CBSE 2024, Set 3

Question

Sketch the graph of $y = x|x|$ and hence find the area bounded by this curve, X-axis and the ordinates $x = -2$ and $x = 2$, using integration.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1: Analyze the function and sketch**

The function is $y = x|x|$. By the definition of absolute value, we can write it as a piecewise function:
$$y = \begin{cases} x(x) = x^2, & 	ext{if } x \ge 0 \ x(-x) = -x^2, & 	ext{if } x < 0 \end{cases}$$

The graph consists of an upward-opening parabola $y = x^2$ in the first quadrant (for $x \ge 0$) and a downward-opening parabola $y = -x^2$ in the third quadrant (for $x < 0$). The function is an odd function, meaning it is symmetric about the origin.

![Graph of y = x|x|](https://www.edvaya.com/cbse/2024/mathematics/set3/q32-graph-y-x-mod-x.webp)

**Step 2: Setup the definite integral for area**

We need to find the total bounded area between $x = -2$ and $x = 2$.
Area is strictly positive, so we take the absolute value of the integrals for regions below the X-axis:
$$	ext{Total Area} = \left| \int_{-2}^{0} x|x| \,dx ight| + \int_{0}^{2} x|x| \,dx$$

Because the function is symmetric about the origin (odd function), the area bounded in the third quadrant (from $-2$ to $0$) is geometrically equal to the area bounded in the first quadrant (from $0$ to $2$).
Therefore, we can simply evaluate the area in the first quadrant and multiply by $2$:
$$	ext{Total Area} = 2 \int_{0}^{2} x^2 \,dx$$

**Step 3: Integrate and evaluate**

$$	ext{Total Area} = 2 \left[ \frac{x^3}{3} ight]_{0}^{2}$$

$$= 2 \left( \frac{2^3}{3} - 0 ight)$$

$$= 2 \left( \frac{8}{3} ight)$$

$$= \frac{16}{3}$$

**Therefore, the bounded area is $\frac{16}{3}$ square units.**

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Mathematics > Calculus > Application Of Integrals

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