Question

JEE Mains question is in the image.

Answer 1

we have to evaluate $\int\limits^2_0{||x-1|-x|}\,dx$

solution : in interval, 0 < x < 1

|x - 1| = -(x - 1)

in interval 1 < x < 2

|x - 1| = (x - 1)

so, $\int\limits^2_0{||x-1|-x|}\,dx=\int\limits^1_0{|1-x-x|}\,dx+\int\limits^2_1{|x-1-x|}\,dx$

$=\int\limits^1_0|{(1-2x)|}\,dx+\int\limits^2_11\,dx$

in interval 0 < x < 1/2

|(1 - 2x)| = (1 - 2x)

in interval 1/2 < x < 1

|1 - 2x| = (2x - 1)

$=\int\limits^{1/2}_0{(1-2x)}\,dx+\int\limits^1_{1/2}{(2x-1)}\,dx+[2 - 1]$

= [x - x²]½_0 + [x² - x]¹_½ + 1

= [1/2 - 1/4 ] + [1 - 1 - 1/2 + 1/4] + 1

= 1/4 + 1/4 + 1

= 3/2

Therefore the value of given integration is 3/2