What is the integral from 0 to 1 of r^3 ?

The integral from 0 to 1 of r^3 is 1/4

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integral from 0 to 1 of r^3

\int_{0}^{1} r^{3} d r

\frac{1}{4}

Decimal Notation

0.25

Solution steps

\int_{0}^{1} r^{3} d r

Apply the Power Rule

: [\frac{r ^{4}}{4}]_{0}^{1}

= [\frac{r ^{4}}{4}]_{0}^{1}

Compute the boundaries:

\frac{1}{4}

= \frac{1}{4}

\frac{\partial}{\partial x} ((x^{2} + y^{2}) e^{x^{2} y})

\frac{x}{( x ^{2} - 1 )}

\int sin (x) e^{2 x} d x

f (x) = x, (25, 5)

\int (u - 2)^{3} ln (u - 2) d u

What is the integral from 0 to 1 of r^3 ?
The integral from 0 to 1 of r^3 is 1/4