{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }{PSTYLE " Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "R3 Font 0 " -1 256 1 {CSTYLE "" -1 -1 "Helvetica" 1 14 0 0 0 0 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 14 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 47 "Finding integrals by the \+ Method of Substitution" }}{PARA 0 "" 0 "" {TEXT -1 52 "Example 1. Fin d the antiderivative of the function " }}{PARA 0 "" 0 "" {TEXT -1 40 " " }{XPPEDIT 18 0 "x^4*sqrt(x^5 +1)" "6#*&%\"xG\"\"%-%%sqrtG6#,&*$F$\"\"&\"\"\"\"\"\"F,F," }}{PARA 0 " " 0 "" {TEXT -1 64 "Solution: We will show how to integrate this func tion using the" }}{PARA 0 "" 0 "" {TEXT -1 28 "\"Int\" and \"value\" c ommands.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Int(x^4*sqrt(x^5+1),x ): %=value(%) +C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&%\"xG \"\"%,&*$F(\"\"&\"\"\"F-F-#F-\"\"#F(,&*$F*#\"\"$F/#F/\"#:%\"CGF-" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Example 2. Find the antiderivativ e of the function" }}{PARA 0 "" 0 "" {TEXT -1 56 " \+ " }{XPPEDIT 18 0 "x^3*exp(x^4+1)" "6#*&%\"xG\"\"$-%$expG6#,&*$F$\"\"%\"\"\"\"\"\"F,F," }{TEXT -1 1 "." } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Int(x^3*exp(x^4+1),x) : % = value( %) + C;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&%\"xG\"\"$-%$ex pG6#,&*$F(\"\"%\"\"\"F0F0F0F(,&F*#F0F/%\"CGF0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Example 1a. Use \"changevar\" to evaluate the integra l in Example1." }}{PARA 0 "" 0 "" {TEXT -1 106 "Solution: the first e xample above will now be reworked using \"changevar\". Invoke the \" student\" package." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student) :" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Define the integral." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "I1 := \+ Int(x^4*sqrt(x^5+1),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#I1G-%$In tG6$*&%\"xG\"\"%,&*$F)\"\"&\"\"\"F.F.#F.\"\"#F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 181 "If this expression were integrated by hand the sub stitution u=x^5+1 would be used. The proper syntax in this case is x^ 5+1=u. This reduces the integration problem the following: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "I2 := student[changevar](x^5+1=u,I1,u);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#I2G-%$IntG6$,$*$%\"uG#\"\"\"\"\"# #F,\"\"&F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 173 "The new integratio n problem is equivalent to the original one. However, the new problem \+ is simpler and can be integrated by the power rule discussed in the pr evious section." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$%\"uG#\"\"$\"\"##F(\"#:" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "The value of the original integral is now obtained by using \"subs\" with u = x^5 +1." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "subs(u = x^5+1,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$,&*$%\"xG\"\"&\"\"\"F)F)#\"\"$\"\"##F,\"#:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "As with any antidifferentiation problem you can check your result by" }}{PARA 0 "" 0 "" {TEXT -1 16 "differentiation. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "diff(%,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"xG\"\"%,&*$F$\"\"&\"\"\"F)F)#F)\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "When evaluating a definite integral it i s usually easier to change the limits of integration defined by the tr ansformation." }}{PARA 0 "" 0 "" {TEXT -1 58 "Example 3. Evaluate the \+ definite integral of the function" }}{PARA 0 "" 0 "" {TEXT -1 64 " \+ x^4 *sqrt(x^5+3)" }} {PARA 0 "" 0 "" {TEXT -1 24 "over the interval [0,1]." }}{PARA 0 "" 0 "" {TEXT -1 48 "Solution: Enter the following Maple V statement." }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "II1 := Int(x^4*sqrt(x^5+1),x=0..1); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Proceed just as you would whe n evaluating an indefinite integral with \"changevar\"." }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 41 "II2 := student[changevar](x^5+1=u,II1,u);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "Observe the new limits of integrat ion. Corresponding to x=0 for the original lower limit is " }}{PARA 0 "" 0 "" {TEXT -1 80 " \+ u = 0^5 +1 = 1," }}{PARA 0 "" 0 "" {TEXT -1 57 " and corresponding to x=1 in the original upper limit is " }}{PARA 0 "" 0 "" {TEXT -1 76 " \+ u = 1^5 +1." }}{PARA 0 "" 0 "" {TEXT -1 107 " One can now e valuate the transformed integral without the need for substituting in \+ the original variables." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 187 "When using the method of substitution either by hand or with the \+ \"changevar\" command the goal is to reformulate an integral into a fo rm in which the integral follows from a basic formula." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "Example 4. Evaluate int(x/sqrt(4-9*x^2),x)." }}{PARA 0 "" 0 "" {TEXT -1 137 "Solution: W ith some practice you should eventually be able to solve this by hand. The integral will be calculating by using \"changevar\"." }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 29 "I3 := Int(x/sqrt(4-9*x^2),x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "What substitution should be used? In th is case if u = 4 - 9x^2, then du = -18xdx. This appears to be worth trying." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "I4 := student[changevar ](4-9*x^2=u,I3,u);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "I5 := simplify(I4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "The goal here \+ is to reduce the original problem to one of the basic formulas. We hav e done it and now its okay to apply \"value\"." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "I6 := value(I5);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "Returning to the original variables one has" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "I7 := subs(u=4-9*x^2,I6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "A check of this is performed by differentiation:# " }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "diff(I7,x);" }}}}{MARK "0 0 0" 0 } {VIEWOPTS 1 1 0 1 1 1803 }