Let f be the function defined by f(x)=3ln(x). If six subintervals of equal length are used, what is the value of the left Riemann sum approximation for int_(1)^(4)3ln(x)dx ? Round to the nearest thousandth if necessary. Answer:

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Let  f be the function defined by  f(x)=3ln(x). If six subintervals of equal length are used, what is the value of the left Riemann sum approximation for  int_(1)^(4)3ln(x)dx ? Round to the nearest thousandth if necessary. Answer:

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Determine Subinterval Width: To approximate the integral of f(x) = 3ln(x) from 1 to 4 using a left Riemann sum with six subintervals, we first need to determine the width of each subinterval. The interval [1, 4] has a length of 4 - 1 = 3. Dividing this interval into six equal parts gives us a subinterval width of 3/6 = 0.5.Identify Left Endpoints: Next, we identify the x-values at the left endpoints of each subinterval. Starting from x = 1, we increase by the subinterval width of 0.5 until we reach just below the upper limit of 4. The left endpoints are therefore x = 1, 1.5, 2, 2.5, 3, and 3.5.Evaluate Function Values: Now we evaluate the function f(x) = 3ln(x) at each of these left endpoints. This gives us the following values: f(1) = 3ln(1) = 0, f(1.5) = 3ln(1.5), f(2) = 3ln(2), f(2.5) = 3ln(2.5), f(3) = 3ln(3), f(3.5) = 3ln(3.5).Calculate Areas of Rectangles: We then multiply each function value by the width of the subintervals (0.5) to get the area of each rectangle in the Riemann sum approximation. This results in the following areas: A1 = 0.5 * f(1) = 0, A2 = 0.5 * f(1.5), A3 = 0.5 * f(2), A4 = 0.5 * f(2.5), A5 = 0.5 * f(3), A6 = 0.5 * f(3.5).Sum All Areas: We calculate the areas using the function values from the previous step: A2 = 0.5 * 3ln(1.5) = 1.5ln(1.5), A3 = 0.5 * 3ln(2) = 1.5ln(2), A4 = 0.5 * 3ln(2.5) = 1.5ln(2.5), A5 = 0.5 * 3ln(3) = 1.5ln(3), A6 = 0.5 * 3ln(3.5) = 1.5ln(3.5).Calculate Left Riemann Sum: Finally, we sum all the areas to get the left Riemann sum approximation: LRS = A1 + A2 + A3 + A4 + A5 + A6 LRS = 0 + 1.5ln(1.5) + 1.5ln(2) + 1.5ln(2.5) + 1.5ln(3) + 1.5ln(3.5).We perform the calculations to find the numerical value of the left Riemann sum: LRS ≈ 1.5ln(1.5) + 1.5ln(2) + 1.5ln(2.5) + 1.5ln(3) + 1.5ln(3.5) LRS ≈ 1.5(0.4055) + 1.5(0.6931) + 1.5(0.9163) + 1.5(1.0986) + 1.5(1.2528) LRS ≈ 0.6083 + 1.0397 + 1.3745 + 1.6479 + 1.8792 LRS ≈ 6.5496

Let $f$ be the function defined by $f(x)=3 \ln (x)$ . If six subintervals of equal length are used, what is the value of the left Riemann sum approximation for $\int_{1}^{4} 3 \ln (x) d x$ ? Round to the nearest thousandth if necessary. $\newline$ Answer:

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