What is the integral from 0 to 1/15 of xln(15x) ?

The integral from 0 to 1/15 of xln(15x) is -1/900

Popular >

integral from 0 to 1/15 of xln(15x)

\int_{0}^{\frac{1}{15}} x ln (15 x) d x

- \frac{1}{900}

Decimal Notation

- 0.00111 \dots

Solution steps

\int_{0}^{\frac{1}{15}} x ln (15 x) d x

Apply Integration By Parts

: [\frac{1}{2} (x^{2} ln (15 x) - 2 \cdot \int \frac{x}{2} d x)]_{0}^{\frac{1}{15}}

= [\frac{1}{2} (x^{2} ln (15 x) - 2 \cdot \int \frac{x}{2} d x)]_{0}^{\frac{1}{15}}

\int \frac{x}{2} d x = \frac{x ^{2}}{4}

= [\frac{1}{2} (x^{2} ln (15 x) - 2 \cdot \frac{x ^{2}}{4})]_{0}^{\frac{1}{15}}

2 \cdot \frac{x ^{2}}{4} = \frac{x ^{2}}{2}

= [\frac{1}{2} (x^{2} ln (15 x) - \frac{x ^{2}}{2})]_{0}^{\frac{1}{15}}

Compute the boundaries:

- \frac{1}{900}

= - \frac{1}{900}

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

\int_{x^{2}}^{2 x} x^{2} + y^{2} d y

\int_{- 1}^{0} \frac{x}{x + 1} d x

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

\int_{0}^{2} \frac{x}{( x + 1 ) ( x + 2 )} d x

What is the integral from 0 to 1/15 of xln(15x) ?
The integral from 0 to 1/15 of xln(15x) is -1/900