Find intersection points, integrate |x - x²| from 0 to 1.

Area Between y = x² and y = x

Question

Find the area of the region enclosed between the curves $y = x$ and $y = x^2$ .

Solution — Step by Step

Set $x = x^2$ and solve: $x^2 - x = 0$ , so $x(x - 1) = 0$ .

The two curves meet at $x = 0$ and $x = 1$ . These are our limits of integration.

Pick any $x$ between 0 and 1 — say $x = 0.5$ . Then $y = x$ gives $0.5$ and $y = x^2$ gives $0.25$ .

Since $x > x^2$ for all $x \in (0, 1)$ , the line $y = x$ lies above the parabola in this interval.

Area = $\int_0^1 (\text{upper} - \text{lower}) \, dx$

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

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

A = \left( \frac{1}{2} - \frac{1}{3} \right) - (0 - 0) = \frac{3 - 2}{6} = \frac{1}{6}

The area enclosed between the two curves is $\boxed{\dfrac{1}{6}}$ square units.

Why This Works

The integral $\int_a^b (f(x) - g(x)) \, dx$ gives the net signed area between two curves, provided $f(x) \geq g(x)$ throughout $[a, b]$ . When the upper curve changes across the interval, we’d need to split — but here $y = x$ stays on top from 0 to 1 the entire time, so one clean integral does the job.

The geometric picture: $y = x^2$ is a parabola opening upward, and $y = x$ is a straight line through the origin with slope 1. They form a “leaf” shape between $(0,0)$ and $(1,1)$ . The area of that leaf is $\frac{1}{6}$ — smaller than you’d intuitively expect, which is why this result is a classic board exam trap.

This exact result ( $\frac{1}{6}$ sq. units) appeared in CBSE 2024 Board Exam and has shown up in multiple JEE Main previous year questions in slightly disguised forms (shifted parabolas, scaled versions).

Alternative Method

Instead of integrating with respect to $x$ , we can integrate with respect to $y$ .

From $y = x$ , we get $x = y$ . From $y = x^2$ , we get $x = \sqrt{y}$ (taking positive root since we’re in the first quadrant).

For $y \in [0, 1]$ , the curve $x = \sqrt{y}$ lies to the right of $x = y$ :

A = \int_0^1 (\sqrt{y} - y) \, dy = \left[ \frac{2y^{3/2}}{3} - \frac{y^2}{2} \right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}

Same answer, as expected. In board exams, stick with the $x$ -integration — it’s faster. The $y$ -method is useful when the problem asks you to integrate w.r.t. $y$ explicitly, or when the region is easier to describe that way.

Common Mistake

Forgetting to subtract — and squaring the wrong way. Many students write the area as $\int_0^1 x^2 \, dx - \int_0^1 x \, dx$ , flipping the order. This gives $\frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ , a negative area. Area is always positive — always subtract (lower) from (upper), not the other way around. If you’re unsure which is “upper,” just test a point inside the region.

Quick sanity check: The area between $y = x^n$ and $y = x$ over $[0,1]$ is always $\frac{1}{2} - \frac{1}{n+1} = \frac{n-1}{2(n+1)}$ . For $n = 2$ : $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}$ . Memorise this pattern — it saves time when the exponent changes in the question.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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