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)

searchoptimal = truefalse (default: true)

These options can be used to control how the problem is displayed:

showrules = truefalse (default: true)

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

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

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.

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

$\mathrm{Diff}\left(x^2\sin \left(x\right)\,x\right)$

$\frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)$

$\mathrm{ShowSolution}\left(\right)$

$\begin{array}{lll}& & \text{Differentiation Steps}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\sin \left(x\right)\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{product}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{product}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(f\left(x\right)g\left(x\right)\right)=\frac{\ⅆ}{\ⅆx}f\left(x\right)g\left(x\right)+f\left(x\right)\frac{\ⅆ}{\ⅆx}g\left(x\right)\\ & & f\left(x\right)=x^2\\ & & g\left(x\right)=\sin \left(x\right)\\ & & \text{This gives:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)\sin \left(x\right)+x^2\frac{\ⅆ}{\ⅆx}sin\left(x\right)\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆx}\left(x^2\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^{\left[\right]}\right)=\left[\right]x^{\left[\right]-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=\left[\right]\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆx}\left(x^2\right)=\left[\right]\\ & & \text{We can rewrite the derivative as:}\\ & & 2x\sin \left(x\right)+x^2\frac{\ⅆ}{\ⅆx}sin\left(x\right)\\ \text{▫}& & \text{3. Evaluate the derivative of}\sin \text{(}x\text{)}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{sin}\text{rule}\\ & & \frac{\ⅆ}{\ⅆx}sin\left(x\right)=\cos \left(x\right)\\ & & \text{This gives:}\\ & & 2x\sin \left(x\right)+x^2\cos \left(x\right)\end{array}$

$\mathrm{Int}\left({\sin \left(x\right)}^2\,x\right)$

$\int {\sin \left(x\right)}^2\ⅆx$

$\mathrm{Hint}\left(\right)$

$\left[\mathrm{rewrite}\,{\sin \left(x\right)}^2=\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\right]$

$\mathrm{Rule}\left[\right]\left(\right)$

$\int {\sin \left(x\right)}^2\ⅆx=\int \left(\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\right)\ⅆx$

$\mathrm{ShowSolution}\left(\right)$

$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int {\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{1. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & {\sin \left(x\right)}^2=\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\\ & & \text{This gives:}\\ & & \int \left(\frac{1}{2}-\frac{\cos \left(2x\right)}{2}\right)\ⅆx\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \int \left(f\left(x\right)+g\left(x\right)\right)\ⅆx=\int f\left(x\right)\ⅆx+\int g\left(x\right)\ⅆx\\ & & f\left(x\right)=\frac{1}{2}\\ & & g\left(x\right)=-\frac{\cos \left(2x\right)}{2}\\ & & \text{This gives:}\\ & & \int \frac{1}{2}\ⅆx+\int -\frac{\cos \left(2x\right)}{2}\ⅆx\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{constant}}\text{rule to the term}\int \frac{1}{2}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \int C\ⅆx=Cx\\ & \text{◦}& \text{This means}\\ & & \int \frac{1}{2}\ⅆx=\frac{x}{2}\\ & & \text{We can now rewrite the integral as:}\\ & & \frac{x}{2}+\int -\frac{\cos \left(2x\right)}{2}\ⅆx\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{\cos \left(2x\right)}{2}\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int \left[\right]f\left(x\right)\ⅆx=\left[\right]\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{\cos \left(2x\right)}{2}\ⅆx=-\frac{\left(\int \cos \left(2x\right)\ⅆx\right)}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{x}{2}-\frac{\left(\int \cos \left(2x\right)\ⅆx\right)}{2}\\ \text{▫}& & \text{5. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=2x\\ & \text{◦}& \text{Isolate equation for}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 \cos \left(2x\right)\ⅆx=\int \frac{\cos \left(u\right)}{2}\ⅆu\\ & & \text{This gives:}\\ & & \frac{x}{2}-\frac{\left(\int \frac{\cos \left(u\right)}{2}\ⅆu\right)}{2}\\ \text{▫}& & \text{6. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int \frac{\cos \left(u\right)}{2}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int \left[\right]f\left(u\right)\ⅆu=\left[\right]\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int \frac{\cos \left(u\right)}{2}\ⅆu=\frac{\left(\int \cos \left(u\right)\ⅆu\right)}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{x}{2}-\frac{\left(\int \cos \left(u\right)\ⅆu\right)}{4}\\ \text{▫}& & \text{7. Evaluate the integral of}\cos \text{(}u\text{)}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{cos}\text{rule}\\ & & \int \cos \left(u\right)\ⅆu=\sin \left(u\right)\\ & & \text{This gives:}\\ & & \frac{x}{2}-\frac{\sin \left(u\right)}{4}\\ \text{▫}& & \text{8. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}5\\ & & u=2x\\ & & \text{This gives:}\\ & & \frac{x}{2}-\frac{\sin \left(2x\right)}{4}\end{array}$

$\mathrm{Understand}\left(\mathrm{Int}\,\mathrm{`c*`}\,\mathrm{revert}\right)$

$\mathrm{Int}=\left[\mathrm{constantmultiple}\,\mathrm{revert}\right]$

$\mathrm{ShowSolution}\left(\mathrm{Int}\left(\frac{1}{4x^2+4x+1}\,x\right)\right)$

$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int \frac{1}{4x^2+4x+1}\ⅆx\\ \text{▫}& & \text{1. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=x+\frac{1}{2}\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=-\frac{1}{2}+u\\ & \text{◦}& \text{Differentiate both sides}\\ & & \left[\right]=\left[\right]\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & \int \frac{1}{4x^2+4x+1}\ⅆx=\frac{\left(\int \frac{1}{u^2}\ⅆu\right)}{4}\\ & & \text{This gives:}\\ & & \frac{\left(\int \frac{1}{u^2}\ⅆu\right)}{4}\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{power}}\text{rule to the term}\int \frac{1}{u^2}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^{\left[\right]}\ⅆu=\left[\right]\\ & \text{◦}& \text{This means:}\\ & & \int \frac{1}{u^2}\ⅆu=\left[\right]\\ & \text{◦}& \text{So,}\\ & & \int \frac{1}{u^2}\ⅆu=-\frac{1}{u}\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{1}{4x+2}\end{array}$

The Student[Calculus1][ShowSolution] command was updated in Maple 2021.

The output and displaystyle options were introduced in Maple 2021.

For more information on Maple 2021 changes, see Updates in Maple 2021.