 0 of 2x^2e^{3x^3+7} is (2(e^{24}-1))/(9e^{17)}