Find the area bounded by y=x² and y=x using integration

Question

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

(CBSE 2023, 4 marks)

Solution — Step by Step

Set the two curves equal: $x^2 = x$ , which gives $x^2 - x = 0$ , so $x(x - 1) = 0$ .

The curves meet at $x = 0$ and $x = 1$ . The corresponding points are $(0, 0)$ and $(1, 1)$ .

Between $x = 0$ and $x = 1$ , pick a test point like $x = 0.5$ :

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

So $y = x$ is the upper curve and $y = x^2$ is the lower curve in this interval.

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) = \frac{3 - 2}{6} = \mathbf{\frac{1}{6} \text{ sq. units}}

Why This Works

The area between two curves is always $\int_a^b (\text{upper} - \text{lower})\,dx$ . We subtract the lower curve from the upper curve to get the height of a thin vertical strip at each $x$ , then sum all these strips from $a$ to $b$ .

Geometrically, $y = x$ is a straight line through the origin and $y = x^2$ is a parabola. The line rises faster than the parabola between 0 and 1, creating a crescent-shaped region. Beyond $x = 1$ , the parabola overtakes the line — but we only care about the enclosed region.

Alternative Method — Integrating Along y-axis

We can also integrate with respect to $y$ . From $y = x^2$ , we get $x = \sqrt{y}$ (right boundary). From $y = x$ , we get $x = y$ (left boundary, since $y \leq \sqrt{y}$ for $0 \leq y \leq 1$ ).

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. Use whichever direction makes the integral simpler.

In CBSE board exams, always sketch the region and shade it. You get 1 mark for the diagram alone. Also write the intersection points clearly — examiners look for them specifically in the marking scheme.

Common Mistake

The most common error is forgetting to check which curve is on top. If you accidentally compute $\int_0^1 (x^2 - x)\,dx$ , you get $-1/6$ . Some students then write $|{-1/6}| = 1/6$ and get lucky — but this won’t work when the curves cross within the interval. Always identify upper and lower curves first, or you risk subtracting the wrong way in multi-region problems.

Question

Solution — Step by Step

Why This Works

Alternative Method — Integrating Along y-axis

Common Mistake

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