2 f(x) dx = 1, 0. 2f(x 1) dx d) 1 4t t6 t. t 2 dt i)
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1 Math PracTest Be sure to review Lab (ad all labs) There are lots of good questios o it a) State the Mea Value Theorem ad draw a graph that illustrates b) Name a importat theorem where the Mea Value Theorem was used i the proof c) I your ow words, state the defiitio of the defiite itegral of a cotiuous fuctio o a closed iterval [a, b] as a limit of some process d) State the Fudametal Theorem of Calculus (FTC I) e) Determie d [ ] si (t) dt What theorem(s) apply? d f) If g(t) dt = si e, what is g()? Eplai Draw ad the estimate Right() for the graph of f o [, ] o the left below Be careful of the scale I the graph of f o the right above, assume that f() d =, f() d =, ad f() d = Evaluate the followig a) f() d b) f() d c) f( ) d d) e) If f() is symmetric about the origi (a odd fuctio), what is f() d? Calculate these look-alike idefiite itegrals t a) dt b) dt t f) t dt g) + t c) t dt h) 9 + t t t dt + t t dt i) d) f() d t t dt e) e e d j) t + t dt + t dt a) If a atiderivative of f() is si si +, what was f()? b) O a recet test Judy said that cos si d = si + c Elaie said that cos si d = cos + c I gave them both full-credit How ca both be correct if their aswers are differet? Did I make a mistake? a) Fill i the table for Right() for ( ) d Be sure to simplify f( i ) f() [a, b] i f( i ) Right() b) If you were to graph f() o [, ] it would be a icreasig fuctio Is Right() a over or uder estimate of ( ) d? Eplai c) Fid ( ) d usig a limit of Riema sums Check your aswer by evaluatig ( ) d usig the FTC d) Thik: O a test oce, I asked studets to compute Right() ad Right() for f() = e o a iterval [a, b] A studet correctly computed both with his calculator but failed to label which is which Which of her two values is Right(): 788 or 7? Eplai clearly how you ca tell
2 7 Determie these idefiite ad defiite itegrals: a) e) 8 sec() ta(), d b) e e e d f) + e e t t dt c) si () d g) z ta(z + ) dz d) sec( ) d h) si(si t) cos t dt sec ( ) d 8 The graph of f(t) is give below o the left Aswer the followig questios about F () = a) O what iterval(s) is F icreasig? Eplai b) At what poit(s), if ay, does F have a local mi? Eplai c) O what iterval(s) is F cocave dow? Eplai d) Does F have ay poits of iflectio? Eplai e) Fill i the the table below Your values should be reasoably accurate 7 8 F () = f(t) dt f(t) dt f) Graph F () o the aes o the right usig your table values ad kowledge about mas ad mis f(t) Thik ad Iterpret: The velocity of a object v(t) i m/s is graphed above o the left (same graph as i previous questio) a) Whe is the object movig backwards? Eplai b) Assume that the iitial positio of the object is s() = Fill i the row i the table for the positio s(t) for the object The graph the positio fuctio o the aes above o the right t 7 8 Positio s(t) Distace Traveled = b v(t) dt c) What is v ave o the iterval [, 8]? (Hit: Use the v ave formula ad your work above) At what time(s) t did v ave occur (ie, whe was v(t) actually equal to v ave? Eplai d) Speed is v(t) Graph the speed fuctio for the object o the aes o the left Distace traveled is the itegral of the speed: b v(t) dt Fill i the row i the table above for the distace traveled by the object
3 Whe makig up this questio, the priter jammed ad oly part of the graph of a differetiable fuctio F () was prited out, as show below Noetheless, the graph still provides eough iformatio for you to precisely evaluate 8 F () d What is the value of this itegral (Look carefully at the itegrad) 7 8 Graph of F () a) Suppose that the acceleratio of a object is a(t) = e t ad that v() = ad s() =, where v ad s have their usual meaig Fid s(t) b) Suppose istead that a(t) = t + t ad that s() = ad s() = 8 Fid s(t) Evaluate the followig epressio: d t e t dt d O Mars the acceleratio due to gravity equals 9 that of earth or ft/sec/sec Suppose that a Martia calculus studet throws his calculus tet upward at ft/sec off the roof of a buildig a) If it takes secods to hit the groud, how high is the buildig? b) What was the velocity of the book whe it the groud? Let f() = o the iterval [, ] Suppose we wat to fid Right() for this fuctio a) Determie the epressios for, k, ad f( k ) Be sure to simplify where possible b) Determie ad simplify the epressio for Right() c) Evaluate d as a limit of Riema sums d) Check your aswer by usig the Fudametal Theorem of Calculus Oe of the itegral properties states that if we switch the limits of itegratio, the sig of the itegral is switched: a b f() d = b a f() d Give a brief eplaatio of why this should be so Hit: Thik about Use the graph f below to draw ad the estimate the left-had sum Left() for the give fuctio o [, ]
4 7 Three similar itegrals + a) d b) d c) + + d 8 My Hoda Accord accelerates from to 88 ft/sec ( mph) i secods Assume that acceleratio is a costat, a a) Fid the velocity fuctio of the car (usig the velocities at the two times you ca elimiate ay costats from this fuctio) b) How far does it travel i this secod period? 9 A balloo, risig vertically with a velocity of 8 ft/s, releases a sadbag at the istat it is ft above the groud a) How may secods after its release will the bag strike the groud? Remember a(t) = ft/s b) At what velocity does it hit the groud? a) Fid the average value of f(t) = t +t o [, ] b) Fid the average value of f(t) = t t t 7 7t + cos(t) o [/, /] Why is this EZ? a) Page 7 #9 (Aswers i tet) b) Page 7 # (Aswers i tet) Area betwee curves: See Assigmet from Day ad the associated WeBWork problems (ad all of the eamples o lie i the Notes for Day ) Determie a) cos () d b) si () d Set up the itegrals usig the fuctios f(), g(), ad h() ad their poits of itersectio that would be used to fid the shaded areas i the three regios below y = g() Shaded y = f() y = g() Shaded h() f() Sketch the regios for each of the followig problems before fidig the areas a) (Do Later) Fid the area eclosed by the curves y = ad y = (As: /) b) Fid the area eclosed by the curves y = + ad y = (As: /) c) Fid the area eclosed by = y + ad = y + 9 Itegrate alog the y-ais (As: ) g() Sh f() Sh d) Fid the area i the first quadrat eclosed by the curves y =, y =, the -ais, ad the y-ais by usig defiite itegrals alog the y-ais (As: /, if you get 9/, you have the wrog regio) e) Fid the area of the wedge-shaped regio below the curves y =, y =, ad above the -ais Itegrate alog either ais: your choice! (Note: Not the same as (d) As: 7/) f) The curves si ad ta itersect at = Fid b, where b is the smallest positive value of where the curves meet (Hit: Write ta i terms of sies ad cosies) Fially, determie the area betwee si ad ta o the iterval [, b] (As: l ) g) Fid the area betwee the curves f() = cos + si ad g() = cos si over [, ] (As: 8)
5 a) Sketch the regio (use your calculator?) ad fid the area uder y = arcsi o the iterval [, ] Hit: switch aes (As: / ) b) *From a test last year: Cosider the regio bouded by y = l, y =, ad y = show below Fid the area of this regio Hit: Tur your head! (As: e ) y = y = l e 7 Fid the regio eclosed by the three curves y =, y = + 8, ad y = You will eed to fid the itersectios (As: 8) 8 Fid the area of the regio i the first quadrat eclosed by y = 9, y = +, ad the y-ais Hit: The two curves meet at the poit (, )
6 Math : Aswers to Practest a) Mea Value Theorem: See page 9 i your tet b) FTC II or Mea Value Theorem for Itegrals c) Partitio the iterval [a, b] ito equal width subitervals usig poits a = < < < < = b The defiite itegral of a cotiuous fuctio f o a closed iterval [a, b] is b a f() d = lim f(c i), where = b a ad ci is some poit i the i i-th subiterval [ ] [ d d) si (t) dt = d ] si (t) dt = si ( ), where we used FTC II ad the chai rule d d [ e) FTC II implies g() = d ] g(t) dt = d [ ] si = e cos e si So g() = e cos +e si = d d e e e i= = = So Right() = ( 9) + ( 7) + ( ) + () + () = Use basic itegral properties a) b) d) f() d = f() d = f() d f() d f() d = () = f() d = = c) f() d = + ( )[ ()] = e) f( ) d = f() d = f() d = f() d = (horizotal shift) a) u = t, du = dt u / du = u/ + c = (t )/ + c b) Use a u du = arcsi u + c with a a = so a =, u = t, so u = t, du = dt t arcsi( ) + c c) u = t, du = 8t dt, du = t dt t dt = 8 t 8 du = u / u/ + c = t + c u du = arcsi( u ) + c = d) Method : u = t, so u = t, so u = t so u du = dt t t dt = (u ) u u du = u 8u du = u 8 u + c = u 8 u + c = ( t)/ 8 ( t)/ + c d) Method : u = t, so u = t ad du = dt ( u)u / du = u / u / du = 8 u/ + u/ + c = ( t)/ 8 ( t)/ + c e) u = t, u = t, du = tdt du = arcta u + c = +u arcta(t ) + c f) u = + t, du = t dt du = l u + c = l( + u t ) + c g) a = 9 =, u = t, u = t, du = t dt du = arcta( u ) + c = t arcta( ) + c +u 8 h) +t dt = t + t dt = t + t t + c (Just divide first by t ) i) Use a du = arcsi u +c with u a a = so a =, u = e, so u = e, du = e d e arcsi( ) + c u du = arcsi( u )+c =
7 j) u = t u = t du = dt du = dt So dt = t du = u arcsi u + c = arcsi t + c k) + t dt = + t dt Use u = t u = t du = dt du = dt So + t dt = + u du = arcta u + c = arcta t + c a) Sice f() d = si si +, the f() = d (si si + ) = cos si cos d d b) No Just take the derivatives of each: d (si + c) = si cos while d ( d cos + c) = cos ( si ) = si cos also The two atidervatives differ by a costat sice si = cos + a) f() [a, b] i f( i) Right() ( ) ( ) ( ) [, ] + i i i + i ( + i ) i= b) Sice the fuctio is icreasig (f () = > o [, ]), Right() is a overestimate of d because f( i) will be greater tha ay value of f() i [ i, i] 7 a) c) Right() = So [ i i= + i = ] = i= i + i= [ ] [ ] i = ( + )( + ) + ( + ) [ ] = = + + ( ) d = lim Right() = ad usig the FTC, d = = (7 9) ( ) = d) Sice e is a icreasig fuctio Right() is a overestimate Sice the estimates improve (smaller overestimate) as gets larger, the Right() > Right() So Right() = sec() + c = sec() + c (Metal adustmet) b) u = t, du = t dt e u du = e u + c = e t + c c) u = z +, du = z dz, du = z dz So z ta(z + ) dz = d) u = si t, du = cos t dt si u du = cos(u) + c = cos(si t) + c ta u du = l sec u + c = l sec(z + ) + c e) u = e + e, du = (e e )d Chage the limits: Whe =, u = e + e = ; whe =, u = e + e So f) si () d = Itegral property g) sec( ) d = l sec( ) + ta sec( ) + c Adjust! h) sec ( ) d = ta( ) + c Adjust! e e e+e e + e d = e+e u du = l u = l e + e l 8 The graph of f() is give below o the left Aswer the followig questios about F () = f() d a) F icresasig, so F = f > : [, ] ad [, 8] b) Local mi whe F = f chages from egative to positive: = c) b f(t) dt is just the et area uder the curve from to b b 7 8 F (b) = b f(t) dt 8 7
8 f(t) F (t) = s(t) a) Whe is the object movig backwards? v < so [,, ] b) Negative acceleratio whe a = v < : so [, ] c) Positio is the et area uder the curve (same table as above) t 7 8 Positio s(t) 8 Distace Traveled d) v ave = 8 v(t) dt = [s(8) s()] = ( ) = vave occurred at c = ad 8 [Use the graph of v(t) = f(t)] From the graph we see that F () = ad F (8) = So 8 F () d = F () 8 = F (8) F () = = a) Itegrate twice: v(t) = e t dt = et + c v() = = + c c = s(t) = et + dt = et + t + c s() = = + c c = So s(t) = et + t + b) v(t) = t + t dt = 8t + t + c So s(t) = 8t + t + c dt = t + t + ct + d Now s() = = d So s() = 8 = + + c + c = So s(t) = t + t + By FTC II ( d t e t dt = d ) t e t dt = ( ) e = 8 7 e d d a) Give s (t) =, s () = Fid s() The s (t) = dt = t + c But s () = = () + c meas c = ad s (t) = t+ Net, s(t) = t+ dt = t +t+d Now set s() = = () +()+d d = 7 ft So s(t) = t + t + 7 Ad s() = 7 b) Fid s () = () + = ft/s Let f() = o the iterval [, ] Suppose we wat to fid Right() for this fuctio a) = possible b) So Right() = c) Right() =, i = + i ( i i= [ i i= i, ad f(i) = ( + ) ( + i ) = + i + i ( + i ) = i + i Be sure to simplify where + i + i ] = ) ( ) = 8 i + i= So d = lim Right() = d) By FTC, d = = ( 9 9 ) ( i= [ ] [ ] i = 8 ( + )( + ) + ( + ) [ ] = = + + ) = 8
9 For a a b f() d, i the Riema sum = which is the egative of the for the Riema sum for b f() d while the b a values of f( k ) remai the same Thus the sums will be the egatives of each other Note = = So L = (8) + (8) + () + (8) + (8) + () = 7 Similar itegrals doe differetly ad oe etra + a) d = + d = l + + c b) u = + du = d d = du = l u + c = l( + + u ) + c c) u = u = du = d d = du = arcta u + c = + +u arcta( ) + c 8 Give: a(t) = a costat v = ft/s s = Ad v() = 88 ft/s a) For costat acceleratio: v(t) = at + v = at But v() = a = 88 a = So v(t) = t b) For costat acceleratio: s(t) = at + v t + s = t Cosequetly, s() = () = 7 ft 9 Give: a(t) = ft/s is costat v = 8 ft/s s = ft So check that v(t) = t + 8 ft/s ad s(t) = t + 8t + ft a) The bag hits the groud whe the positio s(t) = t +8t+ = Usig the quadratic formula the solutio is t = + (ot ) b) The velocity whe it hits the groud is v( + ) = ( + ) + 8 = 8 ft/s a) f ave = t dt u = + t, du = t dt Whe =, u = ; =, u = So +t f ave = u du = l u l = b) Use symmetry, split ito odd ad eve terms: f ave = = / t t t 7 7t dt + / / / t t t 7 7t + cos(t) dt / (/) / cos(t) dt = + / cos(t) dt = si t / = ( ) = a) Note the metal adjustmet cos () d = + cos() d = + si() + c b) si () d = cos( ) d = si() + c 8 Left: f() d+ g() d Middle: h() f() d+ g() f() d Right: 7 g() f() d+ 7 f() g() d 7 9
10 Sketch the curves! I have oly doe the itegrals here a) A = d = = ( ) = b) A = ( +) ( ) d = + d = + + = ( + 8 ) ( + ) = A + A = + = = ( +) = A = ( ) ( +) d = c) A = y + 9 (y + ) dy = y + 8 y dy = y + 8y y = ( + ) ( + 8 ) = d) The regio is bouded i part by the ad y aes i the first quadrat A = y + dy = y + y = ( + ) () = A = y dy = y y = (9 9 ) ( ) = A + A = e) This is a wedge-shaped regio uder the curves y =, y =, ad above the -ais It ca be doe as a sigle regio alog the y-aes The two curves meet at the poit = ad y = A = ( y) (y + ) dy = y y dy = y y y = ( ) () = 7 f) Whe, si = ta si = si cos = cos = = O the iterval from [, ] cos which curve is o top? Plug i : si( ) = while ta( =, so si is o top ( si ta d = cos + l cos = + l ) ( + l ) = l g) A = (cos + si ) (cos si ) d = si d = cos si ) d = si d = cos = () = Total area is 8 = () = A = (cos si ) (cos + a) Itegrate alog the y ais y = arcsi = si y Be careful: The vertical lie = is ow o top A = / si y dy = y + cos y / = b) Agai switch aes: y = l = e y ad y = = y + So A = e y (y + ) dy = e y y y = (e ) = e 7 7 A = ( ) d ( ) d = [ + ] + [ 7 + 9] 7 = ( + ) ( + ) + ( + ) ( + 9) = = A = 9 + d = 9 d + d But 9 d = 9 = 7 9 = For the secod itegral, u = + du = d ad u = Limits: = u =, = u = So we get + d = (u )u/ du = u/ u / du = u/ u/ Total Area: = [ ] = =
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