The area enclosed between the parabola $y = x^{2} - x + 2$ and the line $y = x + 2$ in square unit equals

Question

Dr. Abubacker · Accepted Answer

(d)
Given the equation of parabola can be rewritten as
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$\left( x - \frac{1}{2} ight)^{2} = y - \frac{7}{4}$
$	herefore\ \ Required\ area = \int_{0}^{2}{\left\lbrack (x + 2) - \left( x^{2} - x + 2 ight) ightbrack dx}$
$\int_{0}^{2}{\left( {- x}^{2} + 2x ight)dx}$
$= \left\lbrack - \frac{x^{3}}{3} + x^{2} ightbrack_{0}^{2} = - \frac{8}{3} + 4 = \frac{4}{3}\ sq\ units$

Q. The area enclosed between the parabola $y = x^{2} - x + 2$ and the line $y = x + 2$ in square unit equals

Explanation

Reviewed by Dr. Abubacker

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