Question

Area included between curves $y = x^2 -3 x + 2$ 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

Answer 1

Answer: (D)

Required area



$=2\int\limits^{2}_{1}\,\left(-x^{2}+3x-2\right)dx$

$=2\left[\frac{-x^{3}}{3}+\frac{3x^{2}}{2}-2x\right]^{2}_{1}$

$=2\left[\frac{-8}{3}+6-4+\frac{1}{3}-\frac{3}{2}+2\right]$

$=2\left[\frac{-8}{3}+4-\frac{7}{6}\right]$

$=2\cdot\frac{1}{6}=\frac{1}{3}$ sq unit