What is the integral from 0 to 1 of 2x^8ln(x) ?

The integral from 0 to 1 of 2x^8ln(x) is -2/81

Popular >

integral from 0 to 1 of 2x^8ln(x)

\int_{0}^{1} 2 x^{8} ln (x) d x

- \frac{2}{81}

Decimal Notation

- 0.02469 \dots

Solution steps

\int_{0}^{1} 2 x^{8} ln (x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} x^{8} ln (x) d x

Apply Integration By Parts

: [\frac{1}{9} x^{9} ln (x) - \int \frac{x ^{8}}{9} d x]_{0}^{1}

= 2 [\frac{1}{9} x^{9} ln (x) - \int \frac{x ^{8}}{9} d x]_{0}^{1}

\int \frac{x ^{8}}{9} d x = \frac{x ^{9}}{81}

= 2 [\frac{1}{9} x^{9} ln (x) - \frac{x ^{9}}{81}]_{0}^{1}

Compute the boundaries:

- \frac{1}{81}

= 2 (- \frac{1}{81})

Simplify

= - \frac{2}{81}

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

\int_{0}^{t} sin (x) d x

\int_{0}^{1} x^{4} e^{- x^{5}} d x

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

\int_{0}^{\frac{3 π}{4}} sec^{2} (x) d x

What is the integral from 0 to 1 of 2x^8ln(x) ?
The integral from 0 to 1 of 2x^8ln(x) is -2/81