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

The integral from-1 to 2 of xe^{6x} is (11e^{18}+7)/(36e^6)

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

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

\frac{11 e ^{18} + 7}{36 e ^{6}}

Solution steps

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

Apply u-substitution

: \int_{- 6}^{12} \frac{e ^{u} u}{36} d u

= \int_{- 6}^{12} \frac{e ^{u} u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int_{- 6}^{12} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 6}^{12}

= \frac{1}{36} [e^{u} u - \int e^{u} d u]_{- 6}^{12}

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

= \frac{1}{36} [e^{u} u - e^{u}]_{- 6}^{12}

Compute the boundaries:

11 e^{12} + \frac{7}{e ^{6}}

= \frac{1}{36} (11 e^{12} + \frac{7}{e ^{6}})

Simplify

= \frac{11 e ^{18} + 7}{36 e ^{6}}

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What is the integral from-1 to 2 of xe^{6x} ?
The integral from-1 to 2 of xe^{6x} is (11e^{18}+7)/(36e^6)