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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 trapezoidal sum approximation for 
int_(2)^(3.5)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 trapezoidal sum approximation for 23.53ln(x)dx \int_{2}^{3.5} 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 trapezoidal sum approximation for 23.53ln(x)dx \int_{2}^{3.5} 3 \ln (x) d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Rephrase Problem: Let's first rephrase the "What is the trapezoidal sum approximation for the integral of 3ln(x)3\ln(x) from 22 to 3.53.5 using three subintervals of equal length?"
  2. Divide Interval: To use the trapezoidal rule, we need to divide the interval [2,3.5][2, 3.5] into three equal subintervals. The length of each subinterval is (3.52)/3=0.5(3.5 - 2) / 3 = 0.5.
  3. Evaluate Function: The endpoints of the subintervals are x0=2x_0 = 2, x1=2.5x_1 = 2.5, x2=3x_2 = 3, and x3=3.5x_3 = 3.5. We will evaluate the function f(x)=3ln(x)f(x) = 3\ln(x) at these points.
  4. Apply Trapezoidal Rule: Calculate the function values: f(x0)=3ln(2)f(x_0) = 3\ln(2), f(x1)=3ln(2.5)f(x_1) = 3\ln(2.5), f(x2)=3ln(3)f(x_2) = 3\ln(3), and f(x3)=3ln(3.5)f(x_3) = 3\ln(3.5).
  5. Substitute Values: Now, we apply the trapezoidal rule, which is given by the formula:\newlineT=(Δx/2)[f(x0)+2f(x1)+2f(x2)+f(x3)]T = (\Delta x/2) \cdot [f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3)],\newlinewhere Δx\Delta x is the length of each subinterval.
  6. Perform Calculations: Substitute the values into the trapezoidal rule formula:\newlineT=0.52×[3ln(2)+2×3ln(2.5)+2×3ln(3)+3ln(3.5)]T = \frac{0.5}{2} \times [3\ln(2) + 2\times3\ln(2.5) + 2\times3\ln(3) + 3\ln(3.5)].
  7. Calculate Sum: Perform the calculations:\newlineT=0.25×[3ln(2)+6ln(2.5)+6ln(3)+3ln(3.5)]T = 0.25 \times [3\ln(2) + 6\ln(2.5) + 6\ln(3) + 3\ln(3.5)].
  8. Add Values: Use a calculator to find the numerical values of the logarithms and perform the multiplication:\newlineT0.25×[3×0.6931+6×0.9163+6×1.0986+3×1.2528]T \approx 0.25 \times [3\times0.6931 + 6\times0.9163 + 6\times1.0986 + 3\times1.2528].
  9. Multiply for Approximation: Calculate the sum:\newlineT0.25×[2.0793+5.4978+6.5916+3.7584]T \approx 0.25 \times [2.0793 + 5.4978 + 6.5916 + 3.7584].
  10. Round to Nearest Thousandth: Add the values together:\newlineT0.25×[17.9271]T \approx 0.25 \times [17.9271].
  11. Round to Nearest Thousandth: Add the values together:\newlineT0.25×[17.9271]T \approx 0.25 \times [17.9271].Multiply by 0.250.25 to get the final approximation:\newlineT0.25×17.92714.4818T \approx 0.25 \times 17.9271 \approx 4.4818.
  12. Round to Nearest Thousandth: Add the values together:\newlineT0.25×[17.9271]T \approx 0.25 \times [17.9271].Multiply by 0.250.25 to get the final approximation:\newlineT0.25×17.92714.4818T \approx 0.25 \times 17.9271 \approx 4.4818.Round to the nearest thousandth:\newlineT4.482T \approx 4.482.