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  • Need clarity, kindly explain! - Integral Calculus - JEE Main-8

\int_{-1}^{1}\left ( x-\left [ 2x \right ] \right )dx  is equal to?

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Answers (1)

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I=\int_{-1}^{1}xdx-\int_{-1}^{1}\left [ 2x \right ]dx

   =\left [ \frac{x^{2}}2{} \right ]_{-1}^{1}-\left \{ \int_{-1}^{-1/2}\left [ 2x \right ]dx+\int_{-1/2}^{0}\left [ 2x \right ]dx+\int_{0}^{1/2}\left [ 2x \right ]dx+\int_{1/2}^{1}\left [ 2x \right ]dx\right \}=0-\left \{ \int_{-1}^{-1/2}-2dx+\int_{-1/2}^{0}-1dx+\int_{0}^{1/2}0dx+\int_{1/2}^{1}1dx\right \}

=2\left [ x \right ]_{-1}^{-1/2}+\left [ x \right ]_{-1/2}^{0}-\left [ x \right ]_{1/2}^{1}=2\left ( \frac{-1}{2}+1 \right )+\left ( 0+\frac{1}{2} \right )-\left (1- \frac{1}{2} \right )=1


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