MATH 31B: MIDTERM 2 REVIEW

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1 MATH 3B: MIDTERM REVIEW JOE HUGHES. Evaluate x (x ) (x 3).. Partial Fractios Solutio: The umerator has degree less tha the deomiator, so we ca use partial fractios. Write x (x ) (x 3) = A x + A (x ) + B x 3 After clearig deomiators, we get x = A (x )(x 3) + A (x 3) + B(x ) Now plug i x = to obtai = A, so A =. Similarly, if we plug i x = 3 the we get = B. To determie A, let s look at the coefficiet of x o both sides. O the left-had side, this coefficiet is zero, while o the right-had side the coefficiet is A + B. A = B =. So our itegral becomes x (x ) = (x 3). Evaluate x (x+)(x +). x (x ) + = l x 3 l x + x + C x 3 Solutio: Whe we do partial fractios with a irrecible quadratic factor like x +, the correspodig term i the expasio should have a liear polyomial i the umerator. So write x (x + )(x + ) = A x + + Bx + C x + Now clear the deomiators to get Next, plug i x = : or A =. x = A(x + ) + (Bx + C)(x + ) ( ) = A( + )

2 JOE HUGHES To fid B, look at the coefficiet of x. O the left-had side this coefficiet is zero, while o the right had side it is A + B. B = A =. Fially, to fid C we look at the costat coefficiet o both sides to get = A + C, so C = A =. x (x + )(x = + ) 3. Evaluate sec θ ta θ dθ. x + x + + x = + x + + = l x + + l(x + ) + ta (x) + C x x + + x + Solutio: Oe might be tempted to try to use the trig idetity sec θ = + ta θ, but I do t thik that helps here. So istead let s start with a substitutio: let u = ta θ, so that = sec θ dθ. The sec θ ta θ dθ = u = Now we ca proceed usig partial fractios. Write (u )(u + ) = A u + B u + ad clear deomiators to obtai = A(u + ) + B(u ) (u )(u + ) Pluggig i u = gives = A, so A =. Similarly, pluggig i u = gives B =. (u )(u + ) = u u + = l u l u + + C ad pluggig i u = ta θ gives sec θ ta θ dθ = l ta θ l ta θ + + C. Improper Itegrals. Determie whether x 7 8 coverges, ad if so, evaluate it: Solutio: The fuctio x 7 8 has a ifiite discotiuity at x =, so this is a improper itegral. We ca evaluate it usig the limit defiitio: [ ] = lim = lim 8x 8 = lim 8 8a 8 = 8 a + a + a a + x 7 8 x 7 8 a the itegral coverges, ad the value of the itegral is 8.. Determie whether x + x coverges.

3 MATH 3B: MIDTERM REVIEW 3 Solutio: x x for x, hece x + x x ad x +x x We ca break our itegral up ito two pieces: x + x x + x = x = lim R x + x + x + x for x. The first itegral is just some fiite umber, so it does t affect the covergece or divergece of the origial itegral. For the secod itegral, the compariso test gives [ ] R l(x) l(r) l() = diverges, which shows that also diverges. x + x x + x = lim R 3. Numerical Itegratio. Fid N such that cos x T N Solutio: The formula for the error boud for the trapezoid rule is give by where K = max x [a,b] f (x). Error(T N ) K (b a) 3 N If f(x) = cos x, the f (x) = si x ad f (x) = cos x. hece K = max x [, ] cos x = Error(T N ) ( ( ))3 N = N We wat the error to be at most. Thus N ad solvig for N gives N 9.

4 JOE HUGHES so ay N 9 will do. How could we hadle the last part without usig a calculator? Usig the fact that = we see that ay N = = suffices.. Arc legth For t [, ], let f(t) be the arclegth of the curve y = x t ( t) xt+ (t + ) o [, ]. Fid the maximum of f(t) o [, ] ad the poit at which it is attaied. Solutio: Recall that the arclegth formula is ( dy ) + By the power rule, hece + dy = x t xt ( dy ) x t = + + xt = x t + + xt ( x t = + xt ) Pluggig this ito the arc legth formula gives f(t) = x t + x t = ( + t) + ( t) = t f is icreasig o [, ], so has a maximum of 3 at t =.. Fluid Force. Fid the fluid force o a metal plate bouded by the parabola y = x ad the lie y =. The surface of the liquid is the lie y =, ad the liquid has desity ρ. Solutio: The formula for fluid force is F = ρg yf(y) dy

5 MATH 3B: MIDTERM REVIEW where f(y) is the width at depth y. I this case, f(y) = y. F = ρg y y dy = ρg y 3 dy = ρg [ y ] = ρg 6. Taylor Polyomials. Let f(x) = xe x. Fid a geeral formula for T (x) cetered at a =. Solutio: Start by makig a table of derivatives of the fuctio, ad their values at : xe x a = f(x) = xe x f (x) = xe x + e x f (x) = xe x + e x f (3) (x) = xe x + 3e x 3 f () (x) = xe x + e x f () (x) = xe x + e x Hopefully at this poit the patter is clear: f () (x) = xe x + e x, ad f () () =. T (x) = f() + f ()x + f () = x + x + + x ( )! x + + f () () x!. Let f(x) = e x. Fid the third Taylor polyomial T 3 (x) for f cetered at a =, ad use the error boud to fid the maximum value for f(.) T 3 (.). Solutio: Recall that T 3 (x) = f() + f ()x + f () we eed to take 3 derivatives: Pluggig i x =, we get x + f (3) () x 3 6 f (x) = e x f (x) = e x f (3) (x) = e x T 3 (x) = x + x x3 6 Next, recall that the error boud formula is f(.) T 3 (.) K.! = K

6 6 JOE HUGHES where K is a upper boud for f () (x) o [,.]. To get the best possible error boud, we take K = max f () (x) x [,.] Now f () (x) = e x, which is a decreasig fuctio. hece K = f(.) T 3 (.) max f () (x) = e = x [,.].6 6 O the midterm, I thik that it would suffice to leave the aswer as. 7. The formal defiitio of covergece. Use the formal defiitio of the limit of a sequece to show that lim + =. Solutio: We eed to show the followig: for all ε >, there exists N (depedig o ε) such that if N the + < ε. Fix ε >. The + = + = Choose N such that N > ε. If N, the + = N < ε As ε was arbitrary, this shows that the sequece coverges to.. Determie lim e. 8. Sequeces Solutio: Let f(x) = x e x. The f() = e, so lim e = lim x x e x. To evaluate this limit, use L Hopital s rule:. Evaluate lim!. lim x x e x x = lim x e x = lim x x e x = lim x e x = Solutio: Let b =!. We wat to use the squeeze theorem to show that lim b =. Set a =. The a b for all.

7 MATH 3B: MIDTERM REVIEW 7 To fid c, ote that! ad are both a proct of terms. So write! = Each of these terms is at most. we ca remove every term except for the last oe to get! = So let c =. The lim a = lim c =, hece lim b = as well.. Determie whether the series sum. = 9. Summig a series = = + coverges, ad if so, fid the value of the Solutio: Rearrage the series to get + ( = ) + ( ) These are geometric series with ratios,, so both series coverge, ad therefore the origial series coverges. For the first series, we have = = = = = = = = ad for the secod, ( ) ( = ) ( ) ( ) ( = = + = + 6 = 37 = = = ) = 6