JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 11)
The integral $$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $$
is equal to______.
is equal to______.
Answer
1.50
Explanation
$$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $$
= $$\int\limits_0^1 {\left| { - \left( {x - 1} \right) - x} \right|} dx$$ + $$ + \int\limits_1^2 {\left| {\left( {x - 1} \right) - x} \right|} dx$$
= $$\int\limits_0^1 {\left| {1 - x - x} \right|} dx + \int\limits_1^2 {dx} $$
= $$\int\limits_0^1 {\left| {1 - 2x} \right|} dx + \int\limits_1^2 {dx} $$
= $$\int\limits_0^{{1 \over 2}} {\left( {1 - 2x} \right)} dx + \int\limits_{{1 \over 2}}^1 {\left( {2x - 1} \right)dx} + \int\limits_1^2 {dx} $$
= $$\left[ {x - {x^2}} \right]_0^{{1 \over 2}} + \left[ {x - {x^2}} \right]_{{1 \over 2}}^1 + \left( {2 - 1} \right)$$
= $$\left( {{1 \over 2} - {1 \over 4}} \right) + \left[ {\left( {1 - 1} \right) - \left( {{1 \over 4} - {1 \over 2}} \right)} \right] + 1$$
= $${3 \over 2}$$ = 1.5
= $$\int\limits_0^1 {\left| { - \left( {x - 1} \right) - x} \right|} dx$$ + $$ + \int\limits_1^2 {\left| {\left( {x - 1} \right) - x} \right|} dx$$
= $$\int\limits_0^1 {\left| {1 - x - x} \right|} dx + \int\limits_1^2 {dx} $$
= $$\int\limits_0^1 {\left| {1 - 2x} \right|} dx + \int\limits_1^2 {dx} $$
= $$\int\limits_0^{{1 \over 2}} {\left( {1 - 2x} \right)} dx + \int\limits_{{1 \over 2}}^1 {\left( {2x - 1} \right)dx} + \int\limits_1^2 {dx} $$
= $$\left[ {x - {x^2}} \right]_0^{{1 \over 2}} + \left[ {x - {x^2}} \right]_{{1 \over 2}}^1 + \left( {2 - 1} \right)$$
= $$\left( {{1 \over 2} - {1 \over 4}} \right) + \left[ {\left( {1 - 1} \right) - \left( {{1 \over 4} - {1 \over 2}} \right)} \right] + 1$$
= $${3 \over 2}$$ = 1.5
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