1. Interpret the region

The region is

$$A=\{(x,y): x^2 \le y \le \min\{x+2,\,4-3x\}\}.$$

So for each fixed $x$, the upper boundary is the smaller of the two lines:

• $y=x+2$
• $y=4-3x$

and the lower boundary is the parabola

• $y=x^2$.

Thus the vertical thickness is

$$\min\{x+2,4-3x\}-x^2,$$

but only where this is nonnegative.

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2. Find which line is the minimum

We compare the two lines:

$$x+2=4-3x$$ $$4x=2 \implies x=\frac12.$$

Hence:

• for $x\le \frac12$, we have $x+2 \le 4-3x$, so the minimum is $x+2$;
• for $x\ge \frac12$, the minimum is $4-3x$.

So the top boundary is piecewise:

$$y=\begin{cases} x+2, & x\le \frac12,\\ 4-3x, & x\ge \frac12. \end{cases}$$

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3. Find where the region actually exists

We need

$$x^2 \le \min\{x+2,4-3x\}.$$

So solve separately.

(i) For $x\le \frac12$

Need

$$x^2 \le x+2$$ $$x^2-x-2\le 0$$ $$(x-2)(x+1)\le 0.$$

This gives

$$-1\le x\le 2.$$

But in this case also $x\le \frac12$, so we get

$$-1\le x\le \frac12.$$

(ii) For $x\ge \frac12$

Need

$$x^2 \le 4-3x$$ $$x^2+3x-4\le 0$$ $$(x+4)(x-1)\le 0.$$

This gives

$$-4\le x\le 1.$$

But here also $x\ge \frac12$, so we get

$$\frac12\le x\le 1.$$

Therefore the full interval is

$$-1\le x\le 1.$$

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4. Set up the area integral

So the required area is

$$\text{Area}=\int_{-1}^{1/2}\big[(x+2)-x^2\big]dx+\int_{1/2}^{1}\big[(4-3x)-x^2\big]dx.$$

That is,

$$\text{Area}=\int_{-1}^{1/2}(-x^2+x+2)\,dx+\int_{1/2}^{1}(4-3x-x^2)\,dx.$$

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5. Evaluate the first integral

$$I_1=\int_{-1}^{1/2}(-x^2+x+2)dx.$$

An antiderivative is

$$-\frac{x^3}{3}+\frac{x^2}{2}+2x.$$

So

$$I_1=\left[-\frac{x^3}{3}+\frac{x^2}{2}+2x\right]_{-1}^{1/2}.$$

At $x=\frac12$:

$$-\frac{(1/2)^3}{3}+\frac{(1/2)^2}{2}+2\cdot\frac12 =-\frac{1}{24}+\frac{1}{8}+1 =1+\frac{2}{24} =\frac{13}{12}.$$

At $x=-1$:

$$-\frac{(-1)^3}{3}+\frac{(-1)^2}{2}+2(-1) =\frac13+\frac12-2 =\frac{5}{6}-2 =-\frac{7}{6}.$$

Hence

$$I_1=\frac{13}{12}-\left(-\frac{7}{6}\right) =\frac{13}{12}+\frac{14}{12} =\frac{27}{12} =\frac94.$$

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6. Evaluate the second integral

$$I_2=\int_{1/2}^{1}(4-3x-x^2)dx.$$

An antiderivative is

$$4x-\frac{3x^2}{2}-\frac{x^3}{3}.$$

So

$$I_2=\left[4x-\frac{3x^2}{2}-\frac{x^3}{3}\right]_{1/2}^{1}.$$

At $x=1$:

$$4-\frac32-\frac13 =\frac{24-9-2}{6} =\frac{13}{6}.$$

At $x=\frac12$:

$$4\cdot\frac12-\frac{3(1/2)^2}{2}-\frac{(1/2)^3}{3} =2-\frac{3}{8}-\frac{1}{24} =2-\frac{10}{24} =2-\frac{5}{12} =\frac{19}{12}.$$

Hence

$$I_2=\frac{13}{6}-\frac{19}{12} =\frac{26-19}{12} =\frac{7}{12}.$$

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7. Total area

$$\text{Area}=I_1+I_2=\frac94+\frac{7}{12} =\frac{27}{12}+\frac{7}{12} =\frac{34}{12} =\frac{17}{6}.$$

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8. Match with options

$$\frac{17}{6}$$

corresponds to Option B.

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9. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.