Let f be the function given by f(x)=x2e−x. It is known that ∫10f(x)ⅆx=0.160603. If a midpoint Riemann sum with two intervals of equal length is used to approximate ∫10f(x)ⅆx, what is the absolute difference between the approximation and ∫10f(x)ⅆx ?
To approximate the value of ∫10f(x)ⅆx using a midpoint Riemann sum with two intervals of equal length, we need to divide the interval [1, 0] into two subintervals of equal length
To approximate the value of ∫10f(x)ⅆx using a midpoint Riemann sum with two intervals of equal length, we need to divide the interval [1, 0] into two subintervals of equal length. The width of each subinterval will be Δx = (1 – 0) / 2 = 0.5.
The midpoint Riemann sum formula for approximating the definite integral is as follows:
∑(f(cᵢ)Δx)
where f(cᵢ) is the value of the function evaluated at the midpoint of each subinterval.
In this case, we have two subintervals with midpoints at x = 0.25 and x = 0.75. So, we need to evaluate the function f(x) = x^2e^(-x) at these midpoints and multiply by the width of each subinterval Δx = 0.5. Then, we sum up the two terms to get the approximation of the integral.
Approximation = f(0.25)(0.5) + f(0.75)(0.5)
To evaluate f(0.25), we substitute x = 0.25 into the function:
f(0.25) = (0.25)^2e^(-0.25) ≈ 0.03936
To evaluate f(0.75), we substitute x = 0.75 into the function:
f(0.75) = (0.75)^2e^(-0.75) ≈ 0.34068
Calculating the approximation:
Approximation = (0.03936)(0.5) + (0.34068)(0.5) ≈ 0.18952
The absolute difference between the approximation and the known value of ∫10f(x)ⅆx = 0.160603 is:
Absolute difference = |0.18952 – 0.160603| ≈ 0.02892
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