Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let 
f be the function defined by 
f(x)=3ln(x). If three subintervals of equal length are used, what is the value of the left Riemann sum approximation for 
int_(1)^(10)3ln(x)dx ? Round to the nearest thousandth if necessary.
Answer:

Let f f be the function defined by f(x)=3ln(x) f(x)=3 \ln (x) . If three subintervals of equal length are used, what is the value of the left Riemann sum approximation for 1103ln(x)dx \int_{1}^{10} 3 \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)=3ln(x) f(x)=3 \ln (x) . If three subintervals of equal length are used, what is the value of the left Riemann sum approximation for 1103ln(x)dx \int_{1}^{10} 3 \ln (x) d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Calculate Width of Subintervals: To calculate the left Riemann sum, we first need to determine the width of each subinterval. The interval [1,10][1, 10] has a length of 101=910 - 1 = 9. Since we are using three subintervals of equal length, each subinterval will have a width of 93=3\frac{9}{3} = 3.
  2. Find X-Values for Left Riemann Sum: Next, we need to find the xx-values at which we will evaluate the function f(x)=3ln(x)f(x) = 3\ln(x) for the left Riemann sum. These xx-values are the left endpoints of each subinterval. The first subinterval is [1,4][1, 4], the second is [4,7][4, 7], and the third is [7,10][7, 10]. Therefore, the xx-values are 11, 44, and 77.
  3. Evaluate Function at X-Values: Now we evaluate the function f(x)f(x) at each of these x-values. f(1)=3ln(1)=0f(1) = 3\ln(1) = 0, f(4)=3ln(4)f(4) = 3\ln(4), and f(7)=3ln(7)f(7) = 3\ln(7).
  4. Calculate Left Riemann Sum: We can now calculate the left Riemann sum by multiplying each function value by the width of the subintervals and summing them up. The left Riemann sum is (f(1)×3)+(f(4)×3)+(f(7)×3) (f(1) \times 3) + (f(4) \times 3) + (f(7) \times 3) .
  5. Substitute Function Values: Substitute the function values into the sum to get the approximation: 0×30 \times 3 + 3ln(4)×33\ln(4) \times 3 + 3ln(7)×33\ln(7) \times 3.
  6. Perform Multiplication: Perform the multiplication: 0+(3×3×ln(4))+(3×3×ln(7))0 + (3 \times 3 \times \ln(4)) + (3 \times 3 \times \ln(7)).
  7. Calculate Numerical Values: Calculate the numerical values: 0+(9×ln(4))+(9×ln(7))0 + (9 \times \ln(4)) + (9 \times \ln(7)).
  8. Calculate Approximation: Using a calculator, we find that ln(4)1.386\ln(4) \approx 1.386 and ln(7)1.946\ln(7) \approx 1.946. So the approximation becomes 0+(9×1.386)+(9×1.946)0 + (9 \times 1.386) + (9 \times 1.946).
  9. Perform Final Calculations: Now we perform the calculations: 0+(9×1.386)+(9×1.946)=0+12.474+17.5140 + (9 \times 1.386) + (9 \times 1.946) = 0 + 12.474 + 17.514.
  10. Round to Nearest Thousandth: Adding these values together gives us the left Riemann sum approximation: 12.474+17.514=29.98812.474 + 17.514 = 29.988.
  11. Round to Nearest Thousandth: Adding these values together gives us the left Riemann sum approximation: 12.474+17.514=29.98812.474 + 17.514 = 29.988. Round to the nearest thousandth if necessary. The approximation is already at the thousandth place, so rounding is not needed. The final answer is 29.98829.988.