Question

The area included between curves $y=x^2-3x+2$ and $y=-x^2+3x-2$ is

• A. $\frac{1}{6}$ sq. units
• B. $\frac{1}{2}$ sq. units
• C. $1$ sq. unit
• D. $\frac{1}{3}$ sq. units

Answer 1

Answer: (D)

Required area $=2\underset{1}{\overset{2}{\int }}\left(-x^2+3x-2\right)dx$

$=2{\left[-\frac{x^3}{3}+\frac{3x^2}{2}-2x\right]}_1^2$
$=2\left[-\frac{8}{3}+6-4-\left(-\frac{1}{3}+\frac{3}{2}-2\right)\right]$
$=2\left[-\frac{8}{3}+2+\frac{5}{6}\right]$
$=\frac{1}{3}$ sq. units