Riemann Sum True 7Db651

1. **State the problem:** We need to determine which statement about the function $f(x) = 2 - \frac{1}{2}x$ over the interval $[0,10]$ is true. 2. **Understand the function and graph:** The function is a decreasing linear function starting at $f(0) = 2$ and ending at $f(10) = 2 - \frac{1}{2} \times 10 = 2 - 5 = -3$. 3. **Calculate the definite integral (net area):** $$\int_0^{10} \left(2 - \frac{1}{2}x\right) dx = \int_0^{10} 2 \, dx - \int_0^{10} \frac{1}{2}x \, dx$$ Calculate each integral: $$\int_0^{10} 2 \, dx = 2x \Big|_0^{10} = 2 \times 10 - 0 = 20$$ $$\int_0^{10} \frac{1}{2}x \, dx = \frac{1}{2} \times \frac{x^2}{2} \Big|_0^{10} = \frac{1}{4} x^2 \Big|_0^{10} = \frac{1}{4} \times 100 = 25$$ So, $$\int_0^{10} f(x) \, dx = 20 - 25 = -5$$ 4. **Interpret the net area:** The net area under the curve from 0 to 10 is $-5$. 5. **Analyze the Riemann sums:** - Since $f(x)$ is decreasing, the **left Riemann sum** over $[0,10]$ will use function values at the left endpoints, which are larger than the function values at the right endpoints. For a decreasing function, the left Riemann sum **overestimates** the integral. - The **midpoint Riemann sum** tends to be more accurate but for a linear function, it equals the exact integral. However, since the function crosses the x-axis and changes sign, the midpoint sum can be checked but generally it will not exceed the net area for this linear function. 6. **Check the statements:** - a. "The left Riemann Sum for $f$ over $[0,10]$ will be less than the net area." This is false because for a decreasing function, left Riemann sum is greater than the integral. - b. "The midpoint Riemann Sum for $f$ over $[0,10]$ will exceed the net area." For a linear function, midpoint sum equals the integral exactly, so this is false. - c. "$\int_0^{10} f(x) \, dx = 13$." We calculated it as $-5$, so false. - d. "The net area is -5." This matches our calculation, so true. - e. "None of the answers." False because d is true. **Final answer:** d. The net area is -5