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Let 
f be the function defined by 
f(x)=(3)/(x). If five subintervals of equal length are used, what is the value of the left Riemann sum approximation for 
int_(3)^(4)(3)/(x)dx ? Round to the nearest thousandth if necessary.
Answer:

Let f f be the function defined by f(x)=3x f(x)=\frac{3}{x} . If five subintervals of equal length are used, what is the value of the left Riemann sum approximation for 343xdx \int_{3}^{4} \frac{3}{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:

Full solution

Q. Let f f be the function defined by f(x)=3x f(x)=\frac{3}{x} . If five subintervals of equal length are used, what is the value of the left Riemann sum approximation for 343xdx \int_{3}^{4} \frac{3}{x} d x ? Round to the nearest thousandth if necessary.\newlineAnswer:
  1. Determine Width of Subintervals: Determine the width of each subinterval.\newlineThe interval from 33 to 44 is divided into 55 equal subintervals. To find the width of each subinterval, subtract the lower bound of the interval from the upper bound and divide by the number of subintervals.\newlineWidth of each subinterval = (43)/5=1/5(4 - 3) / 5 = 1/5.
  2. Identify Left Endpoint xx-values: Identify the xx-values for the left endpoints of each subinterval. Since we are using the left Riemann sum, we will use the left endpoints of each subinterval to calculate the sum. The left endpoints are 33, 3.23.2, 3.43.4, 3.63.6, and 3.83.8.
  3. Evaluate Function at Left Endpoints: Evaluate the function f(x)f(x) at each left endpoint.\newlineWe need to calculate the value of f(x)=3xf(x) = \frac{3}{x} at each of the left endpoints.\newlinef(3)=33=1f(3) = \frac{3}{3} = 1\newlinef(3.2)=33.2=0.9375f(3.2) = \frac{3}{3.2} = 0.9375\newlinef(3.4)=33.4=0.8824f(3.4) = \frac{3}{3.4} = 0.8824\newlinef(3.6)=33.6=0.8333f(3.6) = \frac{3}{3.6} = 0.8333\newlinef(3.8)=33.8=0.7895f(3.8) = \frac{3}{3.8} = 0.7895
  4. Multiply Function Values by Width: Multiply each function value by the width of the subintervals.\newlineNow we multiply each of the function values by the width of the subintervals 15\frac{1}{5} to get the area of each rectangle.\newlineArea1=f(3)×(15)=1×(15)=0.2\text{Area}_1 = f(3) \times \left(\frac{1}{5}\right) = 1 \times \left(\frac{1}{5}\right) = 0.2\newlineArea2=f(3.2)×(15)=0.9375×(15)=0.1875\text{Area}_2 = f(3.2) \times \left(\frac{1}{5}\right) = 0.9375 \times \left(\frac{1}{5}\right) = 0.1875\newlineArea3=f(3.4)×(15)=0.8824×(15)=0.1765\text{Area}_3 = f(3.4) \times \left(\frac{1}{5}\right) = 0.8824 \times \left(\frac{1}{5}\right) = 0.1765\newlineArea4=f(3.6)×(15)=0.8333×(15)=0.1667\text{Area}_4 = f(3.6) \times \left(\frac{1}{5}\right) = 0.8333 \times \left(\frac{1}{5}\right) = 0.1667\newlineArea5=f(3.8)×(15)=0.7895×(15)=0.1579\text{Area}_5 = f(3.8) \times \left(\frac{1}{5}\right) = 0.7895 \times \left(\frac{1}{5}\right) = 0.1579
  5. Add Up Rectangle Areas: Add up the areas of all rectangles to get the left Riemann sum. The left Riemann sum is the sum of the areas of the rectangles we calculated in the previous step. Left Riemann sum =Area1+Area2+Area3+Area4+Area5= \text{Area}_1 + \text{Area}_2 + \text{Area}_3 + \text{Area}_4 + \text{Area}_5 Left Riemann sum =0.2+0.1875+0.1765+0.1667+0.1579= 0.2 + 0.1875 + 0.1765 + 0.1667 + 0.1579 Left Riemann sum =0.8886= 0.8886

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