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QJEEM20Ma5c72: JEE Main 2020 Mathematics - Class 12 JEEM20Ma5c72 Solution: The integral is equal to______.

The integral ∫02∣∣x−1∣−x∣dx\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx}
is equal to______.

Step-by-Step Verified Solution

∫02∣∣x−1∣−x∣dx\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx}

= ∫01∣−(x−1)−x∣dx\int\limits_0^1 {\left| { - \left( {x - 1} \right) - x} \right|} dx + +∫12∣(x−1)−x∣dx+ \int\limits_1^2 {\left| {\left( {x - 1} \right) - x} \right|} dx

= ∫01∣1−x−x∣dx+∫12dx\int\limits_0^1 {\left| {1 - x - x} \right|} dx + \int\limits_1^2 {dx}

= ∫01∣1−2x∣dx+∫12dx\int\limits_0^1 {\left| {1 - 2x} \right|} dx + \int\limits_1^2 {dx}

= ∫012(1−2x)dx+∫121(2x−1)dx+∫12dx\int\limits_0^{{1 \over 2}} {\left( {1 - 2x} \right)} dx + \int\limits_{{1 \over 2}}^1 {\left( {2x - 1} \right)dx} + \int\limits_1^2 {dx}

= [x−x2]012+[x−x2]121+(2−1)\left[ {x - {x^2}} \right]_0^{{1 \over 2}} + \left[ {x - {x^2}} \right]_{{1 \over 2}}^1 + \left( {2 - 1} \right)

= (12−14)+[(1−1)−(14−12)]+1\left( {{1 \over 2} - {1 \over 4}} \right) + \left[ {\left( {1 - 1} \right) - \left( {{1 \over 4} - {1 \over 2}} \right)} \right] + 1

= 32{3 \over 2} = 1.5

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