Q.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) $\frac{1}{3}$ (C) $\frac{2}{3}$ (D) $\frac{4}{3}$ [Hint : $y = x^2$ if $x > 0$ and $y = -x^2$ if $x < 0$].

Concept: Area Under Curve — the function changes definition across the origin because of $|x|$.

Step 1: Write the piecewise form.

For $x \ge 0$, $y = x \cdot x = x^2$.

For $x < 0$, $y = x \cdot (-x) = -x^2$.

Step 2: The curve lies above the $x$-axis for $x>0$ and below it for $x<0$. Area is always positive, so we take absolute value:

$\text{Area} = \int_{-1}^{0} \bigl| -x^2 \bigr| \, dx + \int_{0}^{1} x^2 \, dx = \int_{-1}^{0} x^2 \, dx + \int_{0}^{1} x^2 \, dx$ …

The function $y = x|x|$ is a classic example of a piecewise-defined function. For $x \ge 0$, $|x| = x$, so $y = x \cdot x = x^2$. For $x < 0$, $|x| = -x$, so $y = x \cdot (-x) = -x^2$.

The graph is symmetric in a special way: it’s an odd function (since $f(-x) = -f(x)$), so the area below the $x$-axis on the left equals the area above the $x$-axis on the right. But area is always positive — we must take the absolute value of each region.

1. Set up the integral for area Area bounded by $y = f(x)$, the $x$-axis, and vertical lines $x = a$, $x = b$ is given by

Here $a = -1$, $b = 1$, and $f(x) = x|x|$.

2. Split at $x = 0$ because the expression changes form there

3. Evaluate the right half ($x \ge 0$) For $x \ge 0$, $x|x| = x^2 \ge 0$, so $|x^2| = x^2$.

4. Evaluate the left half ($x < 0$) For $x < 0$, $x|x| = -x^2 \le 0$, so $| -x^2 | = x^2$. $\int_{-1}^0 x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^0 = 0 - \left( -\frac{1}{3} \right) = \frac{1}{3}$ …

Method: Area of a piecewise/odd function that changes sign

Use this for the area bounded by a function like $y = x|x|$ that is defined by different formulas on each side of a point and changes sign there.

Steps

Step 1: Resolve the modulus into pieces.

For $x \ge 0$, $x|x| = x^2$; for $x < 0$, $x|x| = -x^2$. The behaviour (and sign) switches at $x = 0$.

Step 2: Split the interval at the switch point and take the magnitude of each part.

$\text{Area} = \int_{a}^{0} \big| -x^2 \big|\,dx + \int_{0}^{b} \big| x^2 \big|\,dx = \int_a^0 x^2\,dx + \int_0^b x^2\,dx$ …

Common Mistakes

Mistake 1: Using the signed integral of an odd function.

Why it's wrong: $y = x|x|$ is odd, so $\int_{-1}^{1} x|x|\,dx = 0$ — that is option (A) and it is the net value, not the area. Correct approach: split at $x=0$ and take magnitudes, giving $\tfrac13+\tfrac13 = \tfrac23$.

Mistake 2: Mis-resolving the modulus. …