1. Interpret the region

We need the area of the set

$$\{(x,y): x\ge 0,\ x+y\le 3,\ x^2\le 4y,\ y\le 1+\sqrt{x}\}.$$

Rewrite each inequality in a more useful form:

• $x\ge 0$
• $x+y\le 3 \implies y\le 3-x$
• $x^2\le 4y \implies y\ge \dfrac{x^2}{4}$
• $y\le 1+\sqrt{x}$

So for each $x\ge 0$, the region lies above $y=\frac{x^2}{4}$ and below both $y=3-x \quad \text{and} \quad y=1+\sqrt{x}.$

Hence upper boundary is $\min(3-x,\ 1+\sqrt{x}).$

So area is determined by comparing the two upper curves.

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2. Find intersection of the two upper curves

Solve

$$3-x = 1+\sqrt{x}.$$

Then

$$2-x = \sqrt{x}.$$

Let $t=\sqrt{x}$, so $x=t^2$ with $t\ge 0$. Then

$$2-t^2=t$$ $$t^2+t-2=0$$ $$(t+2)(t-1)=0.$$

Since $t\ge 0$, we get $t=1$, hence

$$$$

At $x=1$, both give $y=2$.

Now check which is smaller:

• At $x=0$: $3-x=3$, $1+\sqrt{x}=1$, so $1+\sqrt{x}$ is smaller.
• For $x>1$, $3-x$ becomes smaller.

Thus,

• for $0\le x\le 1$, upper curve is $y=1+\sqrt{x}$,
• for $x\ge 1$, upper curve is $y=3-x$.

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3. Find where the lower curve meets the relevant upper curves

The lower curve is $y=\frac{x^2}{4}.$

(i) Intersection with $y=1+\sqrt{x}$

Solve

$$\frac{x^2}{4}=1+\sqrt{x}.$$

Let $t=\sqrt{x}$, so $x=t^2$. Then

$$\frac{t^4}{4}=1+t$$ $$t^4=4t+4.$$

Check $t=\sqrt{2}$: $t^4=4, \quad 4t+4>4,$ so not equal. Instead test $x=4$:

$$$$

so intersection is not in the interval near $[0,1]$. Since on $0\le x\le 1$, we have $\frac{x^2}{4}\le \frac14 < 1+\sqrt{x},$ there is no issue there; region exists throughout $[0,1]$.

(ii) Intersection with $y=3-x$

Solve

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

Multiply by $4$:

$$x^2=12-4x$$ $$x^2+4x-12=0$$ $$(x+6)(x-2)=0.$$

Since $x\ge 0$, we get

$$$$

Thus for $x>2$, the lower curve exceeds $3-x$, so no region exists.

Hence the total region spans:

• $0\le x\le 1$: between $y=\dfrac{x^2}{4}$ and $y=1+\sqrt{x}$
• $1\le x\le 2$: between $y=\dfrac{x^2}{4}$ and $y=3-x$

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4. Set up the area integral

Therefore,

$$A=\int_0^1 \left(1+\sqrt{x}-\frac{x^2}{4}\right)dx +\int_1^2 \left(3-x-\frac{x^2}{4}\right)dx.$$

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5. Evaluate the first integral

$$I_1=\int_0^1 \left(1+\sqrt{x}-\frac{x^2}{4}\right)dx$$

Integrate termwise:

$$\int 1\,dx=x, \qquad \int \sqrt{x}\,dx=\frac{2}{3}x^{3/2}, \qquad \int \frac{x^2}{4}\,dx=\frac{x^3}{12}.$$

So

$$I_1=\left[x+\frac{2}{3}x^{3/2}-\frac{x^3}{12}\right]_0^1 =1+\frac23-\frac1{12}.$$

Compute:

1+\frac23-\frac1{12}=rac{12+8-1}{12}=rac{19}{12}.

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6. Evaluate the second integral

$$I_2=\int_1^2 \left(3-x-\frac{x^2}{4}\right)dx$$

Integrate termwise:

$$\int \left(3-x-\frac{x^2}{4}\right)dx =3x-\frac{x^2}{2}-\frac{x^3}{12}.$$

Thus

$$I_2=\left[3x-\frac{x^2}{2}-\frac{x^3}{12}\right]_1^2.$$

At $x=2$:

$$3(2)-\frac{4}{2}-\frac{8}{12}=6-2-\frac23=\frac{10}{3}.$$

At $x=1$:

$$3-\frac12-\frac1{12}=3-\frac{6}{12}-\frac{1}{12}=3-\frac{7}{12}=\frac{29}{12}.$$

Therefore,

I_2=\frac{10}{3}-\frac{29}{12}=rac{40-29}{12}=\frac{11}{12}.

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7. Total area

$$A=I_1+I_2=\frac{19}{12}+\frac{11}{12}=\frac{30}{12}=\frac52.$$

So the required area is

$$\boxed{\frac52}.$$

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8. Check options

• A: $\frac32$ ❌
• B: $\frac73$ ❌
• C: $\frac52$ ✅
• D: $\frac{59}{12}$ ❌

Therefore, the correct option is C.