1. Given curve

We need the area enclosed by $y=3-\left|x-\frac12\right|-|x+1|$ and the $x$-axis.

So we must:

• remove the modulus piecewise,
• find where $y\ge 0$ and the intercepts with the $x$-axis,
• integrate over the interval where the graph lies above the $x$-axis.

2. Critical points for modulus

The modulus expressions change at: $x=-1,\quad x=\frac12$

So consider three intervals:

1. $x<-1$
2. $-1\le x<\frac12$
3. $x\ge \frac12$

3. Write $y$ piecewise

Case 1: $x<-1$

Then $\left|x-\frac12\right|=\frac12-x,\qquad |x+1|=-(x+1)=-x-1$ Hence $y=3-\left(\frac12-x\right)-(-x-1)$ $y=3-\frac12+x+x+1=\frac72+2x$

So for $x<-1$, $y=2x+\frac72$

Case 2: $-1\le x<\frac12$

Then $\left|x-\frac12\right|=\frac12-x,\qquad |x+1|=x+1$ Hence $y=3-\left(\frac12-x\right)-(x+1)$ $y=3-\frac12+x-x-1=\frac32$

So for $-1\le x<\frac12$, $y=\frac32$

Case 3: $x\ge \frac12$

Then $\left|x-\frac12\right|=x-\frac12,\qquad |x+1|=x+1$ Hence $y=3-\left(x-\frac12\right)-(x+1)$ $y=3-x+\frac12-x-1=\frac52-2x$

So for $x\ge \frac12$, $y=\frac52-2x$

4. Find the $x$-intercepts

We solve $y=0$ in the relevant pieces.

Left piece:

$2x+\frac72=0$ $2x=-\frac72$ $x=-\frac74$

Middle piece:

$y=\frac32>0$ No intercept here.

Right piece:

$\frac52-2x=0$ $2x=\frac52$ $x=\frac54$

So the curve is above the $x$-axis between $x=-\frac74 \quad \text{and} \quad x=\frac54$

5. Compute the area

The bounded area is $A=\int_{-7/4}^{-1}\left(2x+\frac72\right)dx+\int_{-1}^{1/2}\frac32\,dx+\int_{1/2}^{5/4}\left(\frac52-2x\right)dx$

First integral

$I_1=\int_{-7/4}^{-1}\left(2x+\frac72\right)dx$ Antiderivative: $x^2+\frac72x$ Thus $I_1=\left[x^2+\frac72x\right]_{-7/4}^{-1}$ At $x=-1$: $1-\frac72=-\frac52$ At $x=-\frac74$: $\left(\frac{49}{16}\right)+\frac72\left(-\frac74\right)=\frac{49}{16}-\frac{49}{8}=\frac{49}{16}-\frac{98}{16}=-\frac{49}{16}$ So $I_1=-\frac52-\left(-\frac{49}{16}\right)= -\frac{40}{16}+\frac{49}{16}=\frac{9}{16}$

Second integral

$I_2=\int_{-1}^{1/2}\frac32\,dx=\frac32\left(\frac12-(-1)\right)=\frac32\cdot\frac32=\frac94$

Third integral

$I_3=\int_{1/2}^{5/4}\left(\frac52-2x\right)dx$ Antiderivative: $\frac52x-x^2$ Thus $I_3=\left[\frac52x-x^2\right]_{1/2}^{5/4}$ At $x=\frac54$: $\frac52\cdot\frac54-\left(\frac54\right)^2=\frac{25}{8}-\frac{25}{16}=\frac{25}{16}$ At $x=\frac12$: $\frac52\cdot\frac12-\frac14=\frac54-\frac14=1$ So $I_3=\frac{25}{16}-1=\frac{9}{16}$

6. Total area

$A=I_1+I_2+I_3=\frac{9}{16}+\frac94+\frac{9}{16}$ Convert to denominator $16$: $A=\frac{9}{16}+\frac{36}{16}+\frac{9}{16}=\frac{54}{16}=\frac{27}{8}$

7. Match with options

$\frac{27}{8}$ corresponds to Option C.

8. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C. So they agree.