Question

Evaluate : ${\int }_{-3}^2\left\{\left|x+1\right|+\left|x+2\right|+\left|x-1\right|\right\}dx$

• A. $\frac{31}{2}$
• B. $\frac{35}{2}$
• C. $\frac{47}{2}$
• D. $\frac{39}{2}$

Answer 1

The correct option is C

$\frac{47}{2}$

Explantion for correct option:

Evaluate the given integral

Let $I={\int }_{-3}^2\left\{\left|x+1\right|+\left|x+2\right|+\left|x-1\right|\right\}dx$

$={\int }_{-3}^2\left|x+1\right|dx+{\int }_{-3}^2\left|x+2\right|dx+{\int }_{-3}^2\left|x-1\right|dx$

$={\int }_{-3}^{-1}-\left(x+1\right)dx+{\int }_{-1}^2\left(x+1\right)dx+{\int }_{-3}^{-2}-\left(x+2\right)dx+{\int }_{-2}^2\left(x+2\right)dx+{\int }_{-3}^1-\left(x-1\right)dx+{\int }_1^2\left(x-1\right)dx$

$={-\left[\left(\frac{x^2}{2}+x\right)\right]}_{-3}^{-1}+{\left[\left(\frac{x^2}{2}+x\right)\right]}_{-1}^2{-\left[\left(\frac{x^2}{2}+2x\right)\right]}_{-3}^{-2}+{\left[\left(\frac{x^2}{2}+2x\right)\right]}_{-2}^2{-\left[\left(\frac{x^2}{2}-x\right)\right]}_{-3}^1+{\left[\left(\frac{x^2}{2}-x\right)\right]}_1^2$

$=-\left[\left(\frac{-1}{2}\right)-\left(\frac{3}{2}\right)\right]+\left[4-\left(-\frac{1}{2}\right)\right]-\left[\left(-2\right)-\left(\frac{3}{2}\right)\right]+\left[6-\left(-2\right)\right]-\left[\left(\frac{1}{2}\right)-\frac{15}{2}\right]+\left[0-\left(-\frac{1}{2}\right)\right]$

Substitute the upper and lower limits and simplify

$=2+\frac{9}{2}+\frac{1}{2}+8+8+\frac{1}{2}$

$=19+\frac{9}{2}$

$=\frac{47}{2}$

Thus, Option C is the correct answer.