What is the integral of 1/2-1/4 |x-3| ?

The integral of 1/2-1/4 |x-3| is 1/2 x-1/8 |x-3|(x-3)+C

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integral of 1/2-1/4 |x-3|

\int \frac{1}{2} - \frac{1}{4} ∣ x - 3 ∣ d x

\frac{1}{2} x - \frac{1}{8} ∣ x - 3 ∣ (x - 3) + C

Solution steps

\int \frac{1}{2} - \frac{1}{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 \frac{1}{2} d x - \int \frac{1}{4} ∣ x - 3 ∣ d x

\int \frac{1}{2} d x = \frac{1}{2} x

\int \frac{1}{4} ∣ x - 3 ∣ d x = \frac{1}{8} ∣ x - 3 ∣ (x - 3)

= \frac{1}{2} x - \frac{1}{8} ∣ x - 3 ∣ (x - 3)

Add a constant to the solution

= \frac{1}{2} x - \frac{1}{8} ∣ x - 3 ∣ (x - 3) + C

What is the integral of 1/2-1/4 |x-3| ?
The integral of 1/2-1/4 |x-3| is 1/2 x-1/8 |x-3|(x-3)+C