Integral Calculus Question 162
Question: The value of the integral $ \int_{-1}^{3}{(| x |+| x-1 |)dx} $ is
Options:
4
9
2
D) $ \frac{9}{2} $
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Answer:
Correct Answer: B
Solution:
[b] We have $ | x |+| x-1 |= \begin{matrix} -x-(x-1)=-2x+1,ifx\le 0 \\ x-(x-1)=1,if0\le x\le 1 \\ x+x-1=2x-1,ifx\ge 1 \\ \end{matrix} . $ $ I_1=\int\limits_{-2n-1}^{-2n}{{{\sin }^{27}}xdx=\int\limits_{2n+1}^{2n}{{{\sin }^{27}}(-x)(-dx)}} $
$ \therefore \int\limits_{-1}^{3}{(| x |+| x-1 |)dx} $ $ =\int_{-1}^{0}{(-2x+1)dx+\int_0^{1}{1dx+\int_1^{3}{(2x-1)dx}}} $ $ =[ -x^{2}+x ]_{-1}^{0}+[x]_0^{1}+[ x^{2}-x ]_1^{3}=9 $
BETA
