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 Notation

166.66666 \dots

Solution steps

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

Simplify

= \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}

y^{^{'}} = - 3 y^{2}

\frac{d}{d x} (\frac{1}{12} (x + 3)^{3} + (x + 3)^{- 1})

\int_{1}^{4} t ln (\frac{t}{2}) d t

\int \frac{4 x}{1 - 16 x ^{4}} d x

y = \frac{x}{a} + \frac{a}{x}, at x = \frac{a}{2}

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