What is the integral from 1 to 2 of 4x^3e^{x^4} ?

The integral from 1 to 2 of 4x^3e^{x^4} is e^{16}-e

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integral from 1 to 2 of 4x^3e^{x^4}

\int_{1}^{2} 4 x^{3} e^{x^{4}} d x

e^{16} - e

Decimal Notation

8886107.80222 \dots

Solution steps

\int_{1}^{2} 4 x^{3} e^{x^{4}} d x

Take the constant out:

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

= 4 \cdot \int_{1}^{2} x^{3} e^{x^{4}} d x

Apply u-substitution

: \int_{1}^{8} \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

= 4 \cdot \int_{1}^{8} \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int_{1}^{8} e^{u^{\frac{4}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int_{1}^{16} \frac{3 e ^{v}}{4} d v

= 4 \cdot \frac{1}{3} \cdot \int_{1}^{16} \frac{3 e ^{v}}{4} d v

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \frac{3}{4} \cdot \int_{1}^{16} e^{v} d v

Use the common integral:

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

= 4 \cdot \frac{1}{3} \cdot \frac{3}{4} [e^{v}]_{1}^{16}

Simplify

4 \cdot \frac{1}{3} \cdot \frac{3}{4} [e^{v}]_{1}^{16} : [e^{v}]_{1}^{16}

= [e^{v}]_{1}^{16}

Compute the boundaries:

e^{16} - e

= e^{16} - e

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What is the integral from 1 to 2 of 4x^3e^{x^4} ?
The integral from 1 to 2 of 4x^3e^{x^4} is e^{16}-e