# To estimate the error in this approximation we need

• Notes
• 37

This preview shows page 13 - 17 out of 37 pages.

To estimate the error in this approximation, we need to estimate the second deriva-tivef′′(x) for allxin the interval [0, π]. First calculate (using the Chain Rule):f(x)=sin2(x)f(x)=2 sin(x) cos(x)=sin(2x)f′′(x)=cos(2x)·2=2 cos(2x)13
Since|cos(2x)| ≤1 for anyx(in particular, anyxin our interval [0, π]), we have|f′′(x)|=|2 cos(2x)|=2|cos(2x)| ≤2for allxin our interval [0, π], we can takeK=2 in the trapezoid error approximationformula|ET| ≤K(ba)312n2=2(π0)312·42=π3962.The Simpson’s rule approximation is given bySn=13n(f(x0)+4f(x1)+2f(x2)+4f(x3)+· · ·+2f(xn2)+4f(xn1)+f(xn))In our case,n=4, and therefore the interval [0,1] is divided into 4 equal subintervals,with sample pointsx0=0,x1=1/4,x2=1/2,x3=3/4, andx4=1. We evaluatef(x)=x5at each of these sample points, and getS4=112(f(0)+4f(1/4)+2f(1/2)+4f(3/4)+f(1))=112((0)5+4(0.25)5+2(0.5)5+4(0.75)5+(1)5)=112(0+.00390625+.0625+.94921875+1)=112(2.015625)0.168To estimate the error in this approximation, we need to estimate the fourth deriva-tivef(4)(x) for allxin the interval [0,1]. It is straightforward to see that the fourthderivative off(x)=x5isf(4)(x)=5·4·3·2x=120x. On the interval [0,1] this functionisincreasing and non-negative, and so achieves its maximum absolute value at the rightendpoint of interval— i.e. atx=1— where it takes the valuef(4)(1)=120·1=120.Therefore we have|f(4)(x)| ≤ |f(4)(1)|=120for allxin our interval [0,1], and we can takeK=120 in the Simpson’s rule errorapproximation formula|ES| ≤K(ba)5180n4=120(10)5180·44=120180·2560.003The actual value of the integral is easy to compute:integraldisplay10x5dx=x66vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglex=1x=0=160=160.16714
Note that this differs by only 0.001 from our approximation; this error is within ourestimate of 0.003.15
Problem 71. Does the following integral converge? If yes, evaluate it:integraldisplay+0etcos2(t)dt2. Does the following integral converge? If yes, evaluate it:integraldisplay+1dtt(t+1)3. Does the following integral converge? If yes, evaluate it:integraldisplay1dxx[lnx]2Solution:1.The integrand is everywhere continuous, so the problem is Type I, and we onlyhave to worry about the convergence at the upper limit.Notice that the integrand is non-negative, and since|cos(t)| ≤1, we have0etcos2(t)etfor allt. Therefore by comparison, sinceintegraltext+0etdtconverges, we know that our inte-gralintegraltext+0etcos2(t)dtalso converges.In order to compute what this integral converges to, it is best to first use the identitycos2(t)=12+12cos(2t)so that we haveintegraldisplay0etcos2(t)=integraldisplay012etdt+integraldisplay012cos(2t)etdtThe first integral on the right hand side is relatively easy to evaluate:12integraldisplay0etdt=12limR→∞integraldisplayR0etdt=12limR→∞parenleftBiget
• • • 