ShowSolution - Maple Help

Student[Calculus1]

 ShowSolution
 show all the steps in the solution of a specified problem

 Calling Sequence ShowSolution(p, opts)

Parameters

 p - (optional) posint or a calculus1 problem; the problem to solve opts - (optional) options of the form keyword=value, where keyword is one of maxsteps, searchoptimal, showrules, output, displaystyle.

Description

 • The ShowSolution command is used to show the solution steps for a Calculus1 problem, that is, a limit, differentiation or integration problem such as can be expected to be encountered in a single-variable calculus course.
 • If p is omitted, the current (most recently referenced) problem is solved.  Otherwise, the problem referenced by p is solved.  A problem can be referenced either by its problem number (see GetProblem and ShowIncomplete) or by the problem itself, for example via a label.
 • These options can be used to control how the problem is solved:
 – maxsteps = posint (default: 25)

This puts a limit on the number of rules which can be applied to solve a problem.

 – searchoptimal = truefalse (default: true)

The default behavior is to try to determine the shortest solution sequence. If this option is given as searchoptimal = false, the first solution discovered will be displayed.

 • These options can be used to control how the problem is displayed:
 – showrules = truefalse (default: true)

Normally, the rule applied at each step of the solution is displayed; if this option is given as showrules=false, the rules are not shown, and the displaystyle is brief.

 – output = canvas,script,record,list,print,printf,typeset,link (default: typeset)

The output options are described in Student:-Basics:-OutputStepsRecord

 – displaystyle= columns,compact,linear,brief (default: linear)

The displaystyle options are described in Student:-Basics:-OutputStepsRecord.  Setting displaystyle = brief shows the steps in a compact form with one or two words at each step to describe the rule being applied.

 • If you set infolevel[Student] := 1 or infolevel[Student[Calculus1]] := 1 (see infolevel), Maple may display some additional, useful information about the state of the problem and its solution.

Examples

 > $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{Calculus1}\right]\right):$
 > $\mathrm{Diff}\left({x}^{2}\mathrm{sin}\left(x\right),x\right)$
 $\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{}{\mathrm{sin}}{}\left({x}\right)\right)$ (1)
 > $\mathrm{ShowSolution}\left(\right)$
 $\begin{array}{lll}{}& {}& \text{Differentiation Steps}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}{}{\mathrm{sin}}{}\left({x}\right)\right)\\ \text{▫}& {}& {\text{1. Apply the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{product}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{product}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({f}{}\left({x}\right){}{g}{}\left({x}\right)\right){=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{f}{}\left({x}\right){}{g}{}\left({x}\right){+}{f}{}\left({x}\right){}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{g}{}\left({x}\right)\\ {}& {}& {f}{}\left({x}\right){=}{{x}}^{{2}}\\ {}& {}& {g}{}\left({x}\right){=}{\mathrm{sin}}{}\left({x}\right)\\ {}& {}& \text{This gives:}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){}{\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{sin}{}\left({x}\right)\\ \text{▫}& {}& {\text{2. Apply the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{power}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule to the term}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right)\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{power}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{\left[{}\right]}\right){=}\left[{}\right]{}{{x}}^{\left[{}\right]{-}{1}}\\ {}& \text{◦}& \text{This means:}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){=}\left[{}\right]\\ {}& \text{◦}& \text{So,}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({{x}}^{{2}}\right){=}\left[{}\right]\\ {}& {}& \text{We can rewrite the derivative as:}\\ {}& {}& {2}{}{x}{}{\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{sin}{}\left({x}\right)\\ \text{▫}& {}& {\text{3. Evaluate the derivative of}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{sin}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{(}}{x}{\text{)}}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{sin}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{sin}{}\left({x}\right){=}{\mathrm{cos}}{}\left({x}\right)\\ {}& {}& \text{This gives:}\\ {}& {}& {2}{}{x}{}{\mathrm{sin}}{}\left({x}\right){+}{{x}}^{{2}}{}{\mathrm{cos}}{}\left({x}\right)\end{array}$ (2)
 > $\mathrm{Int}\left({\mathrm{sin}\left(x\right)}^{2},x\right)$
 ${\int }{{\mathrm{sin}}{}\left({x}\right)}^{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}$ (3)
 > $\mathrm{Hint}\left(\right)$
 $\left[{\mathrm{rewrite}}{,}{{\mathrm{sin}}{}\left({x}\right)}^{{2}}{=}\frac{{1}}{{2}}{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\right]$ (4)
 > $\mathrm{Rule}\left[\right]\left(\right)$
 ${\int }{{\mathrm{sin}}{}\left({x}\right)}^{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }\left(\frac{{1}}{{2}}{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}$ (5)
 > $\mathrm{ShowSolution}\left(\right)$
 $\begin{array}{lll}{}& {}& \text{Integration Steps}\\ {}& {}& {\int }{{\mathrm{sin}}{}\left({x}\right)}^{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ \text{▫}& {}& \text{1. Rewrite}\\ {}& \text{◦}& \text{Equivalent expression}\\ {}& {}& {{\mathrm{sin}}{}\left({x}\right)}^{{2}}{=}\frac{{1}}{{2}}{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\\ {}& {}& \text{This gives:}\\ {}& {}& {\int }\left(\frac{{1}}{{2}}{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ \text{▫}& {}& {\text{2. Apply the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{sum}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{sum}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& {\int }\left({f}{}\left({x}\right){+}{g}{}\left({x}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\int }{g}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ {}& {}& {f}{}\left({x}\right){=}\frac{{1}}{{2}}\\ {}& {}& {g}{}\left({x}\right){=}{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\\ {}& {}& \text{This gives:}\\ {}& {}& {\int }\frac{{1}}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\int }{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ \text{▫}& {}& {\text{3. Apply the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{constant}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule to the term}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\int }\frac{{1}}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{constant}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& {\int }{C}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{C}{}{x}\\ {}& \text{◦}& \text{This means}\\ {}& {}& {\int }\frac{{1}}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}\frac{{x}}{{2}}\\ {}& {}& \text{We can now rewrite the integral as:}\\ {}& {}& \frac{{x}}{{2}}{+}{\int }{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ \text{▫}& {}& {\text{4. Apply the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{constant multiple}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule to the term}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\int }{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{constant multiple}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& {\int }\left[{}\right]{}{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}\left[{}\right]{}\left({\int }{f}{}\left({x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)\\ {}& \text{◦}& \text{This means:}\\ {}& {}& {\int }{-}\frac{{\mathrm{cos}}{}\left({2}{}{x}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{-}\frac{\left({\int }{\mathrm{cos}}{}\left({2}{}{x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{{2}}\\ {}& {}& \text{We can rewrite the integral as:}\\ {}& {}& \frac{{x}}{{2}}{-}\frac{\left({\int }{\mathrm{cos}}{}\left({2}{}{x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\right)}{{2}}\\ \text{▫}& {}& {\text{5. Apply a change of variables to rewrite the integral in terms of}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}\\ {}& \text{◦}& {\text{Let}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{be}}\\ {}& {}& {u}{=}{2}{}{x}\\ {}& \text{◦}& {\text{Isolate equation for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}\\ {}& {}& {x}{=}\frac{{u}}{{2}}\\ {}& \text{◦}& \text{Differentiate both sides}\\ {}& {}& \left[{}\right]{=}\frac{\left[{}\right]}{{2}}\\ {}& \text{◦}& \text{Substitute the values for x and dx back into the original}\\ {}& {}& {\int }{\mathrm{cos}}{}\left({2}{}{x}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }\frac{{\mathrm{cos}}{}\left({u}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\\ {}& {}& \text{This gives:}\\ {}& {}& \frac{{x}}{{2}}{-}\frac{\left({\int }\frac{{\mathrm{cos}}{}\left({u}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\right)}{{2}}\\ \text{▫}& {}& {\text{6. Apply the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{constant multiple}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule to the term}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\int }\frac{{\mathrm{cos}}{}\left({u}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{constant multiple}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& {\int }\left[{}\right]{}{f}{}\left({u}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}\left[{}\right]{}\left({\int }{f}{}\left({u}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\right)\\ {}& \text{◦}& \text{This means:}\\ {}& {}& {\int }\frac{{\mathrm{cos}}{}\left({u}\right)}{{2}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}\frac{\left({\int }{\mathrm{cos}}{}\left({u}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\right)}{{2}}\\ {}& {}& \text{We can rewrite the integral as:}\\ {}& {}& \frac{{x}}{{2}}{-}\frac{\left({\int }{\mathrm{cos}}{}\left({u}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\right)}{{4}}\\ \text{▫}& {}& {\text{7. Evaluate the integral of}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathrm{cos}}{\text{(}}{u}{\text{)}}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{cos}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule}}\\ {}& {}& {\int }{\mathrm{cos}}{}\left({u}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}{\mathrm{sin}}{}\left({u}\right)\\ {}& {}& \text{This gives:}\\ {}& {}& \frac{{x}}{{2}}{-}\frac{{\mathrm{sin}}{}\left({u}\right)}{{4}}\\ \text{▫}& {}& \text{8. Revert change of variable}\\ {}& \text{◦}& {\text{Variable we defined in step}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{5}\\ {}& {}& {u}{=}{2}{}{x}\\ {}& {}& \text{This gives:}\\ {}& {}& \frac{{x}}{{2}}{-}\frac{{\mathrm{sin}}{}\left({2}{}{x}\right)}{{4}}\end{array}$ (6)
 > $\mathrm{Understand}\left(\mathrm{Int},\mathrm{c*},\mathrm{revert}\right)$
 ${\mathrm{Int}}{=}\left[{\mathrm{constantmultiple}}{,}{\mathrm{revert}}\right]$ (7)
 > $\mathrm{ShowSolution}\left(\mathrm{Int}\left(\frac{1}{4{x}^{2}+4x+1},x\right)\right)$
 $\begin{array}{lll}{}& {}& \text{Integration Steps}\\ {}& {}& {\int }\frac{{1}}{{4}{}{{x}}^{{2}}{+}{4}{}{x}{+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}\\ \text{▫}& {}& {\text{1. Apply a change of variables to rewrite the integral in terms of}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}\\ {}& \text{◦}& {\text{Let}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{u}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{be}}\\ {}& {}& {u}{=}{x}{+}\frac{{1}}{{2}}\\ {}& \text{◦}& {\text{Isolate equation for}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{x}\\ {}& {}& {x}{=}{-}\frac{{1}}{{2}}{+}{u}\\ {}& \text{◦}& \text{Differentiate both sides}\\ {}& {}& \left[{}\right]{=}\left[{}\right]\\ {}& \text{◦}& \text{Substitute the values for x and dx back into the original}\\ {}& {}& {\int }\frac{{1}}{{4}{}{{x}}^{{2}}{+}{4}{}{x}{+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}\frac{\left({\int }\frac{{1}}{{{u}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\right)}{{4}}\\ {}& {}& \text{This gives:}\\ {}& {}& \frac{\left({\int }\frac{{1}}{{{u}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\right)}{{4}}\\ \text{▫}& {}& {\text{2. Apply the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{power}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule to the term}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\int }\frac{{1}}{{{u}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}\\ {}& \text{◦}& {\text{Recall the definition of the}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\mathbf{\text{power}}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{rule, for n}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{≠}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{\text{-1}}\\ {}& {}& {\int }{{u}}^{\left[{}\right]}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}\left[{}\right]\\ {}& \text{◦}& \text{This means:}\\ {}& {}& {\int }\frac{{1}}{{{u}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}\left[{}\right]\\ {}& \text{◦}& \text{So,}\\ {}& {}& {\int }\frac{{1}}{{{u}}^{{2}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{u}{=}{-}\frac{{1}}{{u}}\\ {}& {}& \text{We can rewrite the integral as:}\\ {}& {}& {-}\frac{{1}}{{4}{}{x}{+}{2}}\end{array}$ (8)

Compatibility

 • The Student[Calculus1][ShowSolution] command was updated in Maple 2021.
 • The output and displaystyle options were introduced in Maple 2021.