Area Under Curves

Finding the area between two curves rests on the upper-minus-lower principle: integrate the difference of the bounding functions across the interval where they enclose a region. The first task is to locate the intersection points by setting x^2 = x, which factors to x(x-1) = 0 and gives x = 0 and x = 1 as the limits. On the open interval (0,1) the line y = x lies above the parabola y = x^2, because for 0 < x < 1 we always have x > x^2. The enclosed area is therefore \int_0^1 (x - x^2),dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}. Option 1/3 forgets to subtract the parabola's contribution from the line's. Option 1/2 uses only the area under the line and ignores the lower boundary. Option 1/12 halves the correct result, a frequent slip. This is the standard bounded-region method examined repeatedly in JEE Advanced. As a final plausibility check, the entire region sits inside the unit square of area 1, and 1/6 is a sensibly small fraction matching the thin sliver between a line and a parabola that meet at both ends of [0,1].