Question

Find the area bounded on the right by the line x+2=y, on the left by the parabola $y=x^2$ and above the x-axis

• A. $9/2$ sq.units
• B. $2$ sq.units
• C. $3$ sq.units
• D. $5/3$ sq.units

Answer 1

The correct option is A $9/2$ sq.units

$x+2=y$ and $y=x^2$ intersect at $x=2$ and $x=-1$

Therefore, $A={\int }_{-1}^2(x+2-x^2)dx$

$={[\frac{x^2}{2}+2x-\frac{x^3}{3}]}_0^1$

$=\frac{1}{2}+2-\frac{1}{3}-0$

$=\frac{9}{2}$