Find area between y=x² and y=x

Question

Find the area of the region bounded by the curves $y = x^2$ and $y = x$ .

Solution — Step by Step

Set the two curves equal to find where they cross:

x^2 = x

x^2 - x = 0

x(x - 1) = 0

So $x = 0$ and $x = 1$ .

The two curves intersect at $(0, 0)$ and $(1, 1)$ .

We need to know which function is greater between $x = 0$ and $x = 1$ .

Test with $x = 0.5$ :

$y = x = 0.5$
$y = x^2 = 0.25$

Since $0.5 > 0.25$ , the line $y = x$ is above the parabola $y = x^2$ in the interval $[0, 1]$ .

So the area = $\int_0^1 (\text{top} - \text{bottom}) \, dx = \int_0^1 (x - x^2) \, dx$

A = \int_0^1 (x - x^2) \, dx

= \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1

= \left( \frac{1}{2} - \frac{1}{3} \right) - \left( 0 - 0 \right)

= \frac{3}{6} - \frac{2}{6} = \frac{1}{6}

The area of the region bounded by $y = x^2$ and $y = x$ is $\boxed{\dfrac{1}{6}}$ square units.

Why This Works

The formula for area between two curves is always:

A = \int_a^b [\text{upper curve} - \text{lower curve}] \, dx

We subtract the lower curve from the upper curve because the integral of the lower curve would “subtract away” the area between the x-axis and the lower curve, leaving only the area between the two curves. Always check which is upper and which is lower at a sample point between the intersection points — don’t assume.

Alternative Method — Graphical Check

Draw the parabola $y = x^2$ (opening upward, vertex at origin) and the line $y = x$ (slope 1 through origin). They enclose a lens-shaped region in the first quadrant between $x = 0$ and $x = 1$ . The enclosed area should be less than the area of the triangle with vertices $(0,0)$ , $(1,0)$ , $(1,1)$ which equals $\frac{1}{2}$ . Our answer $\frac{1}{6} < \frac{1}{2}$ , which is a good sanity check.

In JEE Main, area-between-curves questions almost always involve one curve being a parabola and the other a line or another parabola. The three-step process is always: (1) find intersection points, (2) identify upper/lower curve, (3) integrate the difference. Never skip step 2 — if the curves cross in the middle of your interval, you must split the integral.

Common Mistake

Students often integrate without subtracting the lower curve: they write $\int_0^1 x^2 \, dx$ or $\int_0^1 x \, dx$ separately and then try to subtract. This gives the wrong sign if done carelessly. Always write the integrand as (upper − lower) in one expression before integrating.

Question

Solution — Step by Step

Why This Works

Alternative Method — Graphical Check

Common Mistake

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