What is the integral from 2 to x of (3-x)/2 ?

The integral from 2 to x of (3-x)/2 is 1/4 (-x^2+6x-8)

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integral from 2 to x of (3-x)/2

\int_{2}^{x} \frac{3 - x}{2} d x

\frac{1}{4} (- x^{2} + 6 x - 8)

Solution steps

\int_{2}^{x} \frac{3 - x}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2}^{x} 3 - 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

= \frac{1}{2} (\int_{2}^{x} 3 d x - \int_{2}^{x} x d x)

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

\int_{2}^{x} x d x = \frac{x ^{2}}{2} - 2

= \frac{1}{2} (3 x - 6 - (\frac{x ^{2}}{2} - 2))

Simplify

\frac{1}{2} (3 x - 6 - (\frac{x ^{2}}{2} - 2)) : \frac{1}{4} (- x^{2} + 6 x - 8)

= \frac{1}{4} (- x^{2} + 6 x - 8)

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What is the integral from 2 to x of (3-x)/2 ?
The integral from 2 to x of (3-x)/2 is 1/4 (-x^2+6x-8)