What is the integral from 0 to 1 of x^3ln(x^8) ?

The integral from 0 to 1 of x^3ln(x^8) is -1/2

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integral from 0 to 1 of x^3ln(x^8)

\int_{0}^{1} x^{3} ln (x^{8}) d x

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

\int_{0}^{1} x^{3} ln (x^{8}) d x

Simplify

ln (x^{8}) : 8 ln (x)

= \int_{0}^{1} 8 x^{3} ln (x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{1} x^{3} ln (x) d x

Apply Integration By Parts

: [\frac{1}{4} x^{4} ln (x) - \int \frac{x ^{3}}{4} d x]_{0}^{1}

= 8 [\frac{1}{4} x^{4} ln (x) - \int \frac{x ^{3}}{4} d x]_{0}^{1}

\int \frac{x ^{3}}{4} d x = \frac{x ^{4}}{16}

= 8 [\frac{1}{4} x^{4} ln (x) - \frac{x ^{4}}{16}]_{0}^{1}

Compute the boundaries:

- \frac{1}{16}

= 8 (- \frac{1}{16})

Simplify

= - \frac{1}{2}

f (x) = \frac{x}{x + 3}, at x = 1

\frac{d}{d x} (cos^{2} (4 π x))

f (x) = x^{2} + 2

\frac{d y}{d t} = y (8 - y), y (0) = 9

\int sec^{2} (2 x) tan^{2} (2 x) d x

What is the integral from 0 to 1 of x^3ln(x^8) ?
The integral from 0 to 1 of x^3ln(x^8) is -1/2