Question

The value of the integral $\underset{1}{\overset{5}{\int }}\left[\left|x-3\right|+\left|1-x\right|\right]dx$ is equal to

• A. 4
• B. 8
• C. 12
• D. 16

Answer 1

Answer: (C)

$\underset{1}{\overset{5}{\int }}[|x-3|+|1-x|]dx$
$=\underset{1}{\overset{5}{\int }}|x-3|dx+\underset{1}{\overset{5}{\int }}|1-x|dx$
$=\underset{1}{\overset{3}{\int }}|x-3|dx+\underset{3}{\overset{5}{\int }}|x-3|dx+\underset{1}{\overset{5}{\int }}|1-x|dx$
$=\underset{1}{\overset{3}{\int }}-(x-3)dx+\underset{3}{\overset{5}{\int }}(x-3)dx$
$+\underset{1}{\overset{5}{\int }}-(1-x)dx$
$=\underset{1}{\overset{3}{\int }}(3-x)dx+\underset{3}{\overset{5}{\int }}(x-3)dx+\underset{1}{\overset{5}{\int }}(x-1)dx$
$={\left[3x-\frac{x^2}{2}\right]}^3+{\left[\frac{x^2}{2}-3x\right]}_3^5+{\left[\frac{x^2}{2}-x\right]}_1^5$
$=\left(3\times 3-\frac{9}{2}\right)-\left(3\times 1-\frac{1}{2}\right)+\left(\frac{5\times 5}{2}-3\times 5\right)$
$-\left(\frac{3\times 3}{2}-3\times 3\right)+\left(\frac{5\times 5}{2}-5\right)-\left(\frac{1}{2}-1\right)$
$=\left(9-\frac{9}{2}\right)-\left(3-\frac{1}{2}\right)+\left(\frac{25}{2}-15\right)-\left(\frac{9}{2}-9\right)$
$+\left(\frac{25}{2}-5\right)-\left(-\frac{1}{2}\right)$
$=\frac{9}{2}-\frac{5}{2}-\frac{5}{2}+\frac{9}{2}+\frac{15}{2}+\frac{1}{2}$
$=\frac{9-5-5+9+15+1}{2}=\frac{24}{2}=12$