Calculus1 Single Step Problems - Maple Help

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Calculus1 Single Stepping

The Calculus1 subpackage of the Student package includes commands that single-step through the basic problems of single-variable calculus, namely limit, differentiation, and integration problems.  This worksheet demonstrates the basics of this functionality.

For further information about any command in the Calculus1 package, see the corresponding help page.  For a general overview, see Calculus1.

Getting Started

While any command in the package can be referred to using the long form, for example, Student[Calculus1][Rule], it is easier, and often clearer, to load the package, and then use the short form command names.

 > $\mathrm{restart}$
 > $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Calculus1}\right]\right):$

Several commands in the Calculus1 package provide helpful information through the Maple userinfo facility.  To see these information messages, set the corresponding infolevel table entry, using Student or Student[Calculus1] as the index.

 > $\mathrm{infolevel}\left[\mathrm{Student}\left[\mathrm{Calculus1}\right]\right]≔1:\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}$

The single stepping functionality consists of sets of rules for each basic operation, limit, differentiation, and integration.  Rules are applied by calling the package command Rule, providing the name of a rule and a problem to which that rule is applied.  Problems are represented using the inert forms  Limit, Diff, and Int.

 > $\underset{x→3}{{lim}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({x}^{2}\right)$
 $\underset{{x}{\to }{3}}{{lim}}{}{{x}}^{{2}}$ (1.1)
 > $\mathrm{Rule}\left[\mathrm{power}\right]\left(\right)$
 Creating problem #1
 $\underset{{x}{\to }{3}}{{lim}}{}{{x}}^{{2}}{=}{\left(\underset{{x}{\to }{3}}{{lim}}{}{x}\right)}^{{2}}$ (1.2)

Notice that the rule is specified as the index to the Rule command, while the problem is given as the argument.

The output from a call to Rule is normally an equation in which the left-hand side is the original problem and the right-hand side is the current state of that problem.  This output can be passed to another application of the Rule command.

 > $\mathrm{Rule}\left[\mathrm{identity}\right]\left(\right)$
 $\underset{{x}{\to }{3}}{{lim}}{}{{x}}^{{2}}{=}{9}$ (1.3)

To see the steps from the initial state of a problem to its current state, use the ShowSteps command.

 > $\mathrm{ShowSteps}\left(\right)$
 $\begin{array}{cccc}\multicolumn{4}{c}{\underset{{x}{\to }{3}}{{lim}}{}{{x}}^{{2}}}\\ \phantom{\rule[-0.0ex]{5.0em}{0.0ex}}& {\text{=}}& {\left(\underset{{x}{\to }{3}}{{lim}}{}{x}\right)}^{{2}}\hfill & \phantom{\rule[-0.0ex]{1.0em}{0.0ex}}\left[{\mathrm{power}}\right]\\ \phantom{\rule[-0.0ex]{5.0em}{0.0ex}}& {\text{=}}& {9}\hfill & \phantom{\rule[-0.0ex]{1.0em}{0.0ex}}\left[{\mathrm{identity}}\right]\end{array}$ (1.4)

A differentiation example.

 > $\mathrm{Rule}\left[\mathrm{product}\right]\left(\frac{{ⅆ}}{{ⅆ}x}\left({x}^{2}\mathrm{sin}\left(x\right)\right)\right)$
 Creating problem #2
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{2}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}\left({\mathrm{Diff}}{}\left({{x}}^{{2}}{,}{x}\right)\right){\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}\left({\mathrm{Diff}}{}\left({\mathrm{sin}}{}\left({x}\right){,}{x}\right)\right)$ (1.5)

The more common rules have short form equivalents: difference = -, product = *, power = ^, quotient = /, constantmultiple = c*.  For completeness, there is also sum = +, though both are 3 characters long.

 > $\mathrm{Rule}\left[\mathrm{^}\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{2}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}{2}{x}{\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}\left({\mathrm{Diff}}{}\left({\mathrm{sin}}{}\left({x}\right){,}{x}\right)\right)$ (1.6)

The name of any univariate function can be given as a rule.

 > $\mathrm{Rule}\left[\mathrm{sin}\right]\left(\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{\mathrm{sin}}{}\left({x}\right)\right){=}{2}{x}{\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}{\mathrm{cos}}{}\left({x}\right)$ (1.7)

The Calculus1 package maintains a record of each problem you have worked on in a session.  You can switch between problems using the GetProblem command.

 > $\mathrm{GetProblem}\left(1\right)$
 $\underset{{x}{\to }{3}}{{lim}}{}{{x}}^{{2}}{=}{9}$ (1.8)

You can display the state of any problem, or of all problems, using the Show command.

 > $\mathrm{Show}\left(2\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{\mathrm{sin}}{}\left({x}\right)\right){=}{2}{x}{\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}{\mathrm{cos}}{}\left({x}\right)$ (1.9)
 > $\mathrm{Show}\left(\mathrm{all}\right)$
 Problem 1:
 $\underset{{x}{\to }{3}}{{lim}}{}{{x}}^{{2}}{=}{9}$
 Problem 2:
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{\mathrm{sin}}{}\left({x}\right)\right){=}{2}{x}{\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}{\mathrm{cos}}{}\left({x}\right)$

If you are uncertain about which rule to apply next to a problem, ask for a hint.

 > $\frac{{ⅆ}}{{ⅆ}x}\mathrm{sin}\left({x}^{2}\right)$
 ${\mathrm{Diff}}{}\left({\mathrm{sin}}{}\left({{x}}^{{2}}\right){,}{x}\right)$ (2.1)
 > $\mathrm{Hint}\left(\right)$
 Creating problem #3
 $\left[{\mathrm{chain}}\right]$ (2.2)

The output from the Hint command can always be used as the index to the Rule command.

 > $\mathrm{Rule}\left[\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{3}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}\left({\mathrm{Eval}}{}\left({\mathrm{Diff}}{}\left({\mathrm{sin}}{}\left({\mathrm{_X0}}\right){,}{\mathrm{_X0}}\right){,}{\mathrm{_X0}}{=}{{x}}^{{2}}\right)\right)\left({\mathrm{Diff}}{}\left({{x}}^{{2}}{,}{x}\right)\right)$ (2.3)
 > ${∫}x\mathrm{cos}\left(x\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x$
 ${\mathrm{Int}}{}\left({x}{\mathrm{cos}}{}\left({x}\right){,}{x}\right)$ (2.4)
 > $\mathrm{Hint}\left(\right)$
 Creating problem #4
 $\left[{\mathrm{parts}}{,}{x}{,}{\mathrm{sin}}{}\left({x}\right){,}{1}{,}{\mathrm{cos}}{}\left({x}\right){,}{x}\right]$ (2.5)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{4}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}{x}{\mathrm{sin}}{}\left({x}\right){-}\left({\mathrm{Int}}{}\left({\mathrm{sin}}{}\left({x}\right){,}{x}\right)\right)$ (2.6)
 > $\underset{x→{0}^{+}}{{lim}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{x}^{x}$
 $\underset{{x}{\to }{0}{+}}{{lim}}{}{{x}}^{{x}}$ (2.7)
 > $\mathrm{Hint}\left(\right)$
 Creating problem #5 Rewrite the expression as an exponential to prepare for using l'Hôpital's rule
 $\left[{\mathrm{rewrite}}{,}{{x}}^{{x}}{=}{{ⅇ}}^{{x}{\mathrm{ln}}{}\left({x}\right)}\right]$ (2.8)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 $\underset{{x}{\to }{0}{+}}{{lim}}{}{{x}}^{{x}}{=}\underset{{x}{\to }{0}{+}}{{lim}}{}{{ⅇ}}^{{x}{\mathrm{ln}}{}\left({x}\right)}$ (2.9)
 > $\mathrm{Rule}\left[\mathrm{exp}\right]\left(\right)$
 $\underset{{x}{\to }{0}{+}}{{lim}}{}{{x}}^{{x}}{=}{{ⅇ}}^{\underset{{x}{\to }{0}{+}}{{lim}}{}{x}{\mathrm{ln}}{}\left({x}\right)}$ (2.10)
 > $\mathrm{Hint}\left(\right)$
 $\left[{\mathrm{lhopital}}{,}{\mathrm{ln}}{}\left({x}\right)\right]{,}\left[{\mathrm{lhopital}}{,}{x}\right]$ (2.11)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 $\underset{{x}{\to }{0}{+}}{{lim}}{}{{x}}^{{x}}{=}{{ⅇ}}^{\underset{{x}{\to }{0}{+}}{{lim}}{}\left({-}{x}\right)}$ (2.12)
 > $\mathrm{Rule}\left[\mathrm{c*}\right]\left(\right)$
 $\underset{{x}{\to }{0}{+}}{{lim}}{}{{x}}^{{x}}{=}{{ⅇ}}^{{-}\left(\underset{{x}{\to }{0}{+}}{{lim}}{}{x}\right)}$ (2.13)
 > $\mathrm{Rule}\left[\mathrm{identity}\right]\left(\right)$
 $\underset{{x}{\to }{0}{+}}{{lim}}{}{{x}}^{{x}}{=}{1}$ (2.14)

Learning with You

As you develop expertise with the elementary rules of single-variable calculus, you can automatically apply those rules so that you can concentrate on learning new material.  You do this with the Understand command.  The first argument to Understand is the name of the operation; the remaining arguments are the rules to apply automatically to a problem using that operation.

 > $\mathrm{Understand}\left(\mathrm{Diff},\mathrm{constant},\mathrm{c*},\mathrm{+}\right)$
 ${\mathrm{Diff}}{=}\left[{\mathrm{constant}}{,}{\mathrm{constantmultiple}}{,}{\mathrm{sum}}\right]$ (3.1)

After the product rule is applied, the system applies each applicable understood rule.

 > $\mathrm{Rule}\left[\mathrm{*}\right]\left(\frac{{ⅆ}}{{ⅆ}x}\left({x}^{2}\left(\mathrm{cos}\left(x\right)-2\mathrm{ln}\left(x\right)\right)\right)\right)$
 Creating problem #6
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{6}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}\left({\mathrm{Diff}}{}\left({{x}}^{{2}}{,}{x}\right)\right)\left({\mathrm{cos}}{}\left({x}\right){-}{2}{\mathrm{ln}}{}\left({x}\right)\right){+}{{x}}^{{2}}\left({\mathrm{Diff}}{}\left({\mathrm{cos}}{}\left({x}\right){,}{x}\right){-}{2}\left({\mathrm{Diff}}{}\left({\mathrm{ln}}{}\left({x}\right){,}{x}\right)\right)\right)$ (3.2)

To apply all applicable understood rules, use an empty index on the Rule command.

 > ${\mathrm{Rule}}_{}\left[\right]\left(\frac{{ⅆ}}{{ⅆ}x}\left(3x+{ⅇ}^{x}\right)\right)$
 Creating problem #7
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{7}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}{3}\left({\mathrm{Diff}}{}\left({x}{,}{x}\right)\right){+}{\mathrm{Diff}}{}\left({\mathrm{exp}}{}\left({x}\right){,}{x}\right)$ (3.3)

A More Complicated Example

Consider this integration problem.

 > ${{∫}}_{a}^{b}\frac{{w}^{\frac{1}{3}}}{{w}^{\frac{1}{2}}+1}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}w$
 ${\mathrm{Int}}{}\left(\frac{{{w}}^{{1}{/}{3}}}{\sqrt{{w}}{+}{1}}{,}{w}{=}{a}{..}{b}\right)$ (4.1)
 > $\mathrm{Hint}\left(\right)$
 Creating problem #8
 $\left[{\mathrm{change}}{,}{w}{=}{{u}}^{{6}}{,}{u}\right]$ (4.2)

When Maple applies this change of variables, the end points of the definite integration must be addressed.  In this case, the new end points are a^(1/6) and b^(1/6).  Because this involves the introduction of radicals, the end points remain in the original variable.

 > $\mathrm{Rule}\left[\right]\left(\right)$
 Applying substitution w = u^6, u = w^(1/6) with dw = 6*u^5*du, du = 1/6/w^(5/6)*dw
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}{\mathrm{Int}}{}\left({6}{{u}}^{{4}}{-}{6}{u}{+}\frac{{6}{u}}{{{u}}^{{3}}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.3)
 > $\mathrm{Understand}\left(\mathrm{int},\mathrm{constant},\mathrm{c*},\mathrm{+}\right)$
 ${\mathrm{Int}}{=}\left[{\mathrm{constant}}{,}{\mathrm{constantmultiple}}{,}{\mathrm{sum}}\right]$ (4.4)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}{6}\left({\mathrm{Int}}{}\left({{u}}^{{4}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right){-}{6}\left({\mathrm{Int}}{}\left({u}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right){+}{6}\left({\mathrm{Int}}{}\left(\frac{{u}}{{{u}}^{{3}}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right)$ (4.5)
 > 
 > $\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}\right]{,}\left[{}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}{6}\left({\mathrm{Eval}}{}\left(\frac{{1}}{{5}}{{u}}^{{5}}{,}{w}{=}{a}{..}{b}\right)\right){-}{6}\left({\mathrm{Int}}{}\left({u}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right){+}{6}\left({\mathrm{Int}}{}\left(\frac{{u}}{{{u}}^{{3}}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right)$ (4.6)
 > $\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}\right]{,}\left[{}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}{6}\left({\mathrm{Eval}}{}\left(\frac{{1}}{{5}}{{u}}^{{5}}{,}{w}{=}{a}{..}{b}\right)\right){-}{6}\left({\mathrm{Eval}}{}\left(\frac{{1}}{{2}}{{u}}^{{2}}{,}{w}{=}{a}{..}{b}\right)\right){+}{6}\left({\mathrm{Int}}{}\left(\frac{{u}}{{{u}}^{{3}}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right)$ (4.7)
 > $\mathrm{Hint}\left(\right)$
 $\left[{\mathrm{partialfractions}}\right]$ (4.8)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}\right]{,}\left[{}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}{6}\left({\mathrm{Eval}}{}\left(\frac{{1}}{{5}}{{u}}^{{5}}{,}{w}{=}{a}{..}{b}\right)\right){-}{6}\left({\mathrm{Eval}}{}\left(\frac{{1}}{{2}}{{u}}^{{2}}{,}{w}{=}{a}{..}{b}\right)\right){+}{2}\left({\mathrm{Int}}{}\left(\frac{{u}{+}{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right){-}{2}\left({\mathrm{Int}}{}\left(\frac{{1}}{{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right)$ (4.9)

At this point, there are two integration subproblems.  The system maintains each of these subproblems separately, and assigns each a label.  To work on a specific subproblem, for example, the more complicated one, obtain its label using the ShowIncomplete command, and then use that label in a call to Rule or Hint.

 > $\mathrm{ShowIncomplete}\left(\right)$
 ${\mathrm{%Int23}}{=}{\mathrm{Int}}{}\left(\frac{{u}{+}{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$
 ${\mathrm{%Int24}}{=}{\mathrm{Int}}{}\left(\frac{{1}}{{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.10)

(Because this worksheet must execute correctly independent of the labels, it uses a programmatic approach to obtain the label corresponding to the more complicated subproblem.)

 > $L:=\mathrm{ShowIncomplete}\left(\mathrm{data}\right):\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}$$\mathrm{lbl}:=\mathrm{if}\left(\mathrm{length}\left({L}_{2}\right)<\mathrm{length}\left({L}_{1}\right),{L}_{1,1},{L}_{2,1}\right)$
 ${\mathrm{lbl}}{≔}{\mathrm{%Int23}}$ (4.11)

If the hint is complicated, a message providing additional information may be provided.  As mentioned above, to see these messages you must assign a number (at least 1) to infolevel[Calculus1] or infolevel[Student].

 > $\mathrm{Hint}\left(\mathrm{lbl}\right)$
 Rewrite the numerator in a form which contains the derivative of the denominator
 $\left[{\mathrm{rewrite}}{,}\frac{{u}{+}{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{=}\frac{{2}{u}{-}{1}}{{2}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}{+}\frac{{3}}{{2}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}\right]$ (4.12)

Warning: These subproblem labels are maintained internally by the Calculus1 package and can change between applications of the Rule command. Therefore, before referencing a subproblem by a label, use the ShowIncomplete command to verify that the label is correct.

 > $\mathrm{Rule}\left[\right]\left(\mathrm{lbl}\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}\frac{{1}}{{2}}{\mathrm{Int}}{}\left(\frac{{2}{u}{-}{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right){+}\frac{{3}}{{2}}{\mathrm{Int}}{}\left(\frac{{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.13)
 > $\mathrm{Hint}\left(\right)$
 Note that the derivative of u^2-u+1 is 2*u-1, so we can make a change of variable.
 $\left[{\mathrm{change}}{,}{v}{=}{{u}}^{{2}}{-}{u}{+}{1}{,}{v}\right]$ (4.14)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 Applying substitution u = 1/2+1/2*(-3+4*v)^(1/2), v = u^2-u+1 with du = 1/(-3+4*v)^(1/2)*dv, dv = (2*u-1)*du
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}\frac{{1}}{{2}}{\mathrm{Int}}{}\left(\frac{{1}}{{v}}{,}{v}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right){+}\frac{{3}}{{2}}{\mathrm{Int}}{}\left(\frac{{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.15)
 > $\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}\frac{{1}}{{2}}{\mathrm{Eval}}{}\left({\mathrm{ln}}{}\left({v}\right){,}{w}{=}{a}{..}{b}\right){+}\frac{{3}}{{2}}{\mathrm{Int}}{}\left(\frac{{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.16)

Since the variable u1 does not appear in any incomplete subproblems, revert the corresponding substitution.

 > $\mathrm{Hint}\left(\right)$
 $\left[{\mathrm{revert}}\right]$ (4.17)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 Reverting substitution using v = u^2-u+1
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}\frac{{1}}{{2}}{\mathrm{Eval}}{}\left({\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right){,}{w}{=}{a}{..}{b}\right){+}\frac{{3}}{{2}}{\mathrm{Int}}{}\left(\frac{{1}}{{{u}}^{{2}}{-}{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.18)
 > $\mathrm{Hint}\left(\right)$
 Complete the square and make a change of variable.
 $\left[{\mathrm{change}}{,}{v}{=}{u}{-}\frac{{1}}{{2}}{,}{v}\right]$ (4.19)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 Applying substitution u = 1/2+v with du=dv
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}\frac{{1}}{{2}}{\mathrm{Eval}}{}\left({\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right){,}{w}{=}{a}{..}{b}\right){+}{6}\left({\mathrm{Int}}{}\left(\frac{{1}}{{4}{{v}}^{{2}}{+}{3}}{,}{v}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)\right)$ (4.20)
 > $\mathrm{Rule}\left[\mathrm{Hint}\left(\right)\right]\left(\right)$
 Integrals involving expressions of the form (x^2+a^2)^n or sqrt(x^2+a^2)^n can often be simplified using the substitution x = a*tan(u). Applying substitution v = 1/2*3^(1/2)*tan(z), z = arctan(2/3*v*3^(1/2)) with dv = 1/2*3^(1/2)*(1+tan(z)^2)*dz, dz = 2/3*3^(1/2)/(4/3*v^2+1)*dv
 ${{\int }}_{{w}{=}{a}}^{{w}{=}{b}}\frac{{u}{+}{1}}{{{u}}^{{2}}{-}{u}{+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}\frac{\left(\genfrac{}{}{0}{}{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}}{{w}{=}{a}{..}{b}}\right)}{{2}}{+}{6}\left(\genfrac{}{}{0}{}{\frac{\sqrt{{3}}{z}}{{6}}}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{\frac{\sqrt{{3}}{z}}{{6}}}}{{w}{=}{a}{..}{b}}\right)$ (4.21)
 > $\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$
 Reverting substitution using z = arctan(2/3*v*3^(1/2))
 ${{\int }}_{{w}{=}{a}}^{{w}{=}{b}}\frac{{u}{+}{1}}{{{u}}^{{2}}{-}{u}{+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}\frac{\left(\genfrac{}{}{0}{}{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}}{{w}{=}{a}{..}{b}}\right)}{{2}}{+}{6}\left(\genfrac{}{}{0}{}{\frac{\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}{v}\sqrt{{3}}}{{3}}\right)}{{6}}}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{\frac{\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}{v}\sqrt{{3}}}{{3}}\right)}{{6}}}}{{w}{=}{a}{..}{b}}\right)$ (4.22)

After Maple reverts this last u1 substitution, this subproblem is done.  The system then substitutes the result of the subproblem into the larger problem.

 > $\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$
 Reverting substitution using v = u-1/2
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}{,}{4}\right]{,}\left[{4}{,}{4}{,}{4}{,}{4}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}\frac{{1}}{{6}}{\mathrm{Eval}}{}\left({\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right){,}{w}{=}{a}{..}{b}\right){+}{2}\left({\mathrm{Eval}}{}\left(\frac{{1}}{{6}}\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}}{{3}}\left({u}{-}\frac{{1}}{{2}}\right)\sqrt{{3}}\right){,}{w}{=}{a}{..}{b}\right)\right){-}\frac{{1}}{{3}}{\mathrm{Int}}{}\left(\frac{{1}}{{u}{+}{1}}{,}{u}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.23)
 > $\mathrm{Rule}\left[\mathrm{change},\mathrm{u1}=u+1\right]\left(\right)$
 Applying substitution u = -1+u1 with du=du1
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{8}{,}\left[{4}{,}{4}{,}{4}{,}{4}{,}{4}{,}{4}\right]{,}\left[{4}{,}{4}{,}{4}{,}{4}\right]\right]{,}\left\{{a}{,}{b}{,}{w}\right\}\right){=}\frac{{1}}{{6}}{\mathrm{Eval}}{}\left({\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right){,}{w}{=}{a}{..}{b}\right){+}{2}\left({\mathrm{Eval}}{}\left(\frac{{1}}{{6}}\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}}{{3}}\left({u}{-}\frac{{1}}{{2}}\right)\sqrt{{3}}\right){,}{w}{=}{a}{..}{b}\right)\right){-}\frac{{1}}{{3}}{\mathrm{Int}}{}\left(\frac{{1}}{{\mathrm{u1}}}{,}{\mathrm{u1}}{=}\left({w}{=}{a}\right){..}\left({w}{=}{b}\right)\right)$ (4.24)
 > $\mathrm{Rule}\left[\mathrm{^}\right]\left(\right)$
 ${{\int }}_{{w}{=}{a}}^{{w}{=}{b}}\left(\frac{{u}{+}{1}}{{3}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}{-}\frac{{1}}{{3}\left({u}{+}{1}\right)}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}\frac{\left(\genfrac{}{}{0}{}{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}}{{w}{=}{a}{..}{b}}\right)}{{6}}{+}{2}\left(\genfrac{}{}{0}{}{\frac{\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}\left({u}{-}\frac{{1}}{{2}}\right)\sqrt{{3}}}{{3}}\right)}{{6}}}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{\frac{\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}\left({u}{-}\frac{{1}}{{2}}\right)\sqrt{{3}}}{{3}}\right)}{{6}}}}{{w}{=}{a}{..}{b}}\right){-}\frac{\left(\genfrac{}{}{0}{}{{\mathrm{ln}}{}\left({\mathrm{u1}}\right)}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{{\mathrm{ln}}{}\left({\mathrm{u1}}\right)}}{{w}{=}{a}{..}{b}}\right)}{{3}}$ (4.25)
 > $\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$
 Reverting substitution using u1 = u+1
 ${{\int }}_{{a}}^{{b}}\frac{{{w}}^{{1}}{{3}}}}{\sqrt{{w}}{+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{w}{=}{6}\left(\genfrac{}{}{0}{}{\frac{{{u}}^{{5}}}{{5}}}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{\frac{{{u}}^{{5}}}{{5}}}}{{w}{=}{a}{..}{b}}\right){-}{6}\left(\genfrac{}{}{0}{}{\frac{{{u}}^{{2}}}{{2}}}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{\frac{{{u}}^{{2}}}{{2}}}}{{w}{=}{a}{..}{b}}\right){+}\genfrac{}{}{0}{}{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{{\mathrm{ln}}{}\left({{u}}^{{2}}{-}{u}{+}{1}\right)}}{{w}{=}{a}{..}{b}}{+}{12}\left(\genfrac{}{}{0}{}{\frac{\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}\left({u}{-}\frac{{1}}{{2}}\right)\sqrt{{3}}}{{3}}\right)}{{6}}}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{\frac{\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{{2}\left({u}{-}\frac{{1}}{{2}}\right)\sqrt{{3}}}{{3}}\right)}{{6}}}}{{w}{=}{a}{..}{b}}\right){-}{2}\left(\genfrac{}{}{0}{}{{\mathrm{ln}}{}\left({u}{+}{1}\right)}{\phantom{{w}{=}{a}{..}{b}}}{|}\genfrac{}{}{0}{}{\phantom{{\mathrm{ln}}{}\left({u}{+}{1}\right)}}{{w}{=}{a}{..}{b}}\right)$ (4.26)

Finally, revert the original substitution. The problem is done.

 > $\mathrm{Rule}\left[\mathrm{revert}\right]\left(\right)$
 Reverting substitution using u = w^(1/6)
 ${{\int }}_{{a}}^{{b}}\frac{{{w}}^{{1}}{{3}}}}{\sqrt{{w}}{+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{w}{=}\frac{{6}{{b}}^{{5}}{{6}}}}{{5}}{-}\frac{{6}{{a}}^{{5}}{{6}}}}{{5}}{-}{3}{{b}}^{{1}}{{3}}}{+}{3}{{a}}^{{1}}{{3}}}{+}{\mathrm{ln}}{}\left({{b}}^{{1}}{{3}}}{-}{{b}}^{{1}}{{6}}}{+}{1}\right){-}{\mathrm{ln}}{}\left({{a}}^{{1}}{{3}}}{-}{{a}}^{{1}}{{6}}}{+}{1}\right){+}{2}\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{\sqrt{{3}}\left({2}{{b}}^{{1}}{{6}}}{-}{1}\right)}{{3}}\right){-}{2}\sqrt{{3}}{\mathrm{arctan}}{}\left(\frac{\sqrt{{3}}\left({2}{{a}}^{{1}}{{6}}}{-}{1}\right)}{{3}}\right){-}{2}{\mathrm{ln}}{}\left({{b}}^{{1}}{{6}}}{+}{1}\right){+}{2}{\mathrm{ln}}{}\left({{a}}^{{1}}{{6}}}{+}{1}\right)$ (4.27)

Experimenting

Integration problems, in particular, can be quite difficult, and learning by experimentation is a valuable technique.  To facilitate this, the Calculus1 package includes an Undo command, which can be used to reverse to a previous state of a problem.

 > ${∫}\sqrt{1-{x}^{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}x$
 ${\mathrm{Int}}{}\left(\sqrt{{-}{{x}}^{{2}}{+}{1}}{,}{x}\right)$ (5.1)
 > $\mathrm{Rule}\left[\mathrm{change},x=\mathrm{tan}\left(u\right)\right]\left(\right)$
 Creating problem #9 Applying substitution x = tan(u), u = arctan(x) with dx = (1+tan(u)^2)*du, du = 1/(x^2+1)*dx
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{9}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}{\mathrm{Int}}{}\left(\sqrt{{-}{{\mathrm{tan}}{}\left({u}\right)}^{{2}}{+}{1}}{{\mathrm{tan}}{}\left({u}\right)}^{{2}}{,}{u}\right){+}{\mathrm{Int}}{}\left(\sqrt{{-}{{\mathrm{tan}}{}\left({u}\right)}^{{2}}{+}{1}}{,}{u}\right)$ (5.2)
 > $\mathrm{Undo}\left(\right)$
 ${\mathrm{Int}}{}\left(\sqrt{{-}{{x}}^{{2}}{+}{1}}{,}{x}\right)$ (5.3)
 > $\mathrm{Rule}\left[\mathrm{change},x=\mathrm{sin}\left(u\right)\right]\left(\right)$
 Applying substitution x = sin(u), u = arcsin(x) with dx = cos(u)*du, du = 1/(-x^2+1)^(1/2)*dx
 ${\mathrm{CALCULUS1OBJECT}}{}\left(\left[{9}{,}\left[{}\right]{,}\left[{}\right]\right]{,}\left\{{x}\right\}\right){=}{u}{-}\left({\mathrm{Int}}{}\left({{\mathrm{sin}}{}\left({u}\right)}^{{2}}{,}{u}\right)\right)$ (5.4)

Tutors

These tutors provide the same functionality as single stepping through a single problem in a worksheet, in a graphical environment

 > $\mathrm{DiffTutor}\left(\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({x}{\mathrm{sin}}{}\left({x}\right)\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({x}{\mathrm{sin}}{}\left({x}\right)\right)$ (6.1)
 > $\mathrm{IntTutor}\left(\right)$
 ${\mathrm{Int}}{}\left({{\mathrm{sin}}{}\left({x}\right)}^{{2}}{,}{x}\right)$
 ${\mathrm{Int}}{}\left({{\mathrm{sin}}{}\left({x}\right)}^{{2}}{,}{x}\right)$ (6.2)
 > $\mathrm{LimitTutor}\left(\right)$
 $\underset{{x}{\to }{0}}{{lim}}{}{x}{\mathrm{cos}}{}\left({x}\right){\mathrm{ln}}{}\left({x}\right)$
 $\underset{{x}{\to }{0}}{{lim}}{}{x}{\mathrm{cos}}{}\left({x}\right){\mathrm{ln}}{}\left({x}\right)$ (6.3)
 >