Question
$$\int_{ - 1}^1 {\left( {x - \left[ {2x} \right]} \right)dx} $$ is equal to :
A.
1
B.
0
C.
2
D.
4
Answer :
1
Solution :
$$\eqalign{
& I = \int_{ - 1}^1 {x\,dx} - \int_{ - 1}^1 {\left[ {2x} \right]dx} \cr
& \,\,\,\,\, = \left[ {\frac{{{x^2}}}{2}} \right]_{ - 1}^1 - \left\{ {\int_{ - 1}^{ - \frac{1}{2}} {\left[ {2x} \right]dx} + \int_{ - \frac{1}{2}}^0 {\left[ {2x} \right]dx} + \int_0^{\frac{1}{2}} {\left[ {2x} \right]dx} + \int_{\frac{1}{2}}^1 {\left[ {2x} \right]dx} } \right\} \cr
& \,\,\,\,\, = 0 - \left\{ {\int_{ - 1}^{ - \frac{1}{2}} { - 2\,dx} + \int_{ - \frac{1}{2}}^0 { - 1\,dx} + \int_0^{\frac{1}{2}} {0\,dx} + \int_{\frac{1}{2}}^1 {1\,dx} } \right\} \cr
& \,\,\,\,\, = 2\left[ x \right]_{ - 1}^{ - \frac{1}{2}} - \left[ x \right]_{ - \frac{1}{2}}^0 - \left[ x \right]_{ - \frac{1}{2}}^1 \cr
& \,\,\,\,\, = 2\left( { - \frac{1}{2} + 1} \right) + \left( {0 + \frac{1}{2}} \right) - \left( {1 - \frac{1}{2}} \right) \cr
& \,\,\,\,\, = 1 \cr} $$