What is the integral from 0 to 2 of xe^{5x} ?

The integral from 0 to 2 of xe^{5x} is (9e^{10}+1)/(25)

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integral from 0 to 2 of xe^{5x}

\int_{0}^{2} x e^{5 x} d x

\frac{9 e ^{10} + 1}{25}

Decimal Notation

7929.56768 \dots

Solution steps

\int_{0}^{2} x e^{5 x} d x

Apply u-substitution

: \int_{0}^{10} \frac{e ^{u} u}{25} d u

= \int_{0}^{10} \frac{e ^{u} u}{25} d u

Take the constant out:

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

= \frac{1}{25} \cdot \int_{0}^{10} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{10}

= \frac{1}{25} [e^{u} u - \int e^{u} d u]_{0}^{10}

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

= \frac{1}{25} [e^{u} u - e^{u}]_{0}^{10}

Compute the boundaries:

9 e^{10} + 1

= \frac{1}{25} (9 e^{10} + 1)

Simplify

= \frac{9 e ^{10} + 1}{25}

\int_{- 10}^{10} (∣ x ∣ + 5) d x

\int_{0}^{\frac{π}{2}} \frac{sin ( 2 x )}{2} d x

\int_{- 4}^{1} (- x^{2} - 3 x + 4) d x

\int_{- 1}^{7} x 2 + x d x

\int_{0}^{3} 25 - x^{2} d x

What is the integral from 0 to 2 of xe^{5x} ?
The integral from 0 to 2 of xe^{5x} is (9e^{10}+1)/(25)