Math 21B-B - Homework Set 2

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1 Math B-B - Homework Set Sectio 5.:. a) lim P k= c k c k ) x k, where P is a partitio of [, 5. x x ) dx b) lim P k= 4 ck x k, where P is a partitio of [,. 4 x dx c) lim P k= ta c k ) x k, where P is a partitio of [, π/4. π/4 tax) dx. Suppose that f ad g are itegrable ad that: fx) dx = 4, fx) dx = 6, gx) = 8. a) b) c) d) e) gx) dx = Zero Width - Rule gx) dx = 5 gx) dx = 8 Order of Itegratio - Rule fx) dx = fx) dx = 4 =. Costat Multiple - Rule fx) dx = fx) dx Additivity - Rule 5 [fx) gx) dx = fx) dx Sum ad Differece - Rule 4 fx) dx = 6 4) = gx) dx = 6 8 =

2 f) 5 [4fx) gx) dx = 4fx) dx gx) dx = = 4 8 = 6 gx) dx = 4 fx) dx Costat Multiple ad Sum/Differece - Rules ad 4. Suppose that f ad h are itegrable ad that: fx) dx =, fx) dx = 5, hx) dx = 4. a) b) c) d) e) f) fx) dx = Costat Multiple - Rule [fx) + hx) dx = fx) dx = = fx) dx + Sum ad Differece - Rule 4 [fx) hx) dx = fx) dx hx) dx = = 9 hx) dx = 5 4 = = hx) dx = fx) dx Costat Multiple ad Sum/Differece - Rules ad 4 9 fx) dx = fx) dx = ) = Order of Itegratio - Rule fx) dx = fx) dx Additivity - Rule 5 9 [hx) fx) dx = 4 5) = ) = fx) dx = 5 = 6 [ [hx) fx) dx = hx) dx Order of Itegratio ad Sum/Differece - Rules ad 4 fx) dx = 4. x dx

3 x dx = Area of left triagle + Area of right triagle = / + / = + / = + / = 5/ square uits 5. + x ) dx

4 + x ) dx = Area of semicircle + Area of Rectagle = / π + = π + square uits 6. a) fx) = x o [, ) avf) = x ) dx ) [ = x dx ) = = = = ) x x dx ) [ ) ) ) ) ) b) hx) = x o i.) [,, ii.) [,, iii.) [,. 4

5 i. ) avf) = x dx ) = ) = = x x dx x) dx = = ii. ) avf) = x dx = ) = x = x dx 5

6 iii. ) avf) = x dx ) ) [ = x dx + x dx ) [ = x) dx + x dx ) [ = x dx + x dx ) [ = x x ) [ )) ) = ) [ = ) = ) =. a) Cosider the partitio P that subdivides the iterval [a, b ito subitervals of width x = b a ad let c k be the right edpoit of each subiterval. So the partitio is P = {a, a + b a b a) b a), a +,..., a + = b} 6

7 ad c k = a + kb a). We get the Riema sum ) b a fc k ) x = c k k= k= = b a = b a = b a k= k= a + ) kb a) a akb a) + + k b a) ) a + k= = b a a + ab a) ab a) k + k= + ) b a) + ) k k= b a) = b a)a + ab a) + + b a) + + ) 6 = b a)a + ab a) + ) + + b a) + 6 As ad P, this expressio coverges to b a)a + ab a) + b a) 6, which simplifies to b a. Thus b a x dx = b a. b) Cosider the partitio P that subdivides the iterval [, ito subitervals of width x = ad let c k be the right edpoit of each subiterval. So the partitio is P = {, +, +,..., + = } ad c k = + k. We get the Riema sum fc k ) x = + k ) + k )) k= = + k + k ) ) k k= k= = + k k k= k= k= = + ) + ) ) + ) + ) 6 = + + ) + + ) ) + ) 6 ).

8 As ad P, this expressio coverges to + 6 = 5 6. Thus x x ) dx = Use the Max-Mi Iequality to fid upper ad lower bouds for the value of + x dx Notice that the itegrad fx) = +x is a decreasig fuctio f x) = x +x ) for x [, ). Therefore, o the iterval f has a maximum at x = ad has a miimum at x =. max f = f) = mi f = f) = Thus the Max-Mi Iequality gives: + x dx 9. Use the Max-Mi Iequality to fid upper ad lower bouds for the the value of.5 dx ad + x.5 + x dx Usig the observatio from 65) that fx) = +x is decreasig o [,.5 ad [.5,, we get the followig two iequalities: dx.5 + x dx.4. + x Therefore, by addig the two iequalities we get the ew ad improved) estimate:.65 dx.9. + x. We kow that si x x for x. Therefore we ca get the upper boud: 8

9 si x dx x dx = =. Sectio 5.4:. a) b) 4 x + 5) dx = x + 5x) = + ) 4 ) = 6 5 x ) dx = 69 4 = 4 ) 5x x 4 = 4) 5 9 ) = c) π si x dx = cos x π = cosπ) + cos) = + = d) e) f) π/ sec x dx = ta x π/ = taπ/) ta) = = [ ) r +) dr = r +r +) dr = r + r + r [ ) + + ) [ + = + = 8 ) x e x dx = l x + e x) = l + e ) + e ) = l + e e g) h) 4 + x dx = 4 arcta x = 4 arcta) 4 arcta) = 4 π 4 4 = π x 5 dx = x+x ) / dx = + x 5 + x = + x 5 = 6 5 9

10 i) xe x dx Notice that d dx x = x, ad we have a x i the expoet ad a x i the product. Thus, keepig i mid the chai rule we get: fx) = xe x F x) = ex xe x dx = F ) F ) = e = e ). j) l x x dx Notice that d dx l x = x, ad we have a l x ad a x. Thus, keepig i mid the chai rule we get: fx) = l x x F x) = l x). a) d dx i. l x x = F ) F ) = l ) = l ). x cos t dt Therefore we get: ii. Let u = x d dx x d dx x x x cos t dt = si t = si x) si) = si x) cos t dt = d dx si x) cos t dt = = cos x) d du u = cos u du dx = cos x) x x ) cos t dt du dx

11 b) d dθ i. ta θ sec y dy ta θ Therefore we get: ii. Let u = ta θ d dθ d dθ ta θ ta θ sec y dy = ta y ta θ = tata θ) ta) = tata θ) sec y dy = d tata θ) dθ sec y dy = = sec ta θ) sec θ d du u ) sec y dy du dθ = sec u du dθ = sec ta θ) sec θ. y = x si t ) dt = x Let u = x. The we get: si t ) dt dy dx = dy du du dx = siu ) du dx = six) x x = six) x 4. a) y = x x where x First, otice that y = x x = xx+). Therefore i the iterval [,, the fuctio has zeroes at x = ad x =. Area = x x ) dx + x x ) dx x x ) dx = ) x x + ) x x [ ) 8 = 4 9 9) = = 8 [ + ) ) x x 8 4 ) [ 8 4 ) )

12 b) y = x where x First otice that y = x ) = x )x + ). Therefore i the iterval [,, the fuctio has zeroes at x = ±. Area = x ) dx x ) dx + x ) dx = x x ) x x ) + x x ) = [ + ) 8 + 6) [ ) + ) + [8 6) ) = = c) y = x / x where x 8 First otice that y = x / x = x / x /) + x /), so the o [, 8 the fuctio has zeroes at x =, ±. Area = ) x / x dx + = 4 x4/ ) x [ = ) 4 = = 8 4 ) x / x dx 8 ) x / x dx + 4 x4/ ) x 4 x4/ ) 8 x ) [ + 4 ) [ ) ) 4 ) 5. a) v = ds dt = d dt t fx) dx = ft). Therefore v5) = f5) = m/s. b) Recall that the acceleratio is the rate of chage of the velocity, so a = df dt. Sice f has a egative slope at t = 5, the acceleratio is egative at t = 5. c) Sice s = t fx) dx, we kow that the positio at t = ca be represeted as the area uder the graph o [,. Notice that o this iterval, f is a straight lie ad so the value of the itegral is just the area of a triagle formed by fx), the x-axis ad x =.

13 Therefore s) = = 9 d) We wat to kow whe s has a maximum ad we will determie this by lookig at fx), which is the derivative of the s. We kow f is positive o [, 6 ad egative o [6, 9, which meas that s goes from icreasig to decreasig at t = 6. By the first derivative test, we kow that s must have relative max at t = 6. e) At x 4 ad x sice s x) = f x) is zero at these poits. f) The particle is movig towards the origi whe it has egative velocity f is egative. Hece it is movig towards the origi o [6, 9. Similarly the particle is movig away from the origi whe f is positive, hece o [, 6. g) Judgig by the graph of f, we kow that the particle is to the right of the origi o the positive side) because the s9) = the area above the x-axis is greater tha the area below it). x t 6. Fid lim x x t 4 + dt fx) dx > x t Notice that lim x t 4 dt = by the rules of itegrals ad lim + x x =. Thus if you thik of the expressio iside the limit as a fractio, we have that both the umerator ad the deomiator go to. Hece we ca apply l Hôpital s Rule to the limit ad get: lim x x x t t 4 dt = lim + x = lim x [ x t ) t 4 + dt /x [ ) x / x ) x 4 + = lim x x 4 + ) =. Suppose f x) for all values of x, ad that f) =. Defie: gx) = x ft) dt a) True. It follows from the FTC, part I. b) True. Differetiable Cotiuous.

14 c) True. We kow that g ) = f) =, so the taget lie at x = is horizotal. d) False. We kow that f crosses the x-axis at x = ad that f is always icreasig. Therefore g x) = fx) is egative for x < ad g x) = fx) is positive for x >. By the first derivative test, this is a miimum, ot a max. Note that you could also use the fact that g ) = ad g ) > e) True. See d.) f) False. g x) = f x) >, ad hece does ot chage sig. g) True. g x) = fx) ad we kow that the f) = ad is icreasig. 4

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