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Let f be the function defined by f(x)=(6)/(x). If three subintervals of equal length are used, what is the value of the trapezoidal sum approximation for int_(1)^(8.5)(6)/(x)dx ? Round to the nearest thousandth if necessary. Answer:

Let f f be the function defined by f(x)=6x f(x)=\frac{6}{x} . If three subintervals of equal length are used, what is the value of the trapezoidal sum approximation for ∫18.56xdx \int_{1}^{8.5} \frac{6}{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:

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Q. Let f f be the function defined by f(x)=6x f(x)=\frac{6}{x} . If three subintervals of equal length are used, what is the value of the trapezoidal sum approximation for ∫18.56xdx \int_{1}^{8.5} \frac{6}{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Determine subinterval width: To use the trapezoidal rule, we first need to determine the width of each subinterval. The interval from 11 to 8.58.5 has a length of 8.5−1=7.58.5 - 1 = 7.5. Since we are using three subintervals of equal length, each subinterval will have a width of 7.5/3=2.57.5 / 3 = 2.5.
  2. Calculate x-values: Next, we need to calculate the x-values at the endpoints of each subinterval. The first subinterval starts at x=1x = 1 and ends at x=1+2.5=3.5x = 1 + 2.5 = 3.5. The second subinterval ends at x=3.5+2.5=6x = 3.5 + 2.5 = 6. The third subinterval ends at x=6+2.5=8.5x = 6 + 2.5 = 8.5. So, the x-values we will use are 11, 3.53.5, 66, and 8.58.5.
  3. Evaluate function: Now we need to evaluate the function f(x)=6xf(x) = \frac{6}{x} at each of these xx-values. This gives us f(1)=61=6f(1) = \frac{6}{1} = 6, f(3.5)=63.5≈1.714f(3.5) = \frac{6}{3.5} \approx 1.714, f(6)=66=1f(6) = \frac{6}{6} = 1, and f(8.5)=68.5≈0.706f(8.5) = \frac{6}{8.5} \approx 0.706.
  4. Calculate trapezoidal sum: The trapezoidal sum is given by the formula (width2)×(f(x0)+2⋅f(x1)+2⋅f(x2)+…+f(xn))(\frac{\text{width}}{2}) \times (f(x_0) + 2\cdot f(x_1) + 2\cdot f(x_2) + \ldots + f(x_n)), where x0x_0 and xnx_n are the endpoints of the interval and x1x_1 through xn−1x_{n-1} are the points in between. In our case, this becomes (2.52)×(f(1)+2⋅f(3.5)+2⋅f(6)+f(8.5))(\frac{2.5}{2}) \times (f(1) + 2\cdot f(3.5) + 2\cdot f(6) + f(8.5)).
  5. Calculate final value: Plugging in the values we found, the trapezoidal sum is (2.5/2)×(6+2×1.714+2×1+0.706)=(1.25)×(6+3.428+2+0.706)=(1.25)×(12.134)≈15.1675(2.5/2) \times (6 + 2\times1.714 + 2\times1 + 0.706) = (1.25) \times (6 + 3.428 + 2 + 0.706) = (1.25) \times (12.134) \approx 15.1675.
  6. Round to nearest thousandth: Rounding to the nearest thousandth, the value of the trapezoidal sum approximation for the integral is approximately 15.16815.168.

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