CBSE Board Examination 2026 · Section A · 1 mark

Q10: Area bounded by $y = x|x|$ ⁠, the $x$ -axis, $x = -1$ and $x = 1$

Board 2026Ch 8 · Application of Integrals Ch 7 · Integrals

The question

The area bounded by the curve $y = x|x|$ ⁠, $x$ -axis and the ordinates $x = -1$ and $x = 1$ is given by

(A) $0$

(B) $\frac{1}{3}$

(D) $\frac{4}{3}$

Exam answer

$y = -x^2$ for $x < 0$ (below the $x$ -axis) and $y = x^2$ for $x \ge 0$ ⁠.

Why?

$|x| = -x$ when $x < 0$ ⁠, and $|x| = x$ when $x \ge 0$ ⁠. So

\begin{aligned} x|x| &= x(-x) = -x^2,\ x < 0 \\ x|x| &= x \cdot x = x^2,\ x \ge 0 \end{aligned}

For $x < 0$ ⁠, $-x^2$ is negative, so that part of the curve lies below the $x$ -axis.

\text{Area} = \left|\int_{-1}^{0} -x^2\,dx\right| + \int_{0}^{1} x^2\,dx

Why?

An area is always positive. The integral of a negative function is negative, so we take its modulus.

Watch out: $\int_{-1}^{1} x|x|\,dx = 0$ ⁠, because $x|x|$ is odd and the two parts cancel. That is option (A), and it is not the area.

\begin{aligned} \text{Area} &= \left|\left[-\frac{x^3}{3}\right]_{-1}^{0}\right| + \left[\frac{x^3}{3}\right]_{0}^{1} \\ &= \left|-\frac{1}{3}\right| + \frac{1}{3} \\ &= \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \end{aligned}

Why?

Both pieces use $\int x^2\,dx = \frac{x^3}{3}$ ⁠. At the limits:

For $-x^2$ the antiderivative is $-\frac{x^3}{3}$ ⁠. At $x = -1$ ⁠: $-\frac{(-1)^3}{3} = \frac{1}{3}$ ⁠.

\begin{aligned} &\left[-\frac{x^3}{3}\right]_{-1}^{0} \\ &= \left(-\frac{0^3}{3}\right) - \left(-\frac{(-1)^3}{3}\right) \\ &= 0 - \left(-\frac{-1}{3}\right) \\ &= 0 - \frac{1}{3} = -\frac{1}{3} \\[4pt] \left|-\frac{1}{3}\right| &= \frac{1}{3} \\[4pt] \left[\frac{x^3}{3}\right]_{0}^{1} &= \frac{1^3}{3} - \frac{0^3}{3} \\ &= \frac{1}{3} - 0 = \frac{1}{3} \end{aligned}

Answer: (C) $\dfrac{2}{3}$

Where this comes from

Chapter 8, Application of Integrals: Area bounded by a curve and the x-axis.

Question type: A modulus inside the integral

How to spot it: click to see

Type 6. A modulus inside the integral: split at the corner

Spot it: bars inside the integral, like $|x + 3|$ or $x|x|$ ⁠.

Here: $x|x|$ has its corner at 0: it is $-x^2$ from $-1$ to 0 (below the axis, so put bars round that part) and $x^2$ from 0 to 1. Each gives $\frac13$ ⁠, so the area is $\frac23$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 7, Integrals.

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Q10: Area bounded by y=x∣x∣y = x|x|y=x∣x∣⁠, the xxx-axis, x=−1x = -1x=−1 and x=1x = 1x=1

Q10: Area bounded by $y = x|x|$ ⁠, the $x$ -axis, $x = -1$ and $x = 1$