We need the area of the region satisfying

$2y \le x^2+3, \qquad y+|x|\le 3, \qquad y\ge |x-1|.$

So the region is

$y \le \frac{x^2+3}{2}, \qquad y\le 3-|x|, \qquad y\ge |x-1|.$

Thus, for each $x$, the allowed $y$ lies between

$|x-1| \le y \le \min\left(\frac{x^2+3}{2},\, 3-|x|\right).$

We now determine where this is possible.

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1. Compare the two upper bounds

We compare

$\frac{x^2+3}{2} \quad \text{and} \quad 3-|x|.$

Solve

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

This gives

$x^2+3 \le 6-2|x|$ $x^2+2|x|-3 \le 0.$

Let $t=|x|\ge 0$. Then

$t^2+2t-3\le 0 = (t+3)(t-1)\le 0.$

Hence

$0\le t\le 1 \implies |x|\le 1.$

So:

• for $|x|\le 1$, upper bound is $\dfrac{x^2+3}{2}$,
• for $|x|\ge 1$, upper bound is $3-|x|$.

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2. Find where lower bound does not exceed upper bound

We need

$|x-1| \le \min\left(\frac{x^2+3}{2}, 3-|x|\right).$

We check in intervals.

Case I: $x\in[-1,1]$

Here upper bound is $\dfrac{x^2+3}{2}$. Also, since $x\le 1$,

$|x-1|=1-x.$

Need

$1-x \le \frac{x^2+3}{2}.$

Multiply by $2$:

$2-2x \le x^2+3$ $0\le x^2+2x+1=(x+1)^2,$

which is always true. So all $x\in[-1,1]$ work.

Case II: $x\ge 1$

Here upper bound is $3-x$ and

$|x-1|=x-1.$

Need

$x-1 \le 3-x \implies 2x\le 4 \implies x\le 2.$

So $x\in[1,2]$ work.

Case III: $x\le -1$

Here upper bound is $3+x$ (since $|x|=-x$), and

$|x-1|=1-x.$

Need

$1-x \le 3+x \implies -2x\le 2 \implies x\ge -1.$

Together with $x\le -1$, this gives only $x=-1$. So no interval contribution from left side beyond the endpoint.

Hence the region exists for

$x\in[-1,2].$

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3. Set up area integral

So

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

Simplify integrands.

For $x\in[-1,1]$:

$\frac{x^2+3}{2}-(1-x)=\frac{x^2+3-2+2x}{2}=\frac{(x+1)^2}{2}.$

For $x\in[1,2]$:

$(3-x)-(x-1)=4-2x.$

Thus

$A=\int_{-1}^{1}\frac{(x+1)^2}{2}\,dx+\int_{1}^{2}(4-2x)\,dx.$

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4. Evaluate the integrals

First integral

Let $u=x+1$. When $x=-1$, $u=0$; when $x=1$, $u=2$.

=\frac12\left[\frac{u^3}{3}\right]_0^2 =\frac12\cdot \frac{8}{3}=\frac{4}{3}.$$ ### Second integral $$\int_1^2 (4-2x)\,dx = [4x-x^2]_1^2=(8-4)-(4-1)=4-3=1.$$ So $$A=\frac{4}{3}+1=\frac{7}{3}.$$ Therefore, $$6A=6\cdot \frac{7}{3}=14.$$ --- ## 5. Check options - A: $14$ ✅ - B: $18$ ❌ - C: $16$ ❌ - D: $12$ ❌ So the correct option is **A**.