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Let 
f be the function defined by 
f(x)=6ln(x). If four subintervals of equal length are used, what is the value of the left Riemann sum approximation for 
int_(1)^(9)6ln(x)dx ? Round to the nearest thousandth if necessary.
Answer:

Let f f be the function defined by f(x)=6ln(x) f(x)=6 \ln (x) . If four subintervals of equal length are used, what is the value of the left Riemann sum approximation for 196ln(x)dx \int_{1}^{9} 6 \ln (x) d x ? Round to the nearest thousandth if necessary.\newlineAnswer:

Full solution

Q. Let f f be the function defined by f(x)=6ln(x) f(x)=6 \ln (x) . If four subintervals of equal length are used, what is the value of the left Riemann sum approximation for 196ln(x)dx \int_{1}^{9} 6 \ln (x) d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Divide into Subintervals: To approximate the integral using the left Riemann sum, we first need to divide the interval [1,9][1, 9] into four subintervals of equal length. The length of each subinterval is calculated by subtracting the lower bound of the integral from the upper bound and then dividing by the number of subintervals.\newlineLength of each subinterval = (91)/4=8/4=2(9 - 1) / 4 = 8 / 4 = 2.
  2. Calculate Endpoints: The endpoints of the subintervals are then 11, 33, 55, 77, and 99. Since we are using the left Riemann sum, we will use the left endpoints of each subinterval to calculate the sum. These left endpoints are 11, 33, 55, and 77.
  3. Evaluate Function Values: Now we evaluate the function f(x)=6ln(x)f(x) = 6\ln(x) at each of the left endpoints and multiply each function value by the length of the subinterval (which is 22). This gives us the areas of the rectangles that approximate the integral.\newlinef(1)=6ln(1)=6×0=0f(1) = 6\ln(1) = 6 \times 0 = 0\newlinef(3)=6ln(3)f(3) = 6\ln(3)\newlinef(5)=6ln(5)f(5) = 6\ln(5)\newlinef(7)=6ln(7)f(7) = 6\ln(7)
  4. Sum Areas of Rectangles: We calculate the function values:\newlinef(3)=6ln(3)6×1.09866.5916f(3) = 6\ln(3) \approx 6 \times 1.0986 \approx 6.5916\newlinef(5)=6ln(5)6×1.60949.6564f(5) = 6\ln(5) \approx 6 \times 1.6094 \approx 9.6564\newlinef(7)=6ln(7)6×1.945911.6754f(7) = 6\ln(7) \approx 6 \times 1.9459 \approx 11.6754
  5. Calculate Left Riemann Sum: Now we sum up the areas of the rectangles to get the left Riemann sum approximation:\newlineLeft Riemann sum = f(1)×2f(1) \times 2 + f(3)×2f(3) \times 2 + f(5)×2f(5) \times 2 + f(7)×2f(7) \times 2\newlineLeft Riemann sum = (0×2)+(6.5916×2)+(9.6564×2)+(11.6754×2)(0 \times 2) + (6.5916 \times 2) + (9.6564 \times 2) + (11.6754 \times 2)
  6. Perform Multiplication and Addition: Perform the multiplication and addition to get the final approximation:\newlineLeft Riemann sum = (0)+(13.1832)+(19.3128)+(23.3508)(0) + (13.1832) + (19.3128) + (23.3508)\newlineLeft Riemann sum = 55.846855.8468
  7. Round to Nearest Thousandth: Round the result to the nearest thousandth as requested: Left Riemann sum 55.847\approx 55.847