MATH 106A - CALCULUS II
FALL 2005
QUIZ 2
NAME:
Show ALL your work CAREFULLY.
(a) Recall that the error committed by using the Left Hand Sum Lnis less than
or equal to K1·(b−a)2
2nwhere |f(x)|≤K1for some constant K1. Use this result to
give an upper bound for
2
1
exsin(2x2)dx −L10
.
Here, f(x)=exsin(2x2).
First, we must find an upper bound K1for |f(x)|over the interval [1,2].
Since f(x)=exsin(2x2),wehavef(x)=exsin(2x2)+excos(2x2)·4xby th e
product rule and the chain rule. Thus, |f(x)|=ex|sin(2x2)+cos(2x2)·4x|.
Over the interval [1,2], the largest value for exis e2. Neither sin(2x2)nor
cos(2x2)can exceed 1in absolute value. It follows that |f(x)|≤e2(1+1·8) =
9e2. Using the error bound for the Left Hand Sum with K1=9e2,we
conclude that
2
1
exsin(2x2)dx −L10
≤9e2·(2 −1)2
2·10 =9
20e2.
(b) Consider the following given data of a function h(x) on the interval [0,8].
x0 2 4 6 8
h(x) 2 1 -1 4 7
Find T4(trapezoid) AND M2(mid-point). Here the subscript nindicates that
the interval [0,8] 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(0) ·∆x+h(2) ·∆x+h(4) ·∆x+h(6) ·∆x
=(2+1+(−1) + 4) ·∆x= 12;
R4=h(2) ·∆x+h(4) ·∆x+h(6) ·∆x+h(8) ·∆x
=(1+(−1)+4+7)·∆x=22.
Date: September 19, 2005.
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