Key Concepts and Formulas

• Area Between Curves: The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a,b]$, is given by $\int_a^b [f(x) - g(x)] dx$.
• Absolute Value Function: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.
• Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for $x$.

Step-by-Step Solution

Step 1: Analyze the given functions

We are given $y = |x-1|$ and $y = 3 - |x|$. We need to find the area of the region bounded by these two curves. First, we need to understand how these functions behave due to the absolute value signs.

Step 2: Express the functions piecewise

We can rewrite the functions as piecewise functions:

1-x, & \text{if } x < 1 \\ x-1, & \text{if } x \ge 1 \end{cases}$$ $$y = 3 - |x| = \begin{cases} 3+x, & \text{if } x < 0 \\ 3-x, & \text{if } x \ge 0 \end{cases}$$ **Step 3: Find the intersection points** We need to find where the curves intersect. We will consider different cases for the values of $x$: * **Case 1:** $x < 0$. In this case, $|x-1| = 1-x$ and $3-|x| = 3+x$. Setting these equal gives $1-x = 3+x$, so $2x = -2$, which means $x = -1$. Since $-1 < 0$, this is a valid intersection point. At $x=-1$, $y = 3 - |-1| = 3-1 = 2$. So, $(-1, 2)$ is an intersection point. * **Case 2:** $0 \le x < 1$. In this case, $|x-1| = 1-x$ and $3-|x| = 3-x$. Setting these equal gives $1-x = 3-x$, which simplifies to $1=3$. This is impossible, so there are no intersection points in this interval. * **Case 3:** $x \ge 1$. In this case, $|x-1| = x-1$ and $3-|x| = 3-x$. Setting these equal gives $x-1 = 3-x$, so $2x = 4$, which means $x = 2$. Since $2 \ge 1$, this is a valid intersection point. At $x=2$, $y = |2-1| = 1$. So, $(2, 1)$ is an intersection point. Therefore, the intersection points are $(-1, 2)$ and $(2, 1)$. **Step 4: Set up the integral(s) for the area** We need to split the area calculation into integrals based on the x-values of the intersection points and the points where the absolute value functions change definition (x=0 and x=1). We will integrate from $x=-1$ to $x=2$. * From $x=-1$ to $x=0$: The upper curve is $y = 3+x$ and the lower curve is $y = 1-x$. * From $x=0$ to $x=1$: The upper curve is $y = 3-x$ and the lower curve is $y = 1-x$. * From $x=1$ to $x=2$: The upper curve is $y = 3-x$ and the lower curve is $y = x-1$. The total area $A$ is given by: $$A = \int_{-1}^0 [(3+x) - (1-x)] dx + \int_0^1 [(3-x) - (1-x)] dx + \int_1^2 [(3-x) - (x-1)] dx$$ **Step 5: Evaluate the integrals** $$A = \int_{-1}^0 (2+2x) dx + \int_0^1 (2) dx + \int_1^2 (4-2x) dx$$ $$A = [2x + x^2]_{-1}^0 + [2x]_0^1 + [4x - x^2]_1^2$$ $$A = [(0) - (-2 + 1)] + [2 - 0] + [(8-4) - (4-1)]$$ $$A = [0 - (-1)] + 2 + [4 - 3]$$ $$A = 1 + 2 + 1 = 4$$ **Step 6: Re-examine the intervals for simpler calculation** Observe that the region is symmetric about the line $x = \frac{1}{2}$. Let's consider the intersection points. They are $(-1, 2)$ and $(2, 1)$. Let's verify the functions at $x=0$. We have $y=|0-1|=1$ and $y=3-|0|=3$. Let's verify the functions at $x=1$. We have $y=|1-1|=0$ and $y=3-|1|=2$. We can write the area as: $$A = \int_{-1}^{2} (3-|x| - |x-1|)dx = \int_{-1}^{0} (3+x - (1-x))dx + \int_{0}^{1} (3-x - (1-x))dx + \int_{1}^{2} (3-x - (x-1))dx$$ $$A = \int_{-1}^{0} (2+2x)dx + \int_{0}^{1} 2 dx + \int_{1}^{2} (4-2x)dx$$ $$A = [2x+x^2]_{-1}^{0} + [2x]_{0}^{1} + [4x-x^2]_{1}^{2}$$ $$A = (0 - (-2+1)) + (2-0) + (8-4 - (4-1)) = 1 + 2 + 1 = 4$$ The area is 4. **Step 7: Final Answer** The area of the region bounded by the given curves is 4 square units. **Common Mistakes & Tips** * **Forgetting to consider different cases for absolute values:** Always break down absolute value functions into piecewise functions to avoid errors. * **Incorrectly identifying the upper and lower curves:** Sketching the graph can help visualize which function is greater in each interval. * **Not finding all intersection points:** Make sure you've considered all possible intersection points within the relevant domain. **Summary** We found the area bounded by the curves $y=|x-1|$ and $y=3-|x|$ by first expressing the functions piecewise, then finding the intersection points. We then set up and evaluated the definite integrals, splitting the region into subregions where the upper and lower curves were consistently defined. The area of the region is 4 square units. **Final Answer** The final answer is \boxed{4}, which corresponds to option (D).