What is the integral from 0 to r of-pir^2x ?

The integral from 0 to r of-pir^2x is -(pir^4)/2

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integral from 0 to r of-pir^2x

\int_{0}^{r} - π r^{2} x d x

- \frac{π r ^{4}}{2}

Solution steps

\int_{0}^{r} - π r^{2} x d x

Take the constant out:

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

= - π r^{2} \cdot \int_{0}^{r} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{r}

= - π r^{2} [\frac{x ^{2}}{2}]_{0}^{r}

Compute the boundaries:

\frac{r ^{2}}{2}

= - π r^{2} \frac{r ^{2}}{2}

Simplify

= - \frac{π r ^{4}}{2}

\int 8 x^{7} e^{2 x^{8}} d x

arccos (a x)

\frac{\partial}{\partial x} (x^{2} + y^{2} + 10)

\int_{0}^{\infty} 15 e^{- 5 x} d x

f (x) = 6 x - x^{2}, (1, 5)

What is the integral from 0 to r of-pir^2x ?
The integral from 0 to r of-pir^2x is -(pir^4)/2