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

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

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

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

\frac{6 x - x ^{2} - 8}{4}

Solution steps

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

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 - t d t

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 t - \int_{2}^{x} t d t)

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

\int_{2}^{x} t d t = \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{6 x - x ^{2} - 8}{4}

= \frac{6 x - x ^{2} - 8}{4}

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