What is the integral from 0 to 1 of x(4x^4-(x^4)/4) ?

The integral from 0 to 1 of x(4x^4-(x^4)/4) is 5/8

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integral from 0 to 1 of x(4x^4-(x^4)/4)

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

\frac{5}{8}

Decimal

0.625

Solution steps

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

Simplify

4 x^{4} - \frac{x ^{4}}{4} : \frac{15 x ^{4}}{4}

= \int_{0}^{1} x \frac{15 x ^{4}}{4} d x

Take the constant out:

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

= \frac{15}{4} \cdot \int_{0}^{1} x x^{4} d x

Simplify

x x^{4} : x^{5}

= \frac{15}{4} \cdot \int_{0}^{1} x^{5} d x

Apply the Power Rule

: [\frac{x ^{6}}{6}]_{0}^{1}

= \frac{15}{4} [\frac{x ^{6}}{6}]_{0}^{1}

Compute the boundaries:

\frac{1}{6}

= \frac{15}{4} \cdot \frac{1}{6}

Simplify

= \frac{5}{8}

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

\int_{a}^{\infty} \frac{1}{x ^{3}} d x

\int_{0}^{r} r^{2} - r^{2} r d r

\int_{1}^{3} \frac{8 x ^{2} - 7 x + 5}{x} d x

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

What is the integral from 0 to 1 of x(4x^4-(x^4)/4) ?
The integral from 0 to 1 of x(4x^4-(x^4)/4) is 5/8