question_answer

\[\int_{0}^{2}{|x-1}|dx\]is equal to:

A) \[0\]                                     

B) \[1/2\]

C) \[1\]                                     

D) \[2\]

Correct Answer: C

Solution :

Let\[I=\int_{0}^{2}{|x-1|}\,\,dx\]         \[=\int_{0}^{1}{-(x-1)}\,dx+\int_{1}^{2}{(x-1)}\,dx\]        \[=\left[ -\frac{{{x}^{2}}}{2}+x \right]_{0}^{1}+\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}\]        \[=\left[ -\frac{1}{2}+1 \right]+\left[ 2-2-\left( \frac{1}{2}-1 \right) \right]\]       \[=\frac{1}{2}+\frac{1}{2}\]       \[=1\] Alternative Solution: Let          \[I=\int_{0}^{2}{|x-1}|\,\,dx\] It is clear from the figure, Required area = Area of\[\Delta OAD+\]Area of \[\Delta ABC\]                 \[=\frac{1}{2}\times 1\times 1+\frac{1}{2}\times 1\times 1\]                 \[=\frac{1}{2}+\frac{1}{2}=1\]