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Understanding Calculus: Riemann Sums and Integrals Explained

School
Imperial College**We aren't endorsed by this school
Course
MATH 50002
Subject
Mathematics
Date
Dec 12, 2024
Pages
9
Uploaded by PrivateFreedomAlpaca56
Mathematics Year 1, Calculus and Applications ID.T. PapageorgiouProblem Sheet 41.(a) LetS2:=∑ni=1i2. ComputenXi=1[(i+ 1)3-i3] =nXi=1[i3+ 3i2+ 3i+ 1-i3] = 3nX1i2+ 3nX1i+nX11= 3S2+32n(n+ 1) +n.The first sum in the equation above is a telescoping series and is equal to 23-13+ 33-23+. . .+ (n+ 1)3-n3= (n+ 1)3-1. Hence, after a little algebra,S2=16n(n+ 1)(2n+ 1).(b) Partition [0,1] intoxi=ihwhereh= 1/n(note,x0= 0 andxn= 1).Theupper Riemann sum isnX1h(ih)2=h3nX1i3=1n3.16n(n+ 1)(2n+ 1)→13asn→ ∞.2. Need to show limn→∞∑ni=11nei/n=e-1. Letr=e1/n. ThennXi=11nei/n=1nnXi=1ri=1nr1-rn1-r=(1/n)e1/n(e1/n-1)(e-1)→(e-1)asn→ ∞.[E.g. use L’Hˆopital’s rule on limx→0xex-1.]3. Take any partition of [0,1] and calculate the lower Riemann sumL, and upper Rie-mann sumU. Since any interval of real numbers, however small, contains an infinitenumber of rationals and irrationals, it follows thatL= 0,U= 1.This is the case foranypartition, henceL6=Ufor any partition and so the functionis not Riemann integrable.4. First part straightforward. Using this result we see thatddxlog(secx+ tanx) = secx,hence we have the antiderivative of secxas required.The integral result can be applied on intervals that satisfy cosx6= 0 and secx+tanx >0, i.e.1+sinxcosx>0 which is only possible if cosx >0. Hence the appropriate intervalis (-π/2, π/2) since cos(±π/2) = 0.5.(i) Substitutex= tanθ, i.e.dx= sec2θdθandx2+ 1 =tan2θ+ 1 = sec2θ:Z1(x2+ 1)3dx=Zsec2θsec6θdθ=Zcos4θdθ=Z(1 + cos 2θ)24dθ=Z1 + 2 cos 2θ4+1 + cos 4θ8dθ=3θ8+sin 2θ4+sin 4θ32+K.1
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(ii) Use partial fractions1x3-1=1(x-1)(x2+x+1)≡Ax-1+Bx+Cx2+x+1.Solving we findA= 1/3,B=-1/3,C=-2/3, and the integral is13Z1x-1-x+ 2x2+x+ 1dx=13Z1x-1-12(2x+ 1)(x2+x+ 1)-(3/2)(x+ 1/2)2+ (3/4)dx,where I have completed the square in the last term. Now we can integrate tofind13log(x-1)-12log(x2+x+ 1)-(3/2)1√3/2tan-1x+ 1/2√3/2+K.(iii) Writex3+1x3-1= 1 +2x3-1, and the integral of the second piece has just been done.(iv) Use integration by parts. WriteZx3px2+ 1dx=Zx2ddx13(x2+ 1)3/2dx=x213(x2+ 1)3/2-Z2x13(x2+ 1)3/2dx=13x2(x2+ 1)3/2-215(x2+ 1)5/2+K.(v) Use the trigonometric substitutionu= sinxso thatdu= cosxdx, and theintegral becomesZπ/2π/6cosxsinx+ sin3xdx=Zduu(1 +u2)=Z1u-u1 +u2du= log(u)-12log(1+u2) ="logsinxp1 + sin2x#π/2π/6= log(1/√2)-log(1/√5) =12log(5/2)6. Ifn= 1,I1= tan-1x. Start withIn-1In-1=Zdx(x2+ 1)n-1=x(x2+ 1)n-1+ 2(n-1)Zx2(x2+ 1)ndx=x(x2+ 1)n-1+ 2(n-1)Zx2+ 1-1(x2+ 1)ndx=x(x2+ 1)n-1+ 2(n-1)In-1-2(n-1)In,therefore, as required,2(n-1)In=x(x2+ 1)n-1+ (2n-3)In-1.7. We will need the double angle formulassinAcosB=12[sin(A+B) + sin(A-B)],sinAsinB=12[cos(A-B)-cos(A+B)],cosAcosB=12[cos(A-B) + cos(A+B)].2
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(a) For any two integersm,nZπ-πsinmxcosnxdx=Zπ-π12[sin(m+n)x+ sin(m-n)x]dx= 0.Zπ-πsinmxsinnxdx=Zπ-π12[cos(m-n)x-cos(m+n)x] =0ifm6=nπifn=m,Zπ-πcosmxcosnxdx=Zπ-π12[cos(m-n)x+ cos(m+n)x] =0ifm6=nπifn=m.(b) We assumef(x) =a0+NXk=1akcoskx+bksinkx,(*)and need to calculate the a’s and b’s. The constanta0can be found immediatelyby integrating over [-π, π],Zπ-πf(x)dx= 2πa0+NXk=1Zπ-π(abcoskx+bksinkx)dx= 2πa0⇒a0=12πZπ-πf(x)dx,i.e. it is the average (or mean) value of the function over the domain.Next, take any integerm≥1, multiply (*) by cosmxand integrate between-πandπ. Using theorthogonalityresults from part (a) we see that only the termcontainingamin the sum will survive to giveZπ-πf(x) cosmxdx=amZπ-πcos2mxdx=πam⇒am=1πZπ-πf(x) cosmxdx.Now do the same calculation but multiply by sinmxand integrate to findbm=1πZπ-πf(x) sinmxdx.(c) Forf(x) defined byf(x) =1if|x| ≤π/20otherwise.we have by direct application of the formulas in part (b)ak=1πZπ/2-π/2coskxdx=1πsinkxkπ/2-π/2=2kπsinkπ2,bk=1πZπ/2-π/2sinkxdx=1π-coskxkπ/2-π/2= 0.The last result could have been anticipated because the givenf(x) is an evenfunction ofxon [-π/2, π/2] and sinkxis odd there, hence the productf(x) sinkxis also odd, implying that the integral must be zero.3
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8.(i) WriteR∞0e-x2dx=R10e-x2dx+R∞1e-x2dx. The first integral is a finite number.For the second integral we haveZ∞1e-x2dx <Z∞1xe-x2dx=12e,hence the integral is bounded.(ii) Again, given anyM >0 writeR∞0x3(1+x2)2dx=RM0x3(1+x2)2dx+R∞Mx3(1+x2)2dx;the first integral is a finite number and for the second integral, takingM >1we haveZ∞Mx3(1 +x2)2dx >Z∞Mx3(x2+x2)2dx=14Z∞M1xdx.The last integral diverges so by comparison our original integral also diverges.(iii) ForM >0 writeR∞01√x+x3dx=RM01√x+x3dx+R∞M1√x+x3dx, and considerseparately lima→0+RMa1√x+x3dxand limb→∞RbM1√x+x3dx. We havelima→0+ZMa1√x+x3dx=lima→0+ZMa1√x√1 +x2dx≤lima→0+ZMa1√xdx=lima→0+2(√M-√a) = 2√M.Similarly,Z∞Mdx√x+x3<Z∞Mdxx3/2= limb→∞ZbMdxx3/2= limb→∞2(M-1/2-b-1/2) = 2/√M,i.e. also bounded.(iv)Z10sin2x1 +x2dx <Z1011 +x2dx=π/4.(v) Useful to compare with a function we know how to integrate.Letf(x) =x-log(1 +x). We havef(0) = 0 andf0(x) = 1-11+x=x1+x≥0 forx≥0.Hencef(x) is increasing and as a resultx >log(1 +x)forx >0.Hence, given 0< ε <1 we haveZ1εdxlog(1 +x)>Z1εdxx= log(1/ε),and sendingε→0+ proves that the integral is divergent.(vi) This is an improper integral whose integrand does not decay for largex. In factinfinitely more rapid oscillations take place and we can intuitively expect thatthere is cancellation to give a finite result. [In fact this and its cosine sister areknown as Fresnel integrals and come up generically in wave propagation, opticsetc.]4
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Again we split the integralR∞0sin(x2)dx= (R10+R∞1) sin(x2)dx. The first in-tegral is perfectly fine and is equal to a finite number (we cannot find this inclosed form). For the second integral we integrate by partsZb1sin(x2)dx=Zb1xsin(x2)xdx=-12xcos(x2)b1-12Zb1cos(x2)x2dx≤12cos 1-12bcosb2+12Zb1dxx2dx=12cos 1-12bcosb2+12(1-1b),and asbbecomes large we see that the integral is bounded above by 1.Combining this with the integral over [0,1] proves that the Fresnel integralconverges.9. To prove thatR10x32-sin4xdx≤14log 2 we use the inequality sinx≤xin the interval[0,1] of interest. [In fact we have|x| ≤ |sinx|for allx∈R- you have already provedthis elsewhere by showing that the functionf(x) =x-sinxis increasing.]Thisimplies that 2-sin4x≥2-x4, henceZ10x32-sin4xdx≤Z10x32-x4dx=-14log(2-x4)10=14log(2).For the second integral, since we are on the interval [0, π/2] we have cosx >0, henceZπ/20x-π/22 + cosxdx≤Zπ/20|x-π/2|2dx=12Zπ/20(π2-x)dx=π216.10. Proof of theintegral mean value theorem:Letfandgbe continuous on[a, b]withg(x)≥0forx∈[a, b]. Then there exists a numbercbetweenaandbwithZbaf(x)g(x)dx=f(c)Zbag(x)dx.Proof: Sincefis continuous on [a, b] it must have a maximumMand a minimummon [a, b], i.e.m≤f(x)≤M. Sinceg(x)≥0, we have for allx∈[a, b]mg(x)≤f(x)g(x)≤Mg(x),and by the properties of integrals we have in turnmZbag(x)dx≤Zbaf(x)g(x)dx≤MZbag(x)dx.IfRbag(x)dx= 0 then we also haveRbaf(x)g(x)dx= 0 and the result follows.IfRbag(x)dx6= 0, we havem≤Rbaf(x)g(x)dxRbag(x)dx≤M,5
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and by the Intermediate Value Theorem there must be a numbercbetweenaandbso thatf(c) =Rbaf(x)g(x)dxRbag(x)dx,and the result of the theorem follows.An example that violates the conclusion of the theorem if we dropg(x)≥0 is,f(x) =sinx,g(x) = sinx,a=-π/2,b=π/2. ThenRπ/2-π/2fgdx6= 0 butf(c)Rπ/2-π/2sinxdx=0 for allc∈[-π/2, π/2].11. These are fairly straightforward. I will do a couple of them.x2on [0,1]. Here we haveμ=R10x2dx= 1/3 andσ2=R10(x2-1/3)2dx=445.f(x) =1on [0,1]2on (1,2].Hereμ=12R101dx+R212dx=32. Next, (f(x)-μ)2isequal to (1-3/2)2= 1/4 on [0,1] and (2-3/2)2= 1/4 on (1,2].Henceσ2=12R20(1/4)dx= 1/4.12.(a)μ=1b-aZbaf(x)dx=1b-anXi=1Zxixi-1kidx=1b-anXi=1ki(xi-xi-1)Henceσ2=1b-aZba(f(x)-μ)2dx=1b-anXi=1Zxixi-1(f(x)-μ)2dx=1b-anXi=1(ki-μ)2(xi-xi-1).(b) If the partition consists of equally spaced points, we havexi-xi-1=b-an, hencethe formulas areμ=1nnXi=1ki,σ2=1nnXi=1(ki-μ)2.(c) The standard deviation of a step function is a sum of non-negative terms (inboth cases of uniform and non-uniform partitions). The only way this can bezero is if the function is a constant (and hence equal to its average).(d) For a list of numbersa1, a2, . . . , anthe mean isμ=1n∑ni=1aiandσ2=1n∑ni=1(ai-μ)2, so completely analogous to the uniform partition result forfunctions.(e) If its standard deviation is zero, then the numbers are equal.13.(a) Calculate using definition off(x):Zbag(x)δn(x)dx=Z1/2n-1/2ng(x)ndx=Z1/2-1/2g(y/n)dy→g(0)asn→ ∞,sincegis a continuous function.6
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(b) Using the result above, we havelimn→∞Ztag(x)δn(x)dx=g(0)t >00t <0.(c) Here takeg(x) = 1 and use the results above to findlimn→∞Zt2t1δn(x)dx=0t1< t2<000< t1< t21t1<0< t2.(d) We have shown in (c) that the anti-derivative ofδ(x) is the functionH(x) (calledthe Heaviside function) defined byH(x) =0x <01x >0.14. Begin by constructingf(x): For 0≤x <1 we havef(x) = 1, and for 1≤x <2,f(x) = 2.Hence, for 0≤x≤1 we haveF(x) =Rx0dx=x, and for 1≤x≤2,F(x) =R10dx+Rx12dx= 2x-1:F(x) =xfor 0≤x≤12x-1for 1≤x≤2.The functionF(x) is continuous butF0(1) does not exist. This does not contradict thefundamental theorem of calculus sincef(x) is not continuous. Recall the statementof the FTC:Fundamental Theorem of CalculusIffis Riemann integrable on[a, b]andF(x) =Rxaf(t)dt, thenF(x)is contin-uous on[a, b]. If in additionfis continuous on[a, b], thenFis differentiableon[a, b]andF0=f.15. Can evaluate the integral using integration by parts:Zn1logxdx= [xlogx]n1-Zn1x.1xdx=nlogn-(n-1).Now partition [1, n] in unit intervals as instructed to find the upper Riemann sumUand lower Riemann sumL:U= log 2 + log 3 +. . .+ logn= log[n!],L= log(1) + log(2) +. . .+ log(n-1) = log[(n-1)!],therefore we have the inequalitylog[(n-1)!]≤nlogn-(n-1)≤log[n!],7
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and usingnlogn-(n-1) = log(nn) + log(e-(n-1))= log (nne-ne), along with thefact that log is monotonic increasing we obtain the desired result(n-1)!≤nne-ne≤n!Using this we can boundn!/nnas followse-ne≤n!nn≤ne-ne⇒e-1e1/n≤n!nn1/n≤n1/ne-1e1/n.Asngets large we know thate-1/n→1 andn1/n→1 (why?), hence by the squeezingtheorem the result limn→∞(n!nn)1/n= 1/efollows.16.(a) This question involves integration by parts. CalculateI0=Z∞0e-xdx=-e-x∞0= 1J0=Z∞0e-xcosx dx=e-xsinx∞0+Z∞0e-xsinx dx=-e-xcosx∞0-J0HenceJ0= 1/2.[Alternatively use complex variables and writee-xcosx=R[e(i-1)x] so thatJ0=Rhe(i-1)x(i-1)i∞0=R(1+i2)= 1/2.]In=-e-x(sinx)n∞0+Z∞0ne-x(sinx)n-1cosx dx=nJn-1.For any non-negative integern, letIn=Z∞0e-x(sinx)ndx,Jn=Z∞0e-x(sinx)ncosxdx.To get the last expression we writee-x=d(-e-x) and integrate that first by parts,i.e.Jn=-e-x(sinx)ncosx∞0-Z∞0(-e-x)-(sinx)n+1+n(sinx)n-1cos2xdx=Z∞0e-xn(sinx)n-1-(1 +n)(sinx)n+1dx=nIn-1-(n+ 1)In+1⇒Jn=nIn-1-(n+ 1)In+1(*1)as required.Note:Jncan also be integrated by parts by noting that (sinx)ncosx=d(sinx)n+1n+1and integrating this first, to immediately obtain the alternative expressionJn=e-x(sinx)n+1n+ 1∞0+Z∞0e-x(sinx)n+1n+ 1dx⇒Jn=1n+ 1In+1.(*2)As we will see, this is of course consistent with the recursion formula forInwe findnext.8
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(b) Usen= 1 in the formulaIn=nJn-1to findI1=J0= 1/2.From (*)J1=I0-2I2=I0-2(2J1), hence 5J1= 1 as needed.We haveIn=nJn-1. Evaluate (*1) atn-1 to findJn-1= (n-1)In-2-nInandsubstitute into the expression forInto getIn=n(n-1)In-2-n2In⇒In=n(n-1)(1 +n2)In-2.(*3)Note:This result should of course be consistent with equating (*1) = (*2) whichimpliesIn+1=n(n+1)1+(n+1)2In-1which is identical to (*3) once we shift the indexn+1→n.To find the recursion forJ, we now eliminateIn-1andIn+1in (*1) to findJn=n(n-1)Jn-2-(n+ 1)2Jn⇒Jn=n(n-1)1 + (n+ 1)2Jn-2.(*4)(c) We need to calculate explicit expressions from (*3) and (*4) forn≥2. In both caseswe have starting valuesI0= 1,I1= 1/2 andJ0= 1/2,J1= 1/5, and all other valueswill be in terms of these. Inspection of (*3) and (*4) shows that all even indices willinvolveI0orJ0and odd indices will involveI1andJ1. Lets calculate a few termsand you will get the general formula.I2=2·1(1+22)I0I3=3·2(1+32)I1I4=4·3(1+42)I2=4!(1+22)(1+42)I0,I5=5·4(1+52)I3=5!(1+32)(1+52)I1. . .. . .I2k= (2k)!Qkp=111+(2p)2,I2k+1=12(2k+ 1)!Qkp=111+(2p+1)2Similarly for theJs we haveJ2=2·1(1+32)J0J3=3·2(1+42)J1J4=4!(1+32)(1+52)J0J5=5!(1+42)(1+62)J1. . .. . .J2k=12(2k)!Qkp=11(1+(2p+1)2J2k+1=15(2k+ 1)!Qkp=11(1+(2p+2)2We can compare directlyI2kwithJ2kfound above; the products inJ2kare smallersince1(1+(2p+1)2<11+(2p)2, in addition to the factor of 1/2. HenceI2k> J2k. Analo-gous reasoning shows also thatI2k+1> J2k+1and soIn> Jnfor alln≥0. This isreasonable since comparison of the integrands ofJnandIn, respectively shows thate-x(sinx)ncosx≤e-x(sinx)n.9
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