Question: In the following exercises, evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the x-axis.
71.
Middle graph: three upward triangles above the x-axis, labeled $1 - |x - 1|$ from $x=0$ to $x=2$, $2 - |x - 4|$ from $x=2$ to $x=6$, and $3 - |x - 9|$ from $x=6$ to $x=12$.
1. **State the problem:**
Evaluate the integral of the function composed of three triangles above the x-axis:
- $f(x) = 1 - |x - 1|$ for $x \in [0,2]$
- $f(x) = 2 - |x - 4|$ for $x \in [2,6]$
- $f(x) = 3 - |x - 9|$ for $x \in [6,12]$
2. **Formula for area of a triangle:**
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
3. **Analyze each triangle:**
- For $1 - |x - 1|$ on $[0,2]$:
- Base length = $2 - 0 = 2$
- The function forms a triangle with peak at $x=1$ where $f(1) = 1 - |1-1| = 1$
- Height = 1
- For $2 - |x - 4|$ on $[2,6]$:
- Base length = $6 - 2 = 4$
- Peak at $x=4$ where $f(4) = 2 - |4-4| = 2$
- Height = 2
- For $3 - |x - 9|$ on $[6,12]$:
- Base length = $12 - 6 = 6$
- Peak at $x=9$ where $f(9) = 3 - |9-9| = 3$
- Height = 3
4. **Calculate each area:**
$$\text{Area}_1 = \frac{1}{2} \times 2 \times 1 = 1$$
$$\text{Area}_2 = \frac{1}{2} \times 4 \times 2 = 4$$
$$\text{Area}_3 = \frac{1}{2} \times 6 \times 3 = 9$$
5. **Sum the areas to find the total integral:**
$$\text{Total area} = 1 + 4 + 9 = 14$$
**Final answer:**
$$\boxed{14}$$