What is the integral from-1 to 1 of 3x^2e^{3x^3-7} ?

The integral from-1 to 1 of 3x^2e^{3x^3-7} is (e^6-1)/(3e^{10)}

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integral from-1 to 1 of 3x^2e^{3x^3-7}

\int_{- 1}^{1} 3 x^{2} e^{3 x^{3} - 7} d x

\frac{e ^{6} - 1}{3 e ^{10}}

Decimal

0.00609 \dots

Solution steps

\int_{- 1}^{1} 3 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

= 3 \cdot \int_{- 1}^{1} x^{2} e^{3 x^{3} - 7} d x

Apply u-substitution

= 3 \cdot \int_{- 1}^{1} \frac{e ^{3 u - 7}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{- 1}^{1} e^{3 u - 7} d u

Apply u-substitution

= 3 \cdot \frac{1}{3} \cdot \int_{- 10}^{- 4} e^{v} \frac{1}{3} d v

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \int_{- 10}^{- 4} e^{v} d v

Use the common integral:

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

= 3 \cdot \frac{1}{3} \cdot \frac{1}{3} [e^{v}]_{- 10}^{- 4}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{1}{3} [e^{v}]_{- 10}^{- 4} : \frac{1}{3} [e^{v}]_{- 10}^{- 4}

= \frac{1}{3} [e^{v}]_{- 10}^{- 4}

Compute the boundaries:

\frac{1}{e ^{4}} - \frac{1}{e ^{10}}

= \frac{1}{3} (\frac{1}{e ^{4}} - \frac{1}{e ^{10}})

Simplify

= \frac{e ^{6} - 1}{3 e ^{10}}

What is the integral from-1 to 1 of 3x^2e^{3x^3-7} ?
The integral from-1 to 1 of 3x^2e^{3x^3-7} is (e^6-1)/(3e^{10)}