Let f be the function defined by f(x)=x^(3). If six subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for int_(0)^(3)x^(3)dx ? Round to the nearest thousandth if necessary. Answer:

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Let  f be the function defined by  f(x)=x^(3). If six subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for  int_(0)^(3)x^(3)dx ? Round to the nearest thousandth if necessary. Answer:

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Determine Width of Subintervals: Determine the width of each subinterval. Since we are integrating from 0 to 3 and using six subintervals, the width (Δx) of each subinterval is (3 - 0) / 6 = 0.5.Calculate Midpoints: Calculate the midpoints of each subinterval. The midpoints for the six subintervals are found by adding half of the width of a subinterval to the lower bound of each subinterval. The midpoints (m_i) are: 0.25, 0.75, 1.25, 1.75, 2.25, and 2.75.Evaluate Function at Midpoints: Evaluate the function f(x) = x^3 at each midpoint. We calculate f(m_i) for each midpoint: f(0.25) = 0.25^3 = 0.015625 f(0.75) = 0.75^3 = 0.421875 f(1.25) = 1.25^3 = 1.953125 f(1.75) = 1.75^3 = 5.359375 f(2.25) = 2.25^3 = 11.390625 f(2.75) = 2.75^3 = 20.796875Multiply by Subinterval Width: Multiply each function value by the width of the subinterval. We multiply each f(m_i) by Δx to get the area of the rectangle for each subinterval: A_1 = f(0.25) * Δx = 0.015625 * 0.5 = 0.0078125 A_2 = f(0.75) * Δx = 0.421875 * 0.5 = 0.2109375 A_3 = f(1.25) * Δx = 1.953125 * 0.5 = 0.9765625 A_4 = f(1.75) * Δx = 5.359375 * 0.5 = 2.6796875 A_5 = f(2.25) * Δx = 11.390625 * 0.5 = 5.6953125 A_6 = f(2.75) * Δx = 20.796875 * 0.5 = 10.3984375Sum Rectangle Areas: Sum the areas of all rectangles to find the midpoint Riemann sum approximation. The midpoint Riemann sum approximation is the sum of all the areas: R = A_1 + A_2 + A_3 + A_4 + A_5 + A_6 R = 0.0078125 + 0.2109375 + 0.9765625 + 2.6796875 + 5.6953125 + 10.3984375 R = 19.96875Round Result: Round the result to the nearest thousandth if necessary. R = 19.96875, which is already to the nearest thousandth.

Let $f$ be the function defined by $f(x)=x^{3}$ . If six subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for $\int_{0}^{3} x^{3} d x$ ? Round to the nearest thousandth if necessary. $\newline$ Answer:

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Let $f$ be the function defined by $f(x)=x^{3}$ . If six subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for $\int_{0}^{3} x^{3} d x$ ? Round to the nearest thousandth if necessary. $\newline$ Answer: