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

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

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

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int_{1}^{2} 2 + 2 x - x^{2} 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

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

\int_{1}^{2} 2 d x = 2

\int_{1}^{2} 2 x d x = 3

\int_{1}^{2} x^{2} d x = \frac{7}{3}

= \frac{1}{8} (2 + 3 - \frac{7}{3})

Simplify

\frac{1}{8} (2 + 3 - \frac{7}{3}) : \frac{1}{3}

= \frac{1}{3}

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