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

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

\[ \begin{array}{c|c} \text{Graph} & \text{Integral} \\ \hline \text{y} & \text{y = f(x)} \\ \text{18} & \text{0} \\ \text{18} & \text{18} \\ \text{36} & \text{36} \\ \text{54} & \text{54} \\ \text{72} & \text{72} \\ \end{array} \]

(a) \(\int_{0}^{18} f(x) \, dx\)

324

(b) \(\int_{0}^{45} f(x) \, dx\)

(c) \(\int_{45}^{63} f(x) \, dx\)

(d) \(\int_{0}^{81} f(x) \, dx\)

Enhanced Feedback

Please try again. To evaluate an integral by interpreting it in terms of area, consider the area under the curve.

Solution

To solve the integral questions from the image, we’ll evaluate each integral step by step by interpreting the function \( f(x) \) in terms of the area under its graph. ### (b) \(\int_0^{45} f(x) \, dx\) First, let’s break down the integral into three parts based on the behavior of \( f(x) \): 1. From \( x = 0 \) to \( x = 18 \): \[ \int_0^{18} f(x) \, dx = 324 \] 2. From \( x = 18 \) to \( x = 36 \): \( f(x) \) is constant at 36. \[ \int_{18}^{36} f(x) \, dx = 36 \times (36 - 18) = 36 \times 18 = 648 \] 3. From \( x = 36 \) to \( x = 45 \): \( f(x) \) decreases linearly from 36 at \( x = 36 \) to 24 at \( x = 45 \). This forms a trapezoid. \[ \text{Area} = \frac{1}{2} \times (36 + 24) \times (45 - 36) = \frac{1}{2} \times 60 \times 9 = 270 \] Adding these areas together: \[ \int_0^{45} f(x) \, dx = 324 + 648 + 270 = 1242 \] ### (c) \(\int_{45}^{63} f(x) \, dx\) From \( x = 45 \) to \( x = 63 \), \( f(x) \) decreases linearly from 24 to 0, forming a triangle. \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (63 - 45) \times 24 = \frac{1}{2} \times 18 \times 24 = 216 \] ### (d) \(\int_0^{81} f(x) \, dx\) To evaluate this integral, we’ll consider the periodic behavior of \( f(x) \) beyond \( x = 72 \): 1. From \( x = 0 \) to \( x = 72 \): \[ \int_0^{72} f(x) \, dx = 1620 \] 2. From \( x = 72 \) to \( x = 81 \): Assuming \( f(x) \) repeats its behavior every 72 units, \( f(x) \) from 72 to 81 is the same as from 0 to 9. \[ \int_{72}^{81} f(x) \, dx = \int_0^{9} 2x \, dx = \left[ x^2 \right]0^{9} = 81 \] Adding these together: \[ \int_0^{81} f(x) \, dx = 1620 + 81 = 1701 \] ### Final Answers - (b) \(\int_0^{45} f(x) \, dx = 1242\) - (c) \(\int{45}^{63} f(x) \, dx = 216\) - (d) \(\int_0^{81} f(x) \, dx = 1701\)