Q. The area bounded by $y = x^{2} + 2,\ x$-axis , $x = 1\ and\ x = 2\ is$
A) $\frac{16}{3}\ sq\ units$
B) $\frac{17}{3}\ sq\ units$
C) $\frac{13}{3}\ sq\ units$
D) $\frac{20}{3}\ sq\ units$
Explanation
(c)
Required area$= \$area of curve $ABCD$
$= \int_{1}^{2}{\left( x^{2} + 2 \right)dx = \left\lbrack \frac{x^{3}}{3} + 2x \right\rbrack_{1}^{2}}$
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$= \left( \frac{8}{3} + 4 \right) - \left( \frac{1}{3} + 2 \right)$
$= \frac{13}{3}sq\ units$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.