The area (in square units) bounded by the curve \( y=x^{2}|x| \), \( x \)-axis and the lines \( x=-2 \) and \( x=2 \), is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8
\( y=x^2|x|\ge 0 \) for all x, so the curve lies on or above the x-axis. It is also an even function (symmetric about the y-axis):
\( \displaystyle y=x^3\ \text{for}\ x\ge 0 \)
\( \displaystyle y=-x^3\ \text{for}\ x\lt 0 \)
\( \displaystyle \text{Area}=\int_{-2}^{2}x^2|x|\,dx \)
\( \displaystyle =2\int_0^2 x^3\,dx \)
\( \displaystyle =2\left[\frac{x^4}{4}\right]_0^2 \)
\( \displaystyle =2\times 4 \)
\( \displaystyle =8 \)
Why the other options are incorrect
Option (1) 0 — Incorrect. 0 is what \( \int_{-2}^{2}x^3\,dx \) gives, ignoring the modulus; the curve is never below the axis, so the area cannot be 0.
Option (2) 8/3 — Incorrect. \( \frac{8}{3}=\int_0^2 x^2\,dx \); the factor |x| has been dropped.
Option (3) 4 — Incorrect. 4 is the area on one side only, \( \int_0^2 x^3\,dx \); the part from −2 to 0 is equal and must be added.