Q. The area included between curves $y = x^{2} - 3x + 2$ and $y = - x^{2} + 3x - 2$ is
A) $\frac{1}{6}$ sq unit
B) $\frac{1}{2}$ sq unit
C) 1 sq unit
D) $\frac{1}{3}$ sq unit
Explanation
(d)
Required area $= 2\int_{1}^{2}{\left( - x^{2} + 3x - 2 \right)dx}$ ($\because$ both portions are same)
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$= 2\left\lbrack - \frac{x^{3}}{3} + \frac{3x^{2}}{2} - 2x \right\rbrack_{1}^{2}$
$= 2\left\lbrack - \frac{8}{3} + 6 - 4 - \left( - \frac{1}{3} + \frac{3}{2} - 2 \right) \right\rbrack$
$= 2\left\lbrack - \frac{8}{3} + 2 + \frac{5}{6} \right\rbrack = \frac{1}{3}\ sq\ unit$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.