What is the integral from 0 to 4 of 6xe^{3x^2} ?

The integral from 0 to 4 of 6xe^{3x^2} is e^{48}-1

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integral from 0 to 4 of 6xe^{3x^2}

\int_{0}^{4} 6 x e^{3 x^{2}} d x

e^{48} - 1

Decimal

7.01674 E 20

Solution steps

\int_{0}^{4} 6 x e^{3 x^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{4} x e^{3 x^{2}} d x

Apply u-substitution

: \int_{0}^{48} \frac{e ^{u}}{6} d u

= 6 \cdot \int_{0}^{48} \frac{e ^{u}}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{0}^{48} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} [e^{u}]_{0}^{48}

Simplify

6 \cdot \frac{1}{6} [e^{u}]_{0}^{48} : [e^{u}]_{0}^{48}

= [e^{u}]_{0}^{48}

Compute the boundaries:

e^{48} - 1

= e^{48} - 1

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What is the integral from 0 to 4 of 6xe^{3x^2} ?
The integral from 0 to 4 of 6xe^{3x^2} is e^{48}-1