(a) (x 2) dx (d) (x +1) 2 dx. (e) x 1/2 dx. (h) 1. (3/x) dx. Solution: (a) sin(x) dx. (b) 2 cos(x) dx =4sinx π/4. (c) (x 2) dx = ( x 2 /2 2x ) 3

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1 () Use the Fundamental Theorem of Calculus to compute each of the following integrals. The last few are a little more challenging, and will require special care. Some of these integrals may not eist. Eplain why. (a) sin() d () π cos() d (c) 3 ( ) d (d) ( +) d (e) / d (3/) d 3 /3 d / d (i) d (j) d (k) (3/) d Solution: (a) () (c) (d) (e) π 3 cos() d = 3 sin() d = cos π =. cos() d =sin π/ ( ) d = ( / ) 3 =. ( +) d = /3 d = 3 /3 3 ( +)3 3 = 3. / d = 3 3/ =6/3. (3/) d =3ln =3ln. =/. ( =(3/) ( 3) /3 ( ) /3). =.36 does not eist as / is not defined for <. / d

2 (i) d does not eist. In the course notes we showed that d did not eist (it diverged to +. In eactly the same way d and d oth diverge to + ). (j) d = d = =8. The first equality follows from the symmetry aout = of oth and the the interval [, ]. (k) (3/) d = lim ɛ + ɛ (3/) d = lim ɛ + 3ln ɛ = lim ɛ + 3lnɛ =. () The rate of flow of lood through the heart can e descried approimately as a periodic function of the form F (t) =A( + cos(t)) where t is time in multiples of.5 seconds and A is in units of cuic cm/hr. Find the total volume of lood that flows through the heart over full minute, starting at t =. (Epress your answer in terms of A.) Solution: In one full minute there are units of.5 seconds. Thus we integrate from t =tot =. The units of A are cm 3 /hr, which upon changing hours to multiples of.5 seconds, ecomes cm 3 /. Thus the volume of lood is volume = A ( + cos(t))dt = A ( + sin ) A.66 cm 3. (NOTE:somepeopletookA to e in units of cuic cm/sec, since the units seemed inconsistent as the prolem was worded. In that case the answer is 399. Acm 3.) (3) After a heavy rainfall, the rate of flow of water into a lake is found to satisfy the relationship ( ) t F (t) = where t is time in hours, <t<3 and F is in units of, gallons/hr. (a) Find the time, t at which the rate of flow is greatest. () Find the time, t at which the flow is zero. (c) Find how much water in total has flowed into the lake etween these two times.

3 Solution: (a) F (t) = ( t ) = t =. Since F (t) = 5 < this value of t gives the maimum rate of flow. () The time at which the flow is zero is t =3. (c) The total water flowing into the lake etween these times is ( 3 ( ) ) 3 t dt =t ( ) 3 t 3 3 = 6 3. () The reaction time of a driver (time it takes to notice and react to danger in the road ahead) is aout.5 seconds. When the rakes are applied, it then takes the car some time to decelerate and come to a full stop. If the deceleration rate is a = 8m/sec,howlong would it take the driver to stop from an initial speed of km per hour? (Include oth reaction time and slowing-down time.) Solution: Let v(t) anda(t) denote the velocity and acceleration a t time t in m/sec and m/sec respectively. Let t = e the time when the rakes are applied. Then at any time t = T, up to and including t he time when the car comes to a halt, we have Therefore The initial velocity is v(t ) v() = v() = T = T 6 6 v() v(t ). 8 8dt = 8T. = 36 m/sec, and so the time it takes the car to stop, once the rakes are applied, is 36 T = 3.7sec. 8 Adding on the reaction time we see it takes approimately 3.97sec to come to a full stop. (5) During a gamling session lasting 6 hours, the rate of winnings at a casino in dollars per hour are seen to follow the formula w(t) = t( (t/6)). Find the total winnings during that whole session. 3

4 Solution: The total winnings are 6 ( t t ) 6 dt = t 6 8 t3 6 = 36 =. (6) When a spring is compressed, it eerts a restoring force. If the spring is stretched from its equilirium length y an amount (in cm), the force it eerts (in the opposite direction) is F () = k (in units of dynes). The work done (dynes-cm) in stretching the spring from its equilirium length y an amount is given y W = F (s)ds (a) Find W as a function of, i.e. epress the total work as an epression in which appears. () It was found that when a certain spring was stretched y.5 centimeters, it eerted a restoring force of.5 dynes. How much work was done in stretching the spring y 3 more centimeters? Solution: There is no solution for this eercise. (7) A test-tue contains a solution of glucose which has een prepared so that the concentration of glucose is greatest at the ottom and decreases gradually towards the top of the fluid. (This is called a density gradient). Suppose that the concentration c as a function of the depth is c() = (in units of gm/cm 3 ). The cross-sectional area of the tue is cm and the height of the glucose-containing solution is cm. (a) Determine the total amount of glucose in the tue (in gm). () Now suppose that the cross-sectional area is A, the height is h, and the concentration is c() =c o +. What is the total amount of glucose in this more general setting? Solution: (a) The amount of glucose is given y d = = gm. () In this case the amount is h A(c + )d = A ) (c + h = A ) (c h + h.

5 (8) The figure elow shows a graph of the function y = f(). Use this to draw a sketch of the graph of the function F () = f(t)dt. Solution: There is no solution for this eercise. (9) The rate at which animals migrate into and out of a wildlife reserve is descried y two functions shown in the figure elow. I(t) is the rate at which animals enter the reserve and O(t) is the rate at which they leave ( oth in numer per day). (a) Epress the numer of animals in the reserve as a definite integral. () When is the numer of animals in the reserve greatest and when is it smallest? Solution: There is no solution to this eercise. () The rate of growth for two species of trees (in feet per year) is shown in the figure elow. If the trees start at the same height, which tree is taller after 5 years? After years? Solution: There is no solution for this eercise. () Find sec (t) dt Solution: sec (t) dt =tan(t)+c () Find the following integrals. For the last few cases, special care is needed, and some of these integrals may not eist. Say a few words why that happens. (a) () (c) (d) (e) / 3 e d cos() d ( +)d ( ) d /3 d (/) d ( + ) d 5

6 (i) (j) (k) / d (/) d d e c d Solution: There is no solution for this eercise. (3) The rate of flow of lood through the heart can e descried approimately as a periodic function of the form F (t) =A( + sin(.5t)) where t is time in seconds and A is a constant in units of cuic cm per second. Find the total volume of lood that flows through the heart etween t =andt =. (Epress your answer in terms of A.) Solution: The rate of change of volume is V (t) =A( + sin.5 t) sov (t) is the antiderivative of this, or V (t) =A(t (/.5) cos.5 t)+const. Then V () V () = A( (/.5) cos.5 + (/.5)). () n intravenous infusion delivers a flow rate of y = ( t 3 )cm 3 /hr over a period of hour, where t is time in hours. The infusion contains a drug at concentration.mg/cm 3. Find the total volume of fluid and the total amount of drug delivered to the patient over ahourperiodfromt =tot =. Solution: The volume is the anti-derivative of the flow rate, or In one hour the volume delivered is The amount of drug is. times this = 7.5. V (t) = (t t /). V () V () = 75 (5) Find the following integrals using the Fundamental Theorem of Calculus. (a) e kt dt () a A cos(ks) ds 6

7 (c) Ct m dt (d) aq dq (e) (i) (j) a T c T a sec (5) d +t dt ( 3 s ) ds d / sin(3y) dy 3 dt Solution: [ e (a) e kt kt dt = k ] a () A cos(ks) ds = [ A sin(ks) k ] (c) Ct m dt = [ Ct m+ m + ] (m ), C/t dt =[C log t ] [ log q (d) aq dq = a ] T (e) sec (5) d = c [ tan(5) 5 ] T c +t dt =[arctan(t)] ( 3 [ ] 3 s ) ds = s 7

8 T a d = [ ] T / a (i) sin(3y) dy = [ cos(3y) 3 ] (j) 3 dt =[3t] 8