Q. The area bounded by $y = 4 - x^{2}$ and $y = \left\lbrack 3 + \frac{x^{2}}{4} \right\rbrack$, where $\lbrack \bullet \rbrack$ denotes greatest integer function, is
A) 1 sq unit
B) $\frac{1}{3}$ sq unit
C) $\frac{2}{3}$ sq unit
D) $\frac{4}{3}$ sq unit
Explanation
(d)
$y = \left\lbrack 3 + \frac{x^{2}}{4} \right\rbrack = 3, - 2 < x < 2$
[image:8_image11.jpeg]
Required area $= 2\left\{ \int_{0}^{1}{\left( 4 - x^{2} \right)dx - 3} \right\}$
$= 2\left\{ \left\lbrack 4x - \frac{x^{3}}{3} \right\rbrack_{0}^{1} - 3 \right\}$
$= 2\left\{ 4 - \frac{1}{3} - 3 \right\}$
$= \frac{4}{3}\ sq\ unit$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.