What is the integral from 1 to 4 of xe^{6x^2-9} ?

The integral from 1 to 4 of xe^{6x^2-9} is (e^{90}-1)/(12e^3)

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integral from 1 to 4 of xe^{6x^2-9}

\int_{1}^{4} x e^{6 x^{2} - 9} d x

\frac{e ^{90} - 1}{12 e ^{3}}

Decimal

5.06336 E 36

Solution steps

\int_{1}^{4} x e^{6 x^{2} - 9} d x

Apply u-substitution

: \int_{- 3}^{87} \frac{e ^{u}}{12} d u

= \int_{- 3}^{87} \frac{e ^{u}}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{- 3}^{87} e^{u} d u

Use the common integral:

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

= \frac{1}{12} [e^{u}]_{- 3}^{87}

Compute the boundaries:

e^{87} - \frac{1}{e ^{3}}

= \frac{1}{12} (e^{87} - \frac{1}{e ^{3}})

Simplify

= \frac{e ^{90} - 1}{12 e ^{3}}

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What is the integral from 1 to 4 of xe^{6x^2-9} ?
The integral from 1 to 4 of xe^{6x^2-9} is (e^{90}-1)/(12e^3)