Home › MCQs › SSC Pure Mathematics › Question #11189
Back to Questions
SSC Pure Mathematics QUESTION #11189
Question 1
The area bounded by $y = 4 - x^2$ and the $x$-axis is:
  • $\dfrac{4}{3}$
  • $\dfrac{8}{3}$
  • $\dfrac{16}{3}$
  • $\dfrac{32}{3}$✔️
Correct Answer Explanation
$y = 4-x^2 = 0 \Rightarrow x = \pm 2$. Area $= \displaystyle\int_{-2}^{2}(4-x^2)\,dx = 2\int_0^2(4-x^2)\,dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8-\frac{8}{3}\right) = 2\cdot\frac{16}{3} = \frac{32}{3}$.