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

The integral from 2 to 3 of xe^{3x} is (8e^9-5e^6)/9

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

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

\frac{8 e ^{9} - 5 e ^{6}}{9}

Decimal Notation

6978.61416 \dots

Solution steps

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

Apply u-substitution

: \int_{6}^{9} \frac{e ^{u} u}{9} d u

= \int_{6}^{9} \frac{e ^{u} u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int_{6}^{9} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{6}^{9}

= \frac{1}{9} [e^{u} u - \int e^{u} d u]_{6}^{9}

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

= \frac{1}{9} [e^{u} u - e^{u}]_{6}^{9}

Compute the boundaries:

8 e^{9} - 5 e^{6}

= \frac{1}{9} (8 e^{9} - 5 e^{6})

Simplify

= \frac{8 e ^{9} - 5 e ^{6}}{9}

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

\int_{0}^{1} (2 - x)^{4} d x

\int_{- 1}^{1} (x^{3} - x^{5}) d x

\int_{2}^{3} \frac{2 x}{1 + x ^{2}} d x

\int_{0}^{3} e^{\frac{x}{3}} (x^{2} + 2 x) d x

What is the integral from 2 to 3 of xe^{3x} ?
The integral from 2 to 3 of xe^{3x} is (8e^9-5e^6)/9