If $\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|, \forall \mathrm{x} \in[1,4]$, then…
If $\mathrm{f}(\mathrm{x})=|\mathrm{x}-1|+|\mathrm{x}-2|+|\mathrm{x}-3|, \forall \mathrm{x} \in[1,4]$, then $\int_1^4 \mathrm{f}(\mathrm{x}) \mathrm{dx}=$
$\frac{1}{2}$
7
$\frac{9}{2}$
$\frac{19}{2}$
Solution
$\begin{aligned}
& \int_1^4 f(x) d x=\int_1^4[|x-1|+|x-2|+|x-3| d x \\
& =\int_1^2(x-1)+(2-x)+(3-x) d x+\int_2^3[(x-1)+(x-2)+(3-x)] d x \\
& +\int_3^4[(x-1)+(x-2)+(x-3)] d x \\
& =\int_1^2(4-x) d x+\int_2^3 x d x+\int_3^4(3 x-6) d x
\end{aligned}$
$\begin{aligned}
& =\left[4 x-\frac{x^2}{2}\right]_1^2+\left[\frac{x^2}{2}\right]_2^3+\left[\frac{3 x^2}{2}-6 x\right]_3^4 \\
& =\left[(8-2)-\left(4-\frac{1}{2}\right)\right]+\left[\left(\frac{9}{2}-2\right)\right]+\left[(24-24)-\left(\frac{27}{2}-18\right)\right] \\
& =\left(2+\frac{1}{2}\right)+\left(\frac{5}{2}\right)+\left(\frac{9}{2}\right)=\frac{19}{2}
\end{aligned}$