What is the integral from-1 to 1 of t(8-t)^2 ?

The integral from-1 to 1 of t(8-t)^2 is -32/3

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integral from-1 to 1 of t(8-t)^2

\int_{- 1}^{1} t (8 - t)^{2} d t

- \frac{32}{3}

Decimal

- 10.66666 \dots

Solution steps

\int_{- 1}^{1} t (8 - t)^{2} d t

Expand

t (8 - t)^{2} : 64 t - 16 t^{2} + t^{3}

= \int_{- 1}^{1} 64 t - 16 t^{2} + t^{3} d t

Apply the Sum Rule:

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

= \int_{- 1}^{1} 64 t d t - \int_{- 1}^{1} 16 t^{2} d t + \int_{- 1}^{1} t^{3} d t

\int_{- 1}^{1} 64 t d t = 0

\int_{- 1}^{1} 16 t^{2} d t = \frac{32}{3}

\int_{- 1}^{1} t^{3} d t = 0

= 0 - \frac{32}{3} + 0

Simplify

= - \frac{32}{3}

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What is the integral from-1 to 1 of t(8-t)^2 ?
The integral from-1 to 1 of t(8-t)^2 is -32/3