The area bounded by the curve \( y=x|x| \), x-axis and the ordinates \( x=-3 \) and \( x=3 \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 18 sq. units
Without the modulus, the curve is
\( \displaystyle y=x|x|=\begin{cases}x^2, & x\ge 0\\-x^2, & x\lt 0\end{cases} \)
The curve is below the x-axis for x < 0 and above it for x > 0. Area is always taken as positive, so
\( \displaystyle \text{Area}=\int_{-3}^{0}\displaystyle\left|-x^2\right|dx+\int_{0}^{3}\displaystyle x^2\,dx \)
\( \displaystyle =\left[\frac{x^3}{3}\right]_{-3}^{0}+\left[\frac{x^3}{3}\right]_{0}^{3} \)
\( \displaystyle =9+9=18\ \text{sq. units} \)
Shortcut: \( x|x| \) is an odd function, so the two parts are mirror images and have equal areas:
\( \displaystyle \text{Area}=2\int_{0}^{3}x^2\,dx=2\times 9=18 \)
Why the other options are incorrect
Option (1) 2 sq. units — Incorrect. The area is 18, not 2.
Option (2) 9 sq. units — Incorrect. 9 is the area of only one of the two parts (x from 0 to 3).
Option (3) 6 sq. units — Incorrect. The area is 18, not 6.
Tip: The integral \( \int_{-3}^{3}x|x|\,dx \) is 0 because the parts cancel; for an area, integrate |y| or add the parts separately.