Question

The value of integral ${\int }_0^2(|x^2-3x+2|+|\cos x|)dx$, is equal to

• A. $6+\sin 2$
• B. $3-\sin 2$
• C. $3+\sin 2$
• D. $6-\sin 2$

Answer 1

The correct option is B $3-\sin 2$

$\begin{array}{c}{\int }_0^2\left(\left|x^2-3x+2\right|+\left|\cos x\right|\right)dx\\ ={\int }_0^1\left(x^2-3x+2\right)dx+{\int }_1^2\left(-x^2+3x-2\right)dx+{\int }_0^{\frac{\pi }{2}}\cos xdx+{\int }_{\frac{\pi }{2}}^2-\cos xdx\\ =\left[\frac{1}{3}-\frac{3}{2}+2\right]+{{\left[\frac{-x^3}{3}+\frac{3}{2}{x}^2-2x\right]}^2}_1+{\left[\sin x\right]}_0^{\frac{\pi }{2}}-{\left[\sin x\right]}_{\frac{\pi }{2}}^2\\ =\left(\frac{1}{3}+\frac{1}{2}\right)+\left[\left(\frac{-8}{3}+6-4\right)-\left(\frac{-1}{3}+\frac{3}{2}-2\right)\right]+1+1-\sin 2\\ =\left(\frac{1}{3}+\frac{1}{2}\right)+\left[\frac{-7}{3}+4-\frac{3}{2}\right]+2-\sin 2\\ =3-\sin 2\\ Hence,option\left(B\right)iscorrectanswer.\end{array}$