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
To use a midpoint Riemann sum with two intervals of equal length, we need to first divide the interval [1, 0] into two subintervals of equal length. This gives us the subintervals [0, 0.5] and [0.5, 1].
The midpoint of the first subinterval is (0 + 0.5)/2 = 0.25, and the midpoint of the second subinterval is (0.5 + 1)/2 = 0.75.
Using these midpoints, the midpoint Riemann sum approximation for ∫10f(x)ⅆx is:
f(0.25)*(0.5) + f(0.75)*(0.5)
= (0.252*e^(-0.25))*(0.5) + (0.752*e^(-0.75))*(0.5)
= 0.064208
The absolute difference between this approximation and the actual value of ∫10f(x)ⅆx can be found by taking the absolute value of the difference between the two:
|0.160603 – 0.064208| = 0.096395
Therefore, the absolute difference between the midpoint Riemann sum approximation and the actual value of ∫10f(x)ⅆx is 0.096395.
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