MATH 106A - CALCULUS II
WINTER 2007
QUIZ 2
NAME:
Show ALL your work CAREFULLY.
(a) Recall that the error committed by using the Right Hand Sum Rnis less than
or equal to K1·(b−a)2
2nwhere |f0(x)|≤K1for some constant K1. Use this result to
give an upper bound for
Z3
1
x2sin(x2)dx −R8
.
Here, f(x)=x2sin(x2).
First, we must find an upp er bound K1for |f0(x)|over the interval
[1,3]. Since f(x)=x2sin(x2), we have f0(x)=2xsin(x2)+x2cos(x2)·2xby
the product rule and the chain rule. Thus, |f0(x)|=2|x|| sin(x2)+x2cos(x2)|.
Over the interval [1,3], the largest value for xis 3and that for x2is 9.
Neither sin(x2)nor cos(x2)can exceed 1in absolute value. It follows that
|f0(x)|≤6(1 + 9 ·1) = 60. Using the error bound for the Right Hand Sum
with K1=60, we conclude that
Z3
1
x2sin(x2)dx −R8
≤60 ·(3 −1)2
2·8=15.
(b) Consider the following given data of a function h(x) on the interval [1,9].
x1 3 5 7 9
h(x) -1 2 -1 6 3
Find M2(mid-point) AND T4(trapezoid). Here the subscript nindicates that
the interval [1,9] is to be divided into nequal subintervals.
First, we compute the Left and the Right Hand Sums L4and R4.
Using the given data, we have
L4=h(1) ·∆x+h(3) ·∆x+h(5) ·∆x+h(7) ·∆x
=(−1+2+(−1) + 6) ·∆x= 12;
R4=h(3) ·∆x+h(5) ·∆x+h(7) ·∆x+h(9) ·∆x
=(2+(−1) + 6 + 3) ·∆x=20.
It follows that T4=L4+R4
2=16.ForM2, note that ∆x=4since we only
have two subintervals. Thus,
M2=h(3) ·∆x+h(7) ·∆x
= (2 + 6) ·∆x=32.
Date: January 22, 2007.
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