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Let 
f be the function defined by 
f(x)=(2)/(x). If five subintervals of equal length are used, what is the value of the right Riemann sum approximation for 
int_(1)^(6)(2)/(x)dx ? Round to the nearest thousandth if necessary.
Answer:

Let f f be the function defined by f(x)=2x f(x)=\frac{2}{x} . If five subintervals of equal length are used, what is the value of the right Riemann sum approximation for 162xdx \int_{1}^{6} \frac{2}{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:

Full solution

Q. Let f f be the function defined by f(x)=2x f(x)=\frac{2}{x} . If five subintervals of equal length are used, what is the value of the right Riemann sum approximation for 162xdx \int_{1}^{6} \frac{2}{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Determine Subinterval Width: We are given the function f(x)=2xf(x) = \frac{2}{x} and we need to approximate the integral from x=1x = 1 to x=6x = 6 using a right Riemann sum with five subintervals of equal length. The first step is to determine the width of each subinterval. The total interval length is 61=56 - 1 = 5. Since we are using five subintervals, the width of each subinterval (Δx\Delta x) is 55=1\frac{5}{5} = 1.
  2. Find Evaluation Points: Next, we need to find the xx-values at which we will evaluate the function for the right Riemann sum. Since we are using the right endpoints and the first subinterval starts at x=1x = 1, the xx-values will be 22, 33, 44, 55, and 66.
  3. Evaluate Function: Now we evaluate the function f(x)=2xf(x) = \frac{2}{x} at each of these x-values:\newlinef(2)=22=1f(2) = \frac{2}{2} = 1\newlinef(3)=230.667f(3) = \frac{2}{3} \approx 0.667\newlinef(4)=24=0.5f(4) = \frac{2}{4} = 0.5\newlinef(5)=25=0.4f(5) = \frac{2}{5} = 0.4\newlinef(6)=260.333f(6) = \frac{2}{6} \approx 0.333
  4. Calculate Rectangle Areas: We then multiply each function value by the width of the subintervals (Δx=1\Delta x = 1) to get the area of each rectangle in the Riemann sum:\newlineArea at x=2x=2: 1×Δx=1×1=11 \times \Delta x = 1 \times 1 = 1\newlineArea at x=3x=3: 0.667×Δx=0.667×10.6670.667 \times \Delta x = 0.667 \times 1 \approx 0.667\newlineArea at x=4x=4: 0.5×Δx=0.5×1=0.50.5 \times \Delta x = 0.5 \times 1 = 0.5\newlineArea at x=5x=5: 0.4×Δx=0.4×1=0.40.4 \times \Delta x = 0.4 \times 1 = 0.4\newlineArea at x=6x=6: $\(0\).\(333\) \times \Delta x = \(0\).\(333\) \times \(1\) \approx \(0\).\(333\)
  5. Sum Rectangle Areas: Finally, we sum the areas of all the rectangles to get the right Riemann sum approximation for the integral:\(\newline\)Right Riemann sum = \(1 + 0.667 + 0.5 + 0.4 + 0.333 \approx 2.9\)
  6. Round Final Result: We round the result to the nearest thousandth as instructed: Right Riemann sum \(\approx 2.900\)

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