Write the curves clearly

The given curves are:

• Line: $y - x = 2 \Rightarrow y = x + 2$
• Parabola: $x^2 = y \Rightarrow y = x^2$

Find the points of intersection

At intersection, $x^2 = x + 2$ $x^2 - x - 2 = 0$ $(x-2)(x+1)=0$

So, $x=2 \quad \text{or} \quad x=-1$

Corresponding $y$-values:

• For $x=2$, $y=4$
• For $x=-1$, $y=1$

Hence the intersection points are: $(-1,1), \ (2,4)$

Determine which curve is above the other

Between $x=-1$ and $x=2$, compare: $y=x+2 \quad \text{and} \quad y=x^2$

Take a test point, say $x=0$:

• Line gives $y=2$
• Parabola gives $y=0$

So the line lies above the parabola on $[-1,2]$.

Set up the area integral

Area enclosed is: $A=\int_{-1}^{2} \big[(x+2)-x^2\big] \, dx$

Evaluate the integral

$A=\int_{-1}^{2} (x+2-x^2)\,dx$ $A=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}$

At $x=2$: $\frac{2^2}{2}+2(2)-\frac{2^3}{3}=2+4-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}$

At $x=-1$: $\frac{(-1)^2}{2}+2(-1)-\frac{(-1)^3}{3}=\frac{1}{2}-2+\frac{1}{3}$ $=\frac{3-12+2}{6}=-\frac{7}{6}$

Therefore, $A=\frac{10}{3}-\left(-\frac{7}{6}\right)$ $A=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}$

Compare with options

$\frac{9}{2}$ matches Option C.

Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.