What is the integral from-1 to 9 of (9-x^2+8x) ?

The integral from-1 to 9 of (9-x^2+8x) is 500/3

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integral from-1 to 9 of (9-x^2+8x)

\int_{- 1}^{9} (9 - x^{2} + 8 x) d x

\frac{500}{3}

Decimal

166.66666 \dots

Solution steps

\int_{- 1}^{9} 9 - x^{2} + 8 x d x

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}^{9} 9 d x - \int_{- 1}^{9} x^{2} d x + \int_{- 1}^{9} 8 x d x

\int_{- 1}^{9} 9 d x = 90

\int_{- 1}^{9} x^{2} d x = \frac{730}{3}

\int_{- 1}^{9} 8 x d x = 320

= 90 - \frac{730}{3} + 320

Simplify

90 - \frac{730}{3} + 320 : \frac{500}{3}

= \frac{500}{3}

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\int_{0}^{5} 40 e^{- 0.3 t} d t

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What is the integral from-1 to 9 of (9-x^2+8x) ?
The integral from-1 to 9 of (9-x^2+8x) is 500/3