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 ?
0.003
The formula for a midpoint Riemann sum with n intervals of equal length Δx is:
Δx[f(x1/2) + f(x3/2) + … + f(x(2n-1)/2)],
where xi = iΔx and Δx = (b-a)/n.
In this case, we have a = 0, b = 1, and n = 2. Therefore, Δx = 1/2 and xi = i/2 for i = 1, 2, 3, 4.
The Riemann sum is then:
(1/2)[f(1/4) + f(3/4) + f(5/4) + f(7/4)]
= (1/2)[(1/16)e^(-1/4) + (9/16)e^(-3/4) + (25/16)e^(-5/4) + (49/16)e^(-7/4)]
≈ 0.1518
To find the absolute difference between this approximation and the value of the integral, we subtract the integral from the approximation and take the absolute value:
|0.1518 – 0.160603| ≈ 0.008803
Therefore, the absolute difference between the midpoint Riemann sum approximation and the value of the integral is approximately 0.008803.
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