question_answer

The area enclosed between the parabola\[y={{x}^{2}}-x+2\] and the line \[y=x+2\] in sq unit equals :

A)  \[\frac{8}{3}\]                                  

B)  \[\frac{1}{3}\]

C)  \[\frac{2}{3}\]                                  

D)  \[\frac{4}{3}\]

Correct Answer: D

Solution :

Given equation of parabola is \[y={{x}^{2}}-x+2\]                 or            \[{{\left( x-\frac{1}{2} \right)}^{2}}=y-\frac{7}{4}\] and equation of line is                 \[y=x+2\] \[\therefore \] Required area \[=\int_{0}^{2}{[x+2)-({{x}^{2}}-x+2)]dx}\] \[=\int_{0}^{2}{(-{{x}^{2}}+2x)dx}\]                 \[=\left[ -\frac{{{x}^{3}}}{3}+{{x}^{2}} \right]_{0}^{2}=-\frac{8}{3}+4\]                 \[=\frac{4}{3}sq\]unit