Area bounded by the curve \( y=x^3 \), x-axis and lines \( x=-2 \) and \( x=1 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{17}{4} \) Sq. units
The curve \( y=x^3 \) is below the x-axis for x < 0 and above it for x > 0, so the region splits at x = 0 and the two areas are added as positive quantities.
\( \displaystyle \text{Area}=\left|\int_{-2}^{0}x^3\,dx\right|+\int_{0}^{1}x^3\,dx \)
\( \displaystyle \int_{-2}^{0}x^3\,dx=\left[\frac{x^4}{4}\right]_{-2}^{0} \)
\( \displaystyle =0-\frac{16}{4}=-4 \)
\( \displaystyle \int_{0}^{1}x^3\,dx=\left[\frac{x^4}{4}\right]_{0}^{1}=\frac{1}{4} \)
\( \displaystyle \text{Area}=4+\frac{1}{4}=\frac{17}{4}\ \text{sq. units} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{15}{4} \) Sq. units — Incorrect. This is the size of the single integral \( \int_{-2}^{1}x^3\,dx=\frac{1}{4}-4=-\frac{15}{4} \), where the part below the axis cancels part of the area above it.
Option (3) \( -\displaystyle\frac{15}{4} \) Sq. units — Incorrect. This is that single integral itself; an area can never be negative.
Option (4) -9 Sq. units — Incorrect. An area can never be negative, and the area here is \( \frac{17}{4} \).
NCERT (2026-27 edition):
Application of Integrals
