Answer
6.5
Work Step by Step
$f(x)= 3-|x-3|=\left\{\begin{array}{lll}
3-(x-3), & when & x-3 \geq 0\\
3-[-(x-3)] & when & x-3 < 0
\end{array}\right.$
$f(x)=\left\{\begin{array}{lll}
6-x, & when & x \geq 3\\
x & when & x < 3
\end{array}\right.$
Apply Th.4.6 from section 4-3, Additive Interval Property
$\displaystyle \int_{1}^{4}(3-|x-3|)dx=\int_{1}^{3}xdx+\int_{3}^{4}(6-x)]dx$
Use the table "Basic Integration Rules", p.246
$=[\displaystyle \frac{x^{2}}{2}]_{1}^{3}+[6x-\frac{x^{2}}{2}]_{3}^{4}$
$=(\displaystyle \frac{9}{2}-\frac{1}{2})+[(24-8)-(18-\frac{9}{2})]$
$=4+16-18+\displaystyle \frac{9}{2}$
$=\displaystyle \frac{13}{2}$
Checking graphically - see below (work done in desmos.com).
We see two trapezoids,
one (left) with bases 1 and 3, height 2,
one with bases 3 and 2, height 1
$A=\displaystyle \frac{1+3}{2}(2)+\frac{3+2}{2}(1)=4+\frac{5}{2}=6.5$
