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

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

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

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

\frac{e ^{48} - e ^{3}}{6}

Decimal

1.16946 E 20

Solution steps

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

Apply u-substitution

: \int_{3}^{48} \frac{e ^{u}}{6} d u

= \int_{3}^{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

= \frac{1}{6} \cdot \int_{3}^{48} e^{u} d u

Use the common integral:

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

= \frac{1}{6} [e^{u}]_{3}^{48}

Compute the boundaries:

e^{48} - e^{3}

= \frac{1}{6} (e^{48} - e^{3})

Simplify

= \frac{e ^{48} - e ^{3}}{6}

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