Q #2!
Math 106-C (Salomone)
January 23, 2009
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Score (25 points possible):
Problem 1. (10 points) Approximating !1
0
sin x
2dx in four different ways gives the following sums:
L10 =R10 =T10 =M10 =
0.2208 0.2688 0.2448 0.2449
(a) (5 points) What is the greatest amount of error that L10 can commit for this integral? Answer to four significant
figures, and explain your work.
We can use the error bound
"""I−LN"""≤K(b−a)2
2N
where Kis an upper bound for the derivative f#on [0,1].
But f#(x)=1
2cos x
2and since cosine is never larger than 1 in absolute value, we have −1/2≤f#(x)≤1/2.
Thus taking K=1/2 gives us the bound
"""I−LN"""≤K(b−a)2
2N=
1
2(1 −0)2
2(10) =0.025.
This is the greatest amount of error L10 can commit; it optionally gives us the ”trap”
L10 −0.025 =0.1958 ≤I≤0.2458=L10 +0.025
(b) (5 points) What is the greatest amount of error that M10 can commit for this integral? Answer to four significant
figures, and explain your work.
For the midpoint rule we have a new error bound:
"""I−MN"""≤C(b−a)3
24N2
where Cis an upper bound for the second derivative f## on [0,1].
But f##(x)=−1
4sin x
2and since sine is never larger than 1 in absolute value, we have −1
4≤f##(x)≤1
4. Thus
taking C=1
4gives the bound
"""I−MN"""≤C(b−a)3
24N2=
1
4(1 −0)3
24(10)2≈0.0001042.
This is the greatest amount of error M10 can commit; it optionally gives us the ”trap”
M10 −0.0001042 =0.2448 ≤I≤0.2450=M10 +0.0001042
