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Maple applies these techniques without telling us ab out it and comes up with a final answer. For example," }}{PARA 0 "" 0 "" {TEXT 243 3 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "Int( x^2*exp(2*x),x); value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6 $*&)%\"xG\"\"#\"\"\"-%$expG6#,$*&F)F*F(F*F*F*F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*(#\"\"\"\"\"#F&)%\"xGF'F&-%$expG6#,$*&F'F&F)F&F&F&F& *(F%F&F)F&F*F&!\"\"*&#F&\"\"%F&F*F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Int(x^2/(x^3+2),x);value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&*$)%\"xG\"\"#\"\"\"F+,&*$)F)\"\"$F+F+F*F+!\" \"F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&#\"\"\"\"\"$F&-%#lnG6#,&*$ )%\"xGF'F&F&\"\"#F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 685 "Note that Maple does not add an arbitrary constant C when evaluating indefinite integrals. It does the hard part of the job: provides an antiderivative. It is up to us \+ to remember that an indefinite integral contains an arbitrary constant , as well. We are guessing that Maple found the first integral by inte gration by parts, the second by substitution. Maple didn't bother to t ell us about it, and normally we wouldn't care to know. However, in th is worksheet, since we are supposed to learn the basic methods of inte gration, we shall use Maple to illustrate substitution and integration by parts, and help us understand them. Commands that allow us to do t his are contained in the \"" }{TEXT 203 7 "student" }{TEXT 243 32 "\" \+ package. We load the package." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student);" }}{PARA 206 "" 1 "" {XPPMATH 20 "6#7F%\"DG%%DiffG%*DoubleintG%$IntG%&LimitG%(Linei ntG%(ProductG%$SumG%*TripleintG%*changevarG%(combineG%/completesquareG %)distanceG%'equateG%(extremaG%*integrandG%*interceptG%)intpartsG%(iso lateG%(leftboxG%(leftsumG%)makeprocG%)maximizeG%*middleboxG%*middlesum G%)midpointG%)minimizeG%(powsubsG%)rightboxG%)rightsumG%,showtangentG% (simpsonG%&slopeG%(summandG%*trapezoidG%&valueG" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 0 "" }} {PARA 4 "" 0 "" {TEXT 244 27 "Integration by Substitution" }}{PARA 0 " " 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 243 189 "In this workshe et we are going to consider a lot of integrals and further process out puts of many commands. Hence, in order to avoid retyping, we shall be \+ giving names to most of integrals " }{TEXT 204 9 "Q1, Q2,.." }{TEXT 243 35 ".etc, and names to most of outputs " }{TEXT 205 16 "expr1, exp r2,..." }{TEXT 243 4 "etc." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 243 87 "The command that allows us to see the results o f substitution in a given integral is \"" }{TEXT 206 9 "changevar" } {TEXT 243 69 "\". It works as follows. Suppose we want to perform the \+ substitution " }{XPPEDIT 18 0 "u = x^3+2;" "6#/%\"uG,&*$)%\"xG\"\"$\" \"\"F*\"\"#F*" }{TEXT 243 17 " in the integral " }{XPPEDIT 18 0 "int(x ^2/(x^3+2), x);" "6#-%$intG6$*&)%\"xG\"\"#\"\"\",&*$)F(\"\"$F*F*F)F*! \"\"F(" }{TEXT 243 44 " and see what the integral becomes. We type" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "changevar(u=x^3+2,Int(x^2/ (x^3+2),x),u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*(#\"\"\" \"\"$F)F)F)%\"uG!\"\"F)F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 79 "Observe that first we enter the desired substitution, then the integral, using " }{TEXT 207 3 "no t" }{TEXT 243 3 " \"" }{TEXT 208 3 "int" }{TEXT 243 7 "\" but " }{TEXT 209 18 "the inert version " }{TEXT 243 2 "\"" }{TEXT 210 3 "Int" } {TEXT 243 115 "\" of the integration command, and lastly the name of t he new variable of integration. We used the inert version \"" }{TEXT 211 3 "Int" }{TEXT 243 252 "\" of the integration command, which does \+ not evaluate the integral, because we want to perfom substitution in t he original integral and not in its value. As in the rest of the works heet, we shall give names to the integral and the subsequent outputs. \+ " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Q1:=Int(x^2/(x^3+2),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q1G-%$IntG6$*&*$)%\"xG\"\"#\"\"\"F-,&*$)F+\"\"$F-F-F,F-!\"\"F +" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "expr1:=changevar(u=x^3 +2,Q1,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr1G-%$IntG6$,$*(#\" \"\"\"\"$F+F+F+%\"uG!\"\"F+F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 84 "Although the integral is easy to do by hand, we can ask Maple to eval uate it anyway." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "expr2:=value(expr1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr2G,$*&#\"\"\"\"\"$F(-%#lnG6#%\"uGF(F(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 243 110 "We obtained the integral in term s of u. To have it in terms of x we have to substitute back the formul a for u." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs(u=x^3+2,expr2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&#\"\"\"\"\"$F&-%#lnG6#,&*$)%\"xGF'F&F&\"\"#F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 78 "Observe that Maple returns ln(u), rather than ln|u|, as an ant iderivative of " }{XPPEDIT 18 0 "1/u;" "6#*&\"\"\"F$%\"uG!\"\"" } {TEXT 243 107 " . This limits the antiderivative to the domain u > 0. \+ It is one of a few little imperfections of Maple. " }}{PARA 0 "" 0 " " {TEXT 243 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 15 " The command \+ \"" }{TEXT 212 9 "changevar" }{TEXT 243 67 "\" handles subsitution for definite integrals as well. For example," }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Q2:=Int(cos(x)*e xp(sin(x)),x=0..Pi/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q2G-%$Int G6$*&-%$cosG6#%\"xG\"\"\"-%$expG6#-%$sinGF+F-/F,;\"\"!,$*&#F-\"\"#F-%# PiGF-F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "expr3:=changevar (u=sin(x),Q2,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr3G-%$IntG6$ -%$expG6#%\"uG/F+;\"\"!\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 213 62 "As we see, Maple proper ly adjusted the limits of integration. " }{TEXT 243 97 "Hence, the val ue of the latter integral is equal to the value of the original integr al Q2. Indeed" }}{PARA 0 "" 0 "" {TEXT 243 1 " " }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "value(Q2); value(expr3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%$expG6#\"\"\"F'F'!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%$expG6#\"\"\"F'F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 214 12 "Example 1. " }{TEXT 243 22 "Consider the integr al " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 207 "" 0 "" {XPPEDIT 18 0 "int(x^3*sqrt(4-x^2), x);" "6#-%$intG6$*&)%\"xG\"\"$\"\"\"-%%sqrtG6# ,&\"\"%F**$)F(\"\"#F*!\"\"F*F(" }}{PARA 208 "" 0 "" {TEXT 200 0 "" }} {PARA 0 "" 0 "" {TEXT 243 104 "Find a substitution u = u(x) that chang es the integral into an integral of a polynomial in terms of u. " }} {PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 243 6 "Denote" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Q3:=Int(x^3*sqrt(4-x^2),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%#Q3G-%$IntG6$*&)%\"xG\"\"$\"\"\"-%%sqrtG6#,&\"\"%F,*$)F*\"\"#F,!\" \"F,F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 177 "The rest is pretty mu ch trial and errror. We shall try a few substitutions and see if we ca n generate an integral of a desired form. The first substitution that \+ comes to mind is" }}{PARA 0 "" 0 "" {TEXT 243 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "changevar(u=4-x^2,Q3,u);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#-%$IntG6$,$*(#\"\"\"\"\"#F),&\"\"%F)%\"uG!\"\"F)-%%sq rtG6#F-F)F.F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 179 "We obtained an integral that, after multiplication, consists of powers of u. Hence, \+ it can be done by hand. Still, the integrand is not a polynomial in u. Let's try something else" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "changevar(u=sqrt(4-x^2),Q3,u);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*&,&\"\"%\"\"\"*$)%\"uG\"\" #F*!\"\"F*F,F*F/F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 42 "Done! The \+ integrand is a polynomial in u. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 215 11 "Example 2. " }{TEXT 243 23 " Consider the integrals" }} {PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 209 "" 0 "" {XPPEDIT 18 0 "int( exp(x)/(1+exp(2*x)), x), int(cos(x)/(1+sin(x)^2), x);" "6$-%$intG6$*&- %$expG6#%\"xG\"\"\",&F+F+-F(6#*&\"\"#F+F*F+F+!\"\"F*-F$6$*&-%$cosGF)F+ ,&F+F+*$)-%$sinGF)F0F+F+F1F*" }{TEXT 200 3 " ." }}{PARA 210 "" 0 "" {TEXT 200 0 "" }}{PARA 0 "" 0 "" {TEXT 243 76 "Find a substitution for each of them, so that both integrals are reduced to " }{TEXT 216 8 "t he same" }{TEXT 243 44 " simple and familiar integral in terms of u." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 243 104 "Let's start with the first integral a nd try to find \+ a substitution that reduces it to something simple." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Q4:=Int(e xp(x)/(1+exp(2*x)),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q4G-%$Int G6$*&-%$expG6#%\"xG\"\"\",&F-F--F*6#,$*&\"\"#F-F,F-F-F-!\"\"F," }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "changevar(u=1+exp(2*x),Q4,u) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*(#\"\"\"\"\"#F)F)F)*& -%%sqrtG6#,&F)!\"\"%\"uGF)F)F1F)F0F)F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "changevar(u=exp(2*x),Q4,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*(#\"\"\"\"\"#F)F)F)*&-%%sqrtG6#%\"uGF),&F)F )F/F)F)!\"\"F)F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 68 "None of the \+ two substitutions led us to a simple integral. Let's try" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "chan gevar(u=exp(x),Q4,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\" \"\"F',&F'F'*$)%\"uG\"\"#F'F'!\"\"F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 196 "That's it! The latter integral is simply arctan(u) as you rem ember from last semester. Let's see if we can find a substitution whi ch reduces the second integral to the same simple integral in u. " }} {PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Q5:=Int(cos(x)/(1+(sin(x))^2),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q5G-%$IntG6$*&-%$cosG6#%\"xG\"\"\",&F-F-*$)-%$sinGF+\"\"#F-F- !\"\"F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 62 "A natural thing to tr y is to mimic what we did above, that is " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "changevar(u=cos( x),Q5,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*&%\"uG\"\"\"* &,&\"\"#F)*$)F(F,F)!\"\"F)-%%sqrtG6#,&F)F)F-F/F)F/F/F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 25 "It didn't work. Let's try" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "chang evar(u=sin(x),Q5,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&\" \"\"F',&F'F'*$)%\"uG\"\"#F'F'!\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 318 "Success! We reduced the integral to the same integral in terms of u, which is arc tan(u). You shouldn't think, of course, that the integrals Q4 and Q5 a re equal. To obtain their values in terms of x we have to substitute i nto arctan(u) the corresponding u(x) for each of the integrals. Q4 and Q5 are equal respectively:" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "subs(u=exp(x),arctan(u));" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'arctanG6#-%$expG6#%\"xG" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "subs(u=sin(x),arctan(u));" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'arctanG6#-%$sinG6#%\"xG" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 243 80 "Recall that you can always check \+ if your answers are correct by differentiating." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "diff(arcta n(exp(x)),x); diff(arctan(sin(x)),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%$expG6#%\"xG\"\"\",&F(F(*$)F$\"\"#F(F(!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%$cosG6#%\"xG\"\"\",&F(F(*$)-%$sinGF&\"\"#F(F(!\"\" " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 22 "Everything checks out." }} {PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 217 11 "Example 3. " }{TEXT 243 23 " Consider the integral " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }} {PARA 211 "" 0 "" {XPPEDIT 18 0 "int(5/(x^2+x+5/4), x);" "6#-%$intG6$* &\"\"&\"\"\",(*$)%\"xG\"\"#F(F(F,F(*&F'F(\"\"%!\"\"F(F0F," }{TEXT 200 2 " ." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 243 20 "Find the integral by" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 " " 0 "" {TEXT 243 46 "(a) Completing the square in the denominator." } }{PARA 0 "" 0 "" {TEXT 243 69 "(b) Using a subsitution to reduce the \+ integral to something simple. " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 243 95 "Do you remember how to complete t he square? If not, don't despair. Among the commands in the \"" }{TEXT 218 7 "student" }{TEXT 243 18 "\" package we see " }{TEXT 219 16 "\"c ompletesquare" }{TEXT 243 150 "\". It is easy to guess how it works. W e have to type in a polynomial and tell Maple with respect to which va riable we want it to complete the square." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "completesquare(x ^2+x+5/4,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$),&%\"xG\"\"\"#F( \"\"#F(F*F(F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 20 "Our integra l becomes" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 29 "Q6:=Int(5/((x+(1/2))^2+1),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q6G-%$IntG6$,$*(\"\"&\"\"\"F+F+,&*$),&%\"xGF+#F+\"\" #F+F2F+F+F+F+!\"\"F+F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 31 "We use the obvious substitution" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "changevar(u=x+1/2,Q6,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*(\"\"&\"\"\"F)F),&F)F)*$)%\"uG\" \"#F)F)!\"\"F)F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 41 "Hence, Q6=5a rctan(u)+C=5arctan(x+1/2)+C. " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 4 "" 0 "" {TEXT 244 20 "Integration by Parts" }}{PARA 0 "" 0 "" {TEXT 243 0 "" } }{PARA 0 "" 0 "" {TEXT 243 133 "As you know integration by parts is no thing else but an integral version of the product rule. The formula fo r integration by parts is" }}{PARA 212 "" 0 "" {XPPEDIT 18 0 "int(u*D( v), x) = u*v-int(D(u)*v, x);" "6#/-%$intG6$*&%\"uG\"\"\"-%\"DG6#%\"vGF )%\"xG,&*&F(F)F-F)F)-F%6$*&-F+6#F(F)F-F)F.!\"\"" }{TEXT 200 2 " ," }} {PARA 0 "" 0 "" {TEXT 243 64 "where D denotes the derivative. If we wa nt to find the integral " }{XPPEDIT 18 0 "int(u*D(v), x);" "6#-%$intG6 $*&%\"uG\"\"\"-%\"DG6#%\"vGF(%\"xG" }{TEXT 243 18 " and the integral " }{XPPEDIT 18 0 "int(D(u)*v, x);" "6#-%$intG6$*&-%\"DG6#%\"uG\"\"\"%\" vGF+%\"xG" }{TEXT 243 68 " is simpler, integration by parts helps. Con sider a typical example." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Q7:=Int(x*exp(2*x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q7G-%$IntG6$*&%\"xG\"\"\"-%$expG6#,$*&\"\"#F *F)F*F*F*F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 348 "Observe that, as above, we are using th e inert version of the integral command, as we are going to work with \+ the integral and not its value. Which term do you choose as u and whic h term do you consider as the derivative of something D(v)? Usually, y ou have many possible choices. Maple can show you how each of the choi ces works with the command \"" }{TEXT 220 8 "intparts" }{TEXT 243 12 " \" in the \"" }{TEXT 221 7 "student" }{TEXT 243 32 "\" package. Under \+ the command \"" }{TEXT 222 8 "intparts" }{TEXT 243 206 "\" you have to tell Maple what is your integral, and what term you are choosing as u , that is, the term which during integration by parts will get differe ntiated. For example, in the latter integral choose " }{XPPEDIT 18 0 " exp(2*x);" "6#-%$expG6#*&\"\"#\"\"\"%\"xGF(" }{TEXT 243 125 " as u and consider the term x as the derivative D(v) of some v. Let's see what \+ the formula for integration by parts gives us." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "intparts(Q 7,exp(2*x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(#\"\"\"\"\"#F&-%$e xpG6#,$*&F'F&%\"xGF&F&F&)F-F'F&F&-%$IntG6$*&F(F&F.F&F-!\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 243 51 "According to the formula for inte gration by parts, " }{XPPEDIT 18 0 "exp(2*x);" "6#-%$expG6#*&\"\"#\"\" \"%\"xGF(" }{TEXT 243 56 " is differentiated and x replaced by its ant iderivative " }{XPPEDIT 18 0 "x^2/2;" "6#*&)%\"xG\"\"#\"\"\"F&!\"\"" } {TEXT 243 33 " . The new integral to be found, " }{XPPEDIT 18 0 "int(e xp(2*x)*x^2, x);" "6#-%$intG6$*&-%$expG6#*&\"\"#\"\"\"%\"xGF,F,*$)F-F+ F,F,F-" }{TEXT 243 96 " , is not any easier than the integral that we started from. Let's try a different choice of u." }}{PARA 0 "" 0 "" {TEXT 243 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "expr4:=in tparts(Q7,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr4G,&*(#\"\"\" \"\"#F(%\"xGF(-%$expG6#,$*&F)F(F*F(F(F(F(-%$IntG6$,$*&F'F(F+F(F(F*!\" \"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 138 "This, of course, worked a s the latter integral involving a single exponential is very easy to f ind by hand. We can ask Maple to do it, too" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "value(expr4);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*(#\"\"\"\"\"#F&%\"xGF&-%$expG6#,$* &F'F&F(F&F&F&F&*&#F&\"\"%F&F)F&!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 " " 0 "" {TEXT 223 10 "Example 4." }{TEXT 243 62 " Show by integration b y parts that for any positive integer n " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 213 "" 0 "" {XPPEDIT 18 0 "int(x^n*exp(x), x) = x^n*exp(x )-n*int(x^(n-1)*exp(x), x);" "6#/-%$intG6$*&)%\"xG%\"nG\"\"\"-%$expG6# F)F+F),&F'F+*&F*F+-F%6$*&)F),&F*F+F+!\"\"F+F,F+F)F+F6" }{TEXT 200 3 " \+ . " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 243 78 "T his is a so called reduction formula. It shows that any integral of th e form " }{XPPEDIT 18 0 "int(x^n*exp(x), x);" "6#-%$intG6$*&)%\"xG%\"n G\"\"\"-%$expG6#F(F*F(" }{TEXT 243 89 " can be found after integrating by parts n times. To show that the formula is true define" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Q7 :=Int(x^n*exp(x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q7G-%$IntG6 $*&)%\"xG%\"nG\"\"\"-%$expG6#F*F,F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 25 "Now we integrate by parts" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "expr5:=intparts(Q7,x^n);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr5G,&*&)%\"xG%\"nG\"\"\"-%$expG 6#F(F*F*-%$IntG6$*&*(F'F*F)F*F+F*F*F(!\"\"F(F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 67 "This i sn't exactly the form we want. Let's try the simplify command" }} {PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "simplify(expr5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&)%\"xG% \"nG\"\"\"-%$expG6#F&F(F(*&F'F(-%$IntG6$*&)F&,&F'F(F(!\"\"F(F)F(F&F(F3 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 79 "We have shown that Q7 reduce s to the above formula after integration by parts. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 145 " The next example is a little excercise in processing algebraically exp ressions in Maple. It also introduces you to an important technique ca lled " }{TEXT 224 40 "the method of undetermined coefficients." }} {PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 225 11 "Example 5. " }{TEXT 243 19 " Find the integral " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 214 "" 0 "" {XPPEDIT 18 0 "int(x^3*exp(x), x);" "6#-%$int G6$*&)%\"xG\"\"$\"\"\"-%$expG6#F(F*F(" }{TEXT 200 1 " " }}{PARA 0 "" 0 "" {TEXT 243 36 "without performing any integration. " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 243 159 "Observe that the a bove reduction formula tells us that the integral, which can be obtain ed by integrating by parts three times, will eventually be of the form " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 215 "" 0 "" {XPPEDIT 18 0 "x^3*exp(x)+a*x^2*exp(x)+b*x*exp(x)+c*exp(x);" "6#,**&)%\"xG\"\"$\"\" \"-%$expG6#F&F(F(*(%\"aGF(*$)F&\"\"#F(F(F)F(F(*(%\"bGF(F&F(F)F(F(*&%\" cGF(F)F(F(" }{TEXT 200 2 " ." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }} {PARA 0 "" 0 "" {TEXT 243 196 "for some constants a,b,c. All we have t o do is find the right constants a,b,c. Solving problems by guessing a form of the solution and then working backwards to find the right con stants is called " }{TEXT 226 38 "the method of undetermined coeficien ts" }{TEXT 243 40 ". Let's define an appropriate expression" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "e xpr6:=x^3*exp(x)+a*x^2*exp(x)+b*x*exp(x)+c*exp(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr6G,**&)%\"xG\"\"$\"\"\"-%$expG6#F(F*F**(%\"aGF* )F(\"\"#F*F+F*F**(%\"bGF*F(F*F+F*F**&%\"cGF*F+F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 37 "Since expr6 is an anitiderivative of " } {XPPEDIT 18 0 "x^3*exp(x);" "6#*&)%\"xG\"\"$\"\"\"-%$expG6#F%F'" } {TEXT 243 15 " its derivative" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "expr7:=diff(expr6,x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr7G,0*(\"\"$\"\"\")%\"xG\"\"#F(- %$expG6#F*F(F(*&)F*F'F(F,F(F(**F+F(%\"aGF(F*F(F,F(F(*(F2F(F)F(F,F(F(*& %\"bGF(F,F(F(*(F5F(F*F(F,F(F(*&%\"cGF(F,F(F(" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 243 19 "has to be equal to " }{XPPEDIT 18 0 "x^3*exp(x);" "6#* &)%\"xG\"\"$\"\"\"-%$expG6#F%F'" }{TEXT 243 50 ". We have to find cons tants a,b,c such that expr7=" }{XPPEDIT 18 0 "x^3*exp(x);" "6#*&)%\"xG \"\"$\"\"\"-%$expG6#F%F'" }{TEXT 243 130 ". Let's process expr7 a litt le. There are many commands in Maple which allow you to process algebr aic expressions, among others \"" }{TEXT 227 8 "simplify" }{TEXT 243 6 "\", \"" }{TEXT 228 6 "factor" }{TEXT 243 6 "\", \"" }{TEXT 229 6 "e xpand" }{TEXT 243 6 "\", \"" }{TEXT 230 7 "collect" }{TEXT 243 6 "\", \+ \"" }{TEXT 231 7 "combine" }{TEXT 243 68 "\" etc. We shall slowly lear n by examples how to use them. Let's try" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "expr8:=factor(expr7) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr8G*&-%$expG6#%\"xG\"\"\",0 *&\"\"$F*)F)\"\"#F*F**$)F)F-F*F**(F/F*F)F*%\"aGF*F**&F.F*F3F*F*%\"bGF* *&F)F*F5F*F*%\"cGF*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 39 "Since e xpr7 is supposed to be equal to " }{XPPEDIT 18 0 "exp(x)*x^3;" "6#*&-% $expG6#%\"xG\"\"\"*$)F'\"\"$F(F(" }{TEXT 243 52 " , the polynomial in \+ parentheses has to be equal to " }{XPPEDIT 18 0 "x^3;" "6#*$)%\"xG\"\" $\"\"\"" }{TEXT 243 40 ". Let's get rid of the exponential first" }} {PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "expr9:=expr8/exp(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&expr9 G,0*&\"\"$\"\"\")%\"xG\"\"#F(F(*$)F*F'F(F(*(F+F(F*F(%\"aGF(F(*&F)F(F/F (F(%\"bGF(*&F*F(F1F(F(%\"cGF(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 109 "Now we want to group together terms according to powers of x. Thi s is accomplished by the command \"collect\"" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "expr10:=collect( expr9,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'expr10G,,*$)%\"xG\"\"$ \"\"\"F**&,&%\"aGF*F)F*F*)F(\"\"#F*F**&,&%\"bGF**&F/F*F-F*F*F*F(F*F*F2 F*%\"cGF*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 47 "Since the latter po lynomial has to be equal to " }{XPPEDIT 18 0 "x^3;" "6#*$)%\"xG\"\"$\" \"\"" }{TEXT 243 104 ", the corresponding coefficients must be equal. \+ We solve the system of the following equations for a,b,c" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "solv e(\{a+3=0,b+2*a=0,b+c=0\},\{a,b,c\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<%/%\"aG!\"$/%\"bG\"\"'/%\"cG!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 91 "We have found our constants a,b,c. We substitute them back int o expr6 and obtain our answer" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "an:=subs(a=-3,b=6,c=-6,expr6 );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#anG,**&)%\"xG\"\"$\"\"\"-%$ex pG6#F(F*F**(F)F*)F(\"\"#F*F+F*!\"\"*(\"\"'F*F(F*F+F*F**&F3F*F+F*F1" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 243 50 "To check our answer we can simpl y differentiate it" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "diff(an,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&)%\"xG\"\"$\"\"\"-%$expG6#F%F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 58 "We could also ask Maple to find the integral to begin wi th" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Int(x^3*exp(x),x); value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&)%\"xG\"\"$\"\"\"-%$expG6#F(F*F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&)%\"xG\"\"$\"\"\"-%$expG6#F&F(F(*(F'F()F& \"\"#F(F)F(!\"\"*(\"\"'F(F&F(F)F(F(*&F1F(F)F(F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 27 "but that would be cheating!" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 4 "" 0 "" {TEXT 244 17 " Homework Problems" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 232 15 "Problem 1. " }{TEXT 243 52 "Find a substitution u=u( x) that reduces the integral" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }} {PARA 216 "" 0 "" {XPPEDIT 18 0 "int(x/sqrt(2*x+3), x);" "6#-%$intG6$* &%\"xG\"\"\"-%%sqrtG6#,&*&\"\"#F(F'F(F(\"\"$F(!\"\"F'" }{TEXT 200 1 " \+ " }}{PARA 0 "" 0 "" {TEXT 243 46 "to an integral of a polynomial in te rms of u. " }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 233 13 "Problem 2. " }{TEXT 243 83 " Find a substitution for each of the following two integrals which reduces them to " }{TEXT 234 8 "the same" }{TEXT 243 28 " easy integral in terms of u" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 217 "" 0 "" {XPPEDIT 18 0 "int(sqrt(x+3), x), i nt(sqrt(3+sqrt(x))/(2*sqrt(x)), x);" "6$-%$intG6$-%%sqrtG6#,&%\"xG\"\" \"\"\"$F+F*-F$6$*&-F'6#,&F,F+-F'6#F*F+F+*&\"\"#F+F3F+!\"\"F*" }{TEXT 200 3 " ." }}{PARA 218 "" 0 "" {TEXT 200 0 "" }}{PARA 0 "" 0 "" {TEXT 235 13 "Problem 3. " }{TEXT 243 98 "Let n be a positive integer, a \+ a nonzero constant. Show that the following reduction formula holds" } }{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 219 "" 0 "" {XPPEDIT 18 0 "int (x^n*cos(a*x), x) = x^n*sin(a*x)/a-n*int(x^(n-1)*sin(a*x), x)/a;" "6#/ -%$intG6$*&)%\"xG%\"nG\"\"\"-%$cosG6#*&%\"aGF+F)F+F+F),&*(F(F+-%$sinGF .F+F0!\"\"F+*(F*F+-F%6$*&)F),&F*F+F+F5F+F3F+F)F+F0F5F5" }{TEXT 200 6 " ." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}{PARA 0 "" 0 "" {TEXT 236 13 "Problem 4. " }{TEXT 243 28 "Usi ng the reduction formula " }}{PARA 220 "" 0 "" {XPPEDIT 18 0 "int(x^n* exp(x), x) = x^n*exp(x)-n*int(x^(n-1)*exp(x), x);" "6#/-%$intG6$*&)%\" xG%\"nG\"\"\"-%$expG6#F)F+F),&F'F+*&F*F+-F%6$*&)F),&F*F+F+!\"\"F+F,F+F )F+F6" }{TEXT 200 2 " ," }}{PARA 0 "" 0 "" {TEXT 243 30 "guess the for m of the integral" }}{PARA 221 "" 0 "" {XPPEDIT 18 0 "int(x^5*exp(x), \+ x);" "6#-%$intG6$*&)%\"xG\"\"&\"\"\"-%$expG6#F(F*F(" }{TEXT 200 2 " ." }}{PARA 0 "" 0 "" {TEXT 243 85 "Then find the integral using the meth od of undetermined coefficients as in Example 5." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 243 1 " " }{TEXT 237 10 "Problem 5." }{TEXT 243 19 " Have Maple find " }{XPPEDIT 18 0 "in t(x^n*exp(x^2), x);" "6#-%$intG6$*&)%\"xG%\"nG\"\"\"-%$expG6#*$)F(\"\" #F*F*F(" }{TEXT 243 24 " for several values of " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT 243 7 ", say, " }{XPPEDIT 18 0 "n = 1, 2, 3, 4, 5;" " 6'/%\"nG\"\"\"\"\"#\"\"$\"\"%\"\"&" }{TEXT 243 37 "...12. Note that f or some values of " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "n;" "6#%\"nG" } {TEXT 243 89 " Maple gives an answer entirely in terms of familiar fun ctions, but for other values of " }{XPPEDIT 18 0 "n;" "6#%\"nG" } {TEXT 243 26 " the answer involves some " }}{PARA 0 "" 0 "" {TEXT 243 27 "mysterious function called " }{TEXT 238 6 "erf. " }}{PARA 0 "" 0 "" {TEXT 239 2 " " }{TEXT 243 160 "(a) Which values of n result in on ly familiar functions? Make a conjecture (a guess) about the general s ituation, not just the values of n you actually checked." }}{PARA 0 "" 0 "" {TEXT 243 107 " (b) Explain why your answer to part (a) is rea sonable, in terms of what you know about substitution and " }}{PARA 0 "" 0 "" {TEXT 243 23 "integration by parts. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 240 6 "Note. " }{TEXT 243 15 "If you use the " }{TEXT 241 7 "restart" }{TEXT 243 73 " command in your homework worksheet, do not forget to reload the packa ge " }{TEXT 242 13 "with(student)" }{TEXT 243 1 "." }}{PARA 0 "" 0 "" {TEXT 243 0 "" }}}{EXCHG {PARA 222 "" 0 "" {TEXT 200 77 "MTH 142 Maple Worksheets written by B. Kaskosz and L. Pakula, Copyright 1999." }}} {EXCHG {PARA 223 "" 0 "" {TEXT 200 26 "Last modified August 1999." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }