What is the integral from-2 to 6 of 9-(x/2)^2-(6-x) ?

The integral from-2 to 6 of 9-(x/2)^2-(6-x) is 64/3

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integral from-2 to 6 of 9-(x/2)^2-(6-x)

\int_{- 2}^{6} 9 - (\frac{x}{2})^{2} - (6 - x) d x

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

\int_{- 2}^{6} 9 - (\frac{x}{2})^{2} - (6 - x) d x

Simplify

9 - (\frac{x}{2})^{2} - (6 - x) : - \frac{x ^{2}}{4} + x + 3

= \int_{- 2}^{6} - \frac{x ^{2}}{4} + x + 3 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_{- 2}^{6} \frac{x ^{2}}{4} d x + \int_{- 2}^{6} x d x + \int_{- 2}^{6} 3 d x

\int_{- 2}^{6} \frac{x ^{2}}{4} d x = \frac{56}{3}

\int_{- 2}^{6} x d x = 16

\int_{- 2}^{6} 3 d x = 24

= - \frac{56}{3} + 16 + 24

Simplify

- \frac{56}{3} + 16 + 24 : \frac{64}{3}

= \frac{64}{3}

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What is the integral from-2 to 6 of 9-(x/2)^2-(6-x) ?
The integral from-2 to 6 of 9-(x/2)^2-(6-x) is 64/3