The rate at which water enters the tank t minutes after it starts to rain is 1

The rate at which water enters the tank t minutes

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The rate at which water enters the tanktminutes after it starts to rainis140t(40-t) litres per minute. After 20 minutes, the tank springsa leak and water leaks out at 1 litre per minute. If the tank initiallyheld 200 litres of water, what was the average amount of water in thetank over the 40-minute period from when it started to rain?Solution:LetV(t) be the volume of water in the tank (measuredin litres)tminutes after the rain begins. We are told thatV(0) = 200,that during the first 20 minutes of the storm,V0(t) =t(40-t)40=t-t240,and during the second 20 minutes of the storm,V0(t) =t-t240-1.During the first 20 minutes therefore,V(t) =t22-t3120+ 200.Att= 20 minutes, the amount of water in the tank isV(20) = 1000/3litres. Consequently, during the second 20 minutes of the storm,V(t) =t22-t3120-t-3403+10003=t22-t3120-t+ 220where3403=t22-t3120-tt=20. HenceV(t) =(t22-t3120+ 200if 0t20t22-t3120-t+ 220if 20t40.The problem now is to calculate the average value of this function overthe interval [0,40]. We haveZ200V(t)dt=Z200t22-t3120+ 200dt=t36-t4480+ 200t200= 5000,andZ4020V(t)dt=Z4020t22-t3120-t+ 220dt=t36-t4480-t22+ 220t4020=244003.HenceZ400V(t)dt=394003, giving an average of9853= 32813litresover that 40 minute period.41
Integration by substitution77. Evaluate the following integrals by making the indicated substitutions,giving your answers as functions of the original variables.(a)Zsinxcos4x dx,u=cosx(b)Zex4-e2xdx,u=12ex(c)Zsech2(u) coth(u)du,x=tanh(u).Solution:(a)u= cosx, sodu=-sinxdx. The integral becomesZu4(-du) =-u55+C=-cos5x5+C.(b)u=ex2,du=ex2dx. The integral becomesZ2du4-4u2=du1-u2= arcsin(u) +C= arcsinex2+C.(c)x= tanhu,dx= sech2udu. The integral becomesZx-1dx= ln|x|+C= ln|tanhu|+C.78. Evaluate the following integrals:(a)Z(2x-1)8dx(b)Z11 + (3y+ 4)2dy(c)Zπ/20tanhz dz.Solution:(a) Substituteu= (2x-1),du= 2dx. The integral becomes12Zu8du=118u9+C=118(2x-1)9+C.(b) Substituteu= 3y+ 4,du= 3dy. The integral becomesZdu/31 +y2=13arctan(y) +C=13arctan(3y+ 4) +C.42
(c) tanhz=sinhzcoshz. Subsu= coshz,du= sinhz dz. Whenz= 0,u= cosh 0 = 1.Whenu=π/2,u= cosh(π/2).Hence theintegral becomesZcosh(π/2)1duu= [ln|u|]cosh(π/2)1= ln(cosh(π/2))-ln(1) = ln(cosh(π/2))0.9199554. 79. Use the given substitution to find each of these integrals: (a)Solution:Substitutingu= 4 +x2we havedu= 2x dxsoRx(4 +x2)10dx=12c.(b)Solution:Letu=t+ 1, thendu=dtandt=u-1,soRt2t+1dt=R(u-1)2udu=Ru2-2u+1udu=Ru3/2-2u1/2+u-1/2du=25u5/2-43u3/2+ 2u1/2+c=25(t+ 1)5/2-43(t+ 1)3/2+2(t+ 1)1/2+c.(c)Zx(4 +x2)10dx,u= 4 +x2R2x(4 +x2)10dx=12Ru10du=u1122+c=(4+x2)1122+Zt2t+ 1dt,u=t+ 1Z3t(t2+ 1)2dt,u=t2+ 1Solution:Letu=t2+ 1, thendu= 2t dt.ThusR3t(t2+1)-2dt=32Ru-2du=-32u+c=-32(t2+1)+c.

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