What is the integral from-2 to 6 of (3-(x^2)/4+x) ?

The integral from-2 to 6 of (3-(x^2)/4+x) is 64/3

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integral from-2 to 6 of (3-(x^2)/4+x)

\int_{- 2}^{6} (3 - \frac{x ^{2}}{4} + x) d x

\frac{64}{3}

Decimal Notation

21.33333 \dots

Solution steps

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

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

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

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

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

Simplify

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

= \frac{64}{3}

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What is the integral from-2 to 6 of (3-(x^2)/4+x) ?
The integral from-2 to 6 of (3-(x^2)/4+x) is 64/3