1. **Problem Statement:** Evaluate the integrals \(\int_0^{18} f(x) \, dx\), \(\int_0^{45} f(x) \, dx\), \(\int_{45}^{63} f(x) \, dx\), and \(\int_0^{81} f(x) \, dx\) by interpreting them as areas under the curve of \(f(x)\). 2. **Key Idea:** The definite integral \(\int_a^b f(x) \, dx\) represents the net area between the graph of \(f(x)\) and the x-axis from \(x=a\) to \(x=b\). Areas above the x-axis are positive, and areas below are negative. 3. **Given:** - \(\int_0^{18} f(x) \, dx = 324\) (already provided). - The graph crosses the x-axis at approximately \(x=18\) and \(x=54\). - From \(x=18\) to \(x=36\), \(f(x)=0\), so the integral over this interval is 0. - From \(x=36\) to \(x=54\), the graph is above the x-axis (assumed from description). - From \(x=54\) to \(x=63\), the graph is below the x-axis. 4. **Calculate \(\int_0^{45} f(x) \, dx\):** - This integral covers \(0\) to \(18\) and \(18\) to \(45\). - From \(0\) to \(18\), area is 324 (given). - From \(18\) to \(36\), area is 0 (flat at zero). - From \(36\) to \(45\), the graph is above the x-axis. Assume the area from \(36\) to \(54\) is a triangle or trapezoid. Since exact values are not given, approximate the area from \(36\) to \(54\) as positive and from \(36\) to \(45\) as half of that. - Let the area from \(36\) to \(54\) be \(A\). Then area from \(36\) to \(45\) is approximately \(\frac{9}{18}A = \frac{1}{2}A\). - Without exact values, we cannot compute \(A\), so we leave the answer in terms of \(A\). 5. **Calculate \(\int_{45}^{63} f(x) \, dx\):** - From \(45\) to \(54\), graph is above x-axis (part of \(A\) area). - From \(54\) to \(63\), graph is below x-axis. - The area below x-axis is negative. - Let the area from \(54\) to \(63\) be \(B\) (positive value, but integral is \(-B\)). - So \(\int_{45}^{63} f(x) \, dx = (\text{area above x-axis from }45\text{ to }54) - B\). 6. **Calculate \(\int_0^{81} f(x) \, dx\):** - Sum all areas from \(0\) to \(81\). - From \(0\) to \(18\): 324 - From \(18\) to \(36\): 0 - From \(36\) to \(54\): \(A\) - From \(54\) to \(63\): \(-B\) - From \(63\) to \(72\): graph rises slightly but remains below x-axis, so negative area \(C\). - From \(72\) to \(81\): no data, assume 0. - Total integral: \(324 + 0 + A - B - C\). **Final answers:** - (a) \(\int_0^{18} f(x) \, dx = 324\) - (b) \(\int_0^{45} f(x) \, dx = 324 + \text{area from }36\text{ to }45\) (positive, exact value unknown) - (c) \(\int_{45}^{63} f(x) \, dx = \text{area above x-axis }(45\text{ to }54) - \text{area below x-axis }(54\text{ to }63)\) - (d) \(\int_0^{81} f(x) \, dx = 324 + A - B - C\) (sum of all areas) Since exact numeric values for areas \(A, B, C\) are not provided, the integrals (b), (c), and (d) cannot be numerically evaluated here.