What is the integral from 0 to 8 of y^2(8-y) ?

The integral from 0 to 8 of y^2(8-y) is 1024/3

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integral from 0 to 8 of y^2(8-y)

\int_{0}^{8} y^{2} (8 - y) d y

\frac{1024}{3}

Decimal Notation

341.33333 \dots

Solution steps

\int_{0}^{8} y^{2} (8 - y) d y

Expand

y^{2} (8 - y) : 8 y^{2} - y^{3}

= \int_{0}^{8} 8 y^{2} - y^{3} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{8} 8 y^{2} d y - \int_{0}^{8} y^{3} d y

\int_{0}^{8} 8 y^{2} d y = \frac{4096}{3}

\int_{0}^{8} y^{3} d y = 1024

= \frac{4096}{3} - 1024

Simplify

\frac{4096}{3} - 1024 : \frac{1024}{3}

= \frac{1024}{3}

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\frac{d}{d x} (ln (x^{16}))

What is the integral from 0 to 8 of y^2(8-y) ?
The integral from 0 to 8 of y^2(8-y) is 1024/3