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# UCLA MATH 31A - Practice Problems for Midterm (Solutions)

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Practice Problems for Midterm (Solutions)(1) Show that the function f(x) = ex3+3x+1 is one-to-one and compute(f−1)0(2).Solution. We havef0(x) = 3x2ex3+ 3 > 0since x2≥ 0 and ex3> 0. Therefore f is strictly increasing andthus one-to-one. There is no prayer of inverting the formula for f.But we only need the derivative of f−1at a single point 2. We cancalculate this with the formula for the derivative of inverse functionsprovided we can calculate f−1(2). Noticef(0) = e03+ 3 · 0 + 1 = 2,thus f−1(2) = 0. If you miss this, just try plugging in some valuesof x (usually x = 0, 1, −1 are good places to start). By the formulafor (f−1)0we have(f−1)0(2) =1f0(f−1(2))=1f0(0)=13 · 02· e02+ 3=13.(2) Show that the function f(x) = ln x −1xis one-to-one and compute(f−1)0(−1).Solution. This is just like the previous problem. We havef0(x) =1x+1x2> 0when x > 0 (the function f is only defined for x > 0), so f is strictlyincreasing and therefore one-to-one. Noticef(1) = ln 1 −11= −1,thus f−1(−1) = 1. Therefore(f−1)0(−1) =1f0(f−1(−1))=1f0(1)=111+112=12.12(3) Calculate the following limitlimx→∞(ln x)2xSolution. This is an indeterminate form of type∞∞. By L’hopital’srule we havelimx→∞(ln x)2x= limx→∞2 ln x ·1x1= limx→∞2 ln xx.This is also indeterminate of type∞∞. Using L’hopital again, we findthatlimx→∞2 ln xx= limx→∞2 ·1x1= limx→∞2x= 0.Some students will see this and simply say ”ln x grows slower than x,so the limit is 0”. This is not a good answer (in fact it is tantamountto just asserting that the limit is 0). You have to use L’hopital’s ruleto actually show that the limit is 0.(4) Calculate the following limitlimx→0+(tan x)xSolution. This is an indeterminate form of type 00. We want to useL’hopital’s rule, but first we have to convert the exponentiation intomultiplication by taking the logarithm of the expression. We havelimx→0+ln((tan x)x) = limx→0+x ln(tan x) = limx→0+ln(tan x)1x.3The right-hand side is an indeterminate form of type−∞∞. By L’hopital’srule we havelimx→0+ln(tan x)1x= limx→0+sec2xtan x−1x2= limx→0+1cos2x·cos xsin x−1x2= limx→0+1sin x cos x−1x2= limx→0+−x2sin x cos x.This is an indeterminate form of type00. There are a couple ofoptions at this point. One could remember that limx→0xsin x= 1,one could use L’hopital’s rule again, or one could recognize thatsin x cos x =12sin(2x) and then use L’hopital’s rule. We will gowith the second option, as it is the most straightforward. We havelimx→0+−x2sin x cos x= limx→0+−2xcos2x − sin2x= 0by substitution. Thereforelimx→0+(tan x)x= elimx→0+ln((tan x)x)= e0= 1.If you are wondering why the limit is sided, it is because tan x < 0for negative x close to 0. The exponent of a negative number is notalways defined (for example,√−1 = (−1)1/2is not a real number).(5) EvaluateZln 20e3x1 + e3xdxSolution. There is a factor of e3xin the integrand, which is 1/3times the derivative of 1+e3x. This suggests we make the substitutionu = 1 + e3x. Then du = 3e3xdx and the bounds become u(0) = 2 to4u(ln 2) = 9. We calculateZln 20e3x1 + e3xdx =Z9213duu=13ln |u|92=13(ln 9 − ln 2) = ln 3r92!.(6) EvaluateZ1/√201√1 − u2duSolution. You should recognize that1√1−u2is the derivative of arcsin u,thusZ1/√201√1 − u2du = arcsin u|1/√20= arcsin(1/√2) − arcsin(0) =π4.(7) EvaluateZx sin(2x)dxSolution. Any integral of the form xnsin(ax) where n is a nonnega-tive integer and a is any constant can be evaluated using integrationby parts. Powers of x become simpler after taking the derivative,and sin becomes no worse after antidifferentiating. Let u = x andv0= sin(2x). Then u0= 1 and v = −12cos(2x). By the integrationby parts formulaZx sin(2x) dx = −12x cos(2x) −Z−12cos(2x) dx= −12x cos(2x) +Z12cos(2x) dx= −12x cos(2x) +14sin(2x) + C5(8) EvaluateZtan100x sec4xdxSolution. We have a factor of sec2x in the integrand, which is thederivative of tan x. This suggests the substitution u = tan x, whichgives us du = sec2x dx. The remaining factor of sec2x can be dealtwith using the identity sec2x = 1 + tan2x. We haveZtan100x sec4x dx =Ztan100x sec2x sec2x dx=Ztan100(1 + tan2x) sec2x dx=Zu100(1 + u2) du=Z(u100+ u102) du=1101u101+1103u103+ C=1101tan101x +1103tan103x + C.(9) EvaluateZ10(x − 3)(x2+ 1)dxSolution. We have a rational function and the denoninator is al-ready nicely factored for us, so we should immediately think to usepartial fractions. The polynomial x2+ 1 cannot be factored furtherover the real numbers since it is degree 2 and has no real roots. Thefactor x −3 is linear and the factor x2+ 1 is quadratic and there areno repeats, so our decomposition looks like10(x − 3)(x2+ 1)=Ax − 3+Bx + Cx2+ 16for some constants A, B, C. After gettning a common denominatorand equating numerators (or cross multiplying), we find that10 = A(x2+ 1) + (Bx + C)(x − 3).Substituting x = 3, we find that10 = 10A,so A = 1. Once we know A = 1, substituting x = 0 gives us10 = 1 − 3C,so C = −3. Finally, substituting x = 1 tells us that10 = 2 + (B − 3)(−2) = 8 − 2B,so B = −1. Therefore10(x − 3)(x2+ 1)=1x − 3+−x − 3x2+ 1=1x − 3−xx2+ 1−3x2+ 1.We split up the last fraction for the purposes of integrating. Werecognize1x−3as the derivative of ln |x − 3| and3x2+1as the deriva-tive of 3 arctan(x). The second fraction can be integrated using thesubstitution u = x2+ 1, giving du = 2x dx, soZxx2+ 1dx =Z12duu=12ln |u| =12ln |x2+ 1|.In conclusionZ10(x − 3)(x2+ 1)dx =Z1x − 3dx −Zxx2+ 1dx −Z3x2+ 1dx= ln |x − 3| −12ln |x2+ 1| − 3 arctan(x) + C.(10) EvaluateZ104 arctan xdxSolution. The method to use here is integration by parts. This isbecause differentiating arctan(x) gives us a rational function, whichwe should be able to deal with. Alternatively, one could arrive at thisidea by process of elimination. Nothing about this integral suggests7substitution, and certainly partial fractions doesn’t apply here, so byparts is the only option left. By parts often works well on products offunctions when you don’t see a substitution option. Let u = arctan xand v0= 4. Then u0=11+x2and v = 4x. By the by parts formulaZ104 arctan x dx = 4x arctan x10−Z104x1 + x2.We have4 · 1 · arctan(1) − 4 · 0 · arctan(0) = πand the second integral can be solved with the substitution t = 1 + x2.Then dt = 2x dx and the bounds become t(0) = 1 to t(1) = 2, soZ104x1 + x2=Z212 dtt= 2

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