Question: The graph of \( f(x) \) is shown. Evaluate each integral by interpreting it in terms of areas…

The graph of \( f(x) \) is shown. Evaluate each integral by interpreting it in terms of areas.

\[ \begin{aligned} &\text{(a)} \quad \int_{0}^{18} f(x) \, dx \\ &\text{(b)} \quad \int_{0}^{45} f(x) \, dx \\ &\text{(c)} \quad \int_{45}^{63} f(x) \, dx \\ &\text{(d)} \quad \int_{63}^{81} f(x) \, dx \end{aligned} \]

The graph shows a function \( y = f(x) \) with the following points:

• \( (0, 0) \)
• \( (18, 18) \)
• \( (36, 18) \)
• \( (54, 0) \)
• \( (72, -18) \)
• \( (81, 0) \)

The area under the curve from \( x = 0 \) to \( x = 18 \) is a triangle with base 18 and height 18, so the area is \( \frac{1}{2} \times 18 \times 18 = 162 \).

The area under the curve from \( x = 18 \) to \( x = 36 \) is a rectangle with base 18 and height 18, so the area is \( 18 \times 18 = 324 \).

The area under the curve from \( x = 36 \) to \( x = 54 \) is a triangle with base 18 and height 18, so the area is \( \frac{1}{2} \times 18 \times 18 = 162 \).

The area under the curve from \( x = 54 \) to \( x = 72 \) is a triangle with base 18 and height 18, but it is below the x-axis, so the area is \( -\frac{1}{2} \times 18 \times 18 = -162 \).

The area under the curve from \( x = 72 \) to \( x = 81 \) is a triangle with base 9 and height 18, so the area is \( \frac{1}{2} \times 9 \times 18 = 81 \).

Therefore, the integrals are:

\[ \begin{aligned} &\text{(a)} \quad \int_{0}^{18} f(x) \, dx = 162 \\ &\text{(b)} \quad \int_{0}^{45} f(x) \, dx = 162 + 324 + 162 = 648 \\ &\text{(c)} \quad \int_{45}^{63} f(x) \, dx = -162 \\ &\text{(d)} \quad \int_{63}^{81} f(x) \, dx = 81 \end{aligned} \]

Solution

The problem requires evaluating definite integrals by interpreting them as areas under the graph of \( y = f(x) \). Let’s solve (b) and (c). ### (b) Evaluate \(\int_0^{45} f(x) \, dx\) 1. The area from \( x = 0 \) to \( x = 18 \) is a triangle with base 18 and height 18. \[ \text{Area}_1 = \frac{1}{2} \times 18 \times 18 = 162 \] 2. The area from \( x = 18 \) to \( x = 36 \) is a rectangle with base 18 and height 18. \[ \text{Area}_2 = 18 \times 18 = 324 \] 3. The area from \( x = 36 \) to \( x = 45 \) is a triangle with base 9 and height 18. \[ \text{Area}3 = \frac{1}{2} \times 9 \times 18 = 81 \] 4. Add all the areas for the total area from \( x = 0 \) to \( x = 45 \). \[ \int_0^{45} f(x) \, dx = 162 + 324 + 81 = 567 \] ### (c) Evaluate \(\int{45}^{63} f(x) \, dx\) 1. The area from \( x = 45 \) to \( x = 54 \) is a triangle with base 9 and height 18. \[ \text{Area}4 = \frac{1}{2} \times 9 \times 18 = 81 \] 2. The area from \( x = 54 \) to \( x = 63 \) is a triangle with base 9 and height 18 (negative since it’s below the x-axis). \[ \text{Area}5 = \frac{1}{2} \times 9 \times (-18) = -81 \] 3. Add the areas for the total area from \( x = 45 \) to \( x = 63 \). \[ \int{45}^{63} f(x) \, dx = 81 - 81 = 0 \] Thus, the answers are: (b) \(\int_0^{45} f(x) \, dx = 567\) (c) \(\int{45}^{63} f(x) \, dx = 0\)