Let f be the function defined by f(x)=x^(4). If five subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for int_(1)^(2)x^(4)dx ? Round to the nearest thousandth if necessary. Answer:

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Let  f be the function defined by  f(x)=x^(4). If five subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for  int_(1)^(2)x^(4)dx ? Round to the nearest thousandth if necessary. Answer:

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Calculate Width of Subintervals: To calculate the midpoint Riemann sum, we first need to determine the width of each subinterval. The interval [1, 2] has a length of 2 - 1 = 1. Since we are using five subintervals, the width (Δx) of each subinterval is 1/5.Find Midpoints of Subintervals: Next, we need to find the midpoints of each subinterval. The midpoints will be used to evaluate the function f(x) = x^4. The midpoints are found by adding half of the width of a subinterval to the lower endpoint of each subinterval.Evaluate Function at Midpoints: The midpoints (m_i) for the five subintervals are as follows: m_1 = 1 + (1/5)/2 = 1.1 m_2 = 1.2 + (1/5)/2 = 1.3 m_3 = 1.4 + (1/5)/2 = 1.5 m_4 = 1.6 + (1/5)/2 = 1.7 m_5 = 1.8 + (1/5)/2 = 1.9Perform Function Calculations: Now we evaluate the function f(x) = x^4 at each midpoint: f(m_1) = (1.1)^4 f(m_2) = (1.3)^4 f(m_3) = (1.5)^4 f(m_4) = (1.7)^4 f(m_5) = (1.9)^4Calculate Riemann Sum: Perform the calculations for each f(m_i): f(m_1) = (1.1)^4 = 1.4641 f(m_2) = (1.3)^4 = 2.8561 f(m_3) = (1.5)^4 = 5.0625 f(m_4) = (1.7)^4 = 8.3521 f(m_5) = (1.9)^4 = 13.0321Substitute Values into Formula: The midpoint Riemann sum approximation is the sum of the function values at the midpoints multiplied by the width of the subintervals (Δx): Riemann sum = Δx * (f(m_1) + f(m_2) + f(m_3) + f(m_4) + f(m_5))Calculate Riemann Sum: Substitute the values we have calculated into the Riemann sum formula: Riemann sum = (1/5) * (1.4641 + 2.8561 + 5.0625 + 8.3521 + 13.0321)Round Riemann Sum: Calculate the Riemann sum: Riemann sum = (1/5) * (30.7669) Riemann sum = 6.15338Round the Riemann sum to the nearest thousandth: Riemann sum ≈ 6.153

Let $f$ be the function defined by $f(x)=x^{4}$ . If five subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for $\int_{1}^{2} x^{4} d x$ ? Round to the nearest thousandth if necessary. $\newline$ Answer:

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Let $f$ be the function defined by $f(x)=x^{4}$ . If five subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for $\int_{1}^{2} x^{4} d x$ ? Round to the nearest thousandth if necessary. $\newline$ Answer: