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{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 }
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