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

The integral from-1 to 2 of 2x^2e^{2x^3} is (e^{18}-1)/(3e^2)

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

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

\frac{e ^{18} - 1}{3 e ^{2}}

Decimal

2962036.79505 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- 2}^{16} \frac{e ^{u}}{6} d u

= 2 \cdot \int_{- 2}^{16} \frac{e ^{u}}{6} d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int_{- 2}^{16} e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{6} [e^{u}]_{- 2}^{16}

Simplify

2 \cdot \frac{1}{6} [e^{u}]_{- 2}^{16} : \frac{1}{3} [e^{u}]_{- 2}^{16}

= \frac{1}{3} [e^{u}]_{- 2}^{16}

Compute the boundaries:

e^{16} - \frac{1}{e ^{2}}

= \frac{1}{3} (e^{16} - \frac{1}{e ^{2}})

Simplify

= \frac{e ^{18} - 1}{3 e ^{2}}

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What is the integral from-1 to 2 of 2x^2e^{2x^3} ?
The integral from-1 to 2 of 2x^2e^{2x^3} is (e^{18}-1)/(3e^2)