Question

The value of the integral ${\int }_1^5[|x-3|+|1-x|]dx$ is equal to

• A. $4$
• B. $8$
• C. $12$
• D. $16$

Answer 1

Answer: (C)

$12$

${\int }_1^5[|x-3|+|1-x|]dx$

$={\int }_1^5|x-3|dx+{\int }_1^5|1-x|dx$

$={\int }_1^3|x-3|dx+{\int }_3^5|x-3|dx+{\int }_1^5|1-x|dx$

$={\int }_1^3-(x-3)dx+{\int }_3^5(x-3)dx+{\int }_1^5-(1-x)dx$

$={\int }_1^3(3-x)dx+{\int }_3^5(x-3)dx+{\int }_1^5(x-1)dx$

$={[3x-\frac{x^2}{2}]}_1^3+{[\frac{x^2}{3}-2x]}_3^5+{[\frac{x^2}{2}-x]}_1^5$

$=(3\times 3-\frac{9}{2})-(3\times 1-\frac{1}{2})+(\frac{5\times 5}{2}-3\times 5)-(\frac{3\times 3}{2}-3\times 3)+(\frac{5\times 5}{2}-5)-(\frac{1}{2}-1)$

$=(9-\frac{9}{2})-(3-\frac{1}{2})+(\frac{25}{2}-15)-(\frac{9}{2}-9)+(\frac{25}{2}-5)-(-\frac{1}{2})$

$=\frac{9}{2}-\frac{5}{2}-\frac{5}{2}+\frac{9}{2}+\frac{15}{2}+\frac{1}{2}$

$=\frac{9-5-5+9+15+1}{2}=\frac{24}{2}=12$