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Teaching indefinite integrals that require special-casing
Teaching indefinite integrals that require special-casing
What do you do when you realize mid-lecture that your lesson plan is not working?Should $varphi$ be monotone in the integration by substitution?Teaching measurement to 4th grade students by projects?Students strictly follow the steps and notations in sample problems without understanding themTeaching Models for Mathematics (like 5 E's in Science)Struggles with Teaching HS Mathteach that $frac10$ not defined properlyWhat is the ideal teaching style for Calculus exercises only?“Basic ideas” as a concept in teaching maths?Should young math students be taught an abstract concept of form?
$begingroup$
I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?
Let's consider the following example.
Find the indefinite integral
$$
I=intdfracdxxsqrtx^2-1.
$$
Some of my students gave the following answer.
Let $t=1/x$ then $dx=-1/t^2dt$, so we get
$$
I=intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfrac-dtsqrt1-t^2=-arcsinleft(tright)+C=-arcsinleft(frac1xright)+C.
$$
Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfractrighttsqrt1-t^2.
$$
Then we end up with the answer
$$
intdfracdxxsqrtx^2-1=begincases
-arcsinleft(dfrac1xright)+C & textfor x>1,\
arcsinleft(dfrac1xright)+C & textfor x<-1.
endcases
$$
In your teaching practice, how would you usually proceed?
PS. We may encounter the same issue in many other problems. For example, find $intsqrt1-x^2dx$. Then if we let $x=sinleft(tright)$ then
$sqrt1-sin^2left(tright)$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.
teaching teacher-preparation
$endgroup$
add a comment |
$begingroup$
I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?
Let's consider the following example.
Find the indefinite integral
$$
I=intdfracdxxsqrtx^2-1.
$$
Some of my students gave the following answer.
Let $t=1/x$ then $dx=-1/t^2dt$, so we get
$$
I=intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfrac-dtsqrt1-t^2=-arcsinleft(tright)+C=-arcsinleft(frac1xright)+C.
$$
Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfractrighttsqrt1-t^2.
$$
Then we end up with the answer
$$
intdfracdxxsqrtx^2-1=begincases
-arcsinleft(dfrac1xright)+C & textfor x>1,\
arcsinleft(dfrac1xright)+C & textfor x<-1.
endcases
$$
In your teaching practice, how would you usually proceed?
PS. We may encounter the same issue in many other problems. For example, find $intsqrt1-x^2dx$. Then if we let $x=sinleft(tright)$ then
$sqrt1-sin^2left(tright)$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.
teaching teacher-preparation
$endgroup$
2
$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
$endgroup$
– Acccumulation
yesterday
7
$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
yesterday
2
$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
$endgroup$
– kcrisman
yesterday
$begingroup$
Very nice question. Inattention to this detail is indeed a shortcoming of many texts and many of my own lectures. Probably adding a condition ($x>1$ or $x<-1$) to focus attention on one case is the smart solution to not overwhelm students and yet be true to detail. Some texts are more careful than others...
$endgroup$
– James S. Cook
2 hours ago
add a comment |
$begingroup$
I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?
Let's consider the following example.
Find the indefinite integral
$$
I=intdfracdxxsqrtx^2-1.
$$
Some of my students gave the following answer.
Let $t=1/x$ then $dx=-1/t^2dt$, so we get
$$
I=intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfrac-dtsqrt1-t^2=-arcsinleft(tright)+C=-arcsinleft(frac1xright)+C.
$$
Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfractrighttsqrt1-t^2.
$$
Then we end up with the answer
$$
intdfracdxxsqrtx^2-1=begincases
-arcsinleft(dfrac1xright)+C & textfor x>1,\
arcsinleft(dfrac1xright)+C & textfor x<-1.
endcases
$$
In your teaching practice, how would you usually proceed?
PS. We may encounter the same issue in many other problems. For example, find $intsqrt1-x^2dx$. Then if we let $x=sinleft(tright)$ then
$sqrt1-sin^2left(tright)$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.
teaching teacher-preparation
$endgroup$
I encountered the following concern when teaching indefinite integrals. I believe that many of us may overlook this. May I be wrong?
Let's consider the following example.
Find the indefinite integral
$$
I=intdfracdxxsqrtx^2-1.
$$
Some of my students gave the following answer.
Let $t=1/x$ then $dx=-1/t^2dt$, so we get
$$
I=intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfrac-dtsqrt1-t^2=-arcsinleft(tright)+C=-arcsinleft(frac1xright)+C.
$$
Sometimes, I accept this answer since it gives a quick general antiderivative.
However, the problem here is that we should write
$$
intdfrac-1/t^2dtfrac1tsqrtfrac1t^2-1=intdfractrighttsqrt1-t^2.
$$
Then we end up with the answer
$$
intdfracdxxsqrtx^2-1=begincases
-arcsinleft(dfrac1xright)+C & textfor x>1,\
arcsinleft(dfrac1xright)+C & textfor x<-1.
endcases
$$
In your teaching practice, how would you usually proceed?
PS. We may encounter the same issue in many other problems. For example, find $intsqrt1-x^2dx$. Then if we let $x=sinleft(tright)$ then
$sqrt1-sin^2left(tright)$ should be $left|cosleft(tright)right|$. So now we need to explain a bit here to our naive students. Of course, avoiding these kinds of problems is the quickest way to make our teaching job easier. However, we need to prepare a good way of explanining or handing these types of problems. That's what I want to know.
teaching teacher-preparation
teaching teacher-preparation
edited yesterday
Hoa
asked yesterday
HoaHoa
1416
1416
2
$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
$endgroup$
– Acccumulation
yesterday
7
$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
yesterday
2
$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
$endgroup$
– kcrisman
yesterday
$begingroup$
Very nice question. Inattention to this detail is indeed a shortcoming of many texts and many of my own lectures. Probably adding a condition ($x>1$ or $x<-1$) to focus attention on one case is the smart solution to not overwhelm students and yet be true to detail. Some texts are more careful than others...
$endgroup$
– James S. Cook
2 hours ago
add a comment |
2
$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
$endgroup$
– Acccumulation
yesterday
7
$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
yesterday
2
$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
$endgroup$
– kcrisman
yesterday
$begingroup$
Very nice question. Inattention to this detail is indeed a shortcoming of many texts and many of my own lectures. Probably adding a condition ($x>1$ or $x<-1$) to focus attention on one case is the smart solution to not overwhelm students and yet be true to detail. Some texts are more careful than others...
$endgroup$
– James S. Cook
2 hours ago
2
2
$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
$endgroup$
– Acccumulation
yesterday
$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
$endgroup$
– Acccumulation
yesterday
7
7
$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
yesterday
$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
yesterday
2
2
$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
$endgroup$
– kcrisman
yesterday
$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
$endgroup$
– kcrisman
yesterday
$begingroup$
Very nice question. Inattention to this detail is indeed a shortcoming of many texts and many of my own lectures. Probably adding a condition ($x>1$ or $x<-1$) to focus attention on one case is the smart solution to not overwhelm students and yet be true to detail. Some texts are more careful than others...
$endgroup$
– James S. Cook
2 hours ago
$begingroup$
Very nice question. Inattention to this detail is indeed a shortcoming of many texts and many of my own lectures. Probably adding a condition ($x>1$ or $x<-1$) to focus attention on one case is the smart solution to not overwhelm students and yet be true to detail. Some texts are more careful than others...
$endgroup$
– James S. Cook
2 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.
It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)
$endgroup$
add a comment |
$begingroup$
This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.
Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
$$lim_xtopminfty fracxsqrtx^2+1 = pm 1$$
The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:
$$lim_xtopminfty fracxcdot 1/xsqrt(x^2+1)cdot 1/x^2 = lim_xtopminfty frac1sqrt1+1/x^2=1$$
But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.
In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)
So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac1xsqrtx^2-1$ is definitely negative there.
Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.
Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)
$endgroup$
add a comment |
$begingroup$
You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.
In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!
The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.
$endgroup$
add a comment |
$begingroup$
To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.
I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.
As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.
New contributor
$endgroup$
add a comment |
$begingroup$
It is interesting to note that if we instead write the antiderivative as $I=-arctanleft(frac1sqrtx^2-1right)$, then this form is valid for both $x<1$ and $x>1$. In other words, using this arctan representation, we avoid a need for a piecewise representation for the antiderivative.
This problem is well suited for formative assessment. I would give full credit for an answer of the form $I=-arcsinleft(frac1xright)$, and then use this as a launching point for a discussion of these more involved issues that you have uncovered, depending on the mathematical maturity of the student.
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add a comment |
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5 Answers
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5 Answers
5
active
oldest
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active
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active
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$begingroup$
I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.
It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)
$endgroup$
add a comment |
$begingroup$
I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.
It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)
$endgroup$
add a comment |
$begingroup$
I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.
It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)
$endgroup$
I'd avoid giving problems like that to students first learning indefinite integrals (either by not asking it at all, or specifying the range x>1 in the question). It's a subtle algebraic trap, and if the goal is to teach students the mechanics of integration, it's going to be distracting rather than helpful.
It might be an interesting question in a more advanced class, or as a question which is marked as difficult where students are expected (or told) to investigate their answer more carefully. (For instance, graphing the functions will quickly reveal that there's a problem with the first solution, and looking at the graphs is probably enough to figure out what the fixed solution should look like, though figuring out why might take students a while.)
answered yesterday
Henry TowsnerHenry Towsner
7,1652350
7,1652350
add a comment |
add a comment |
$begingroup$
This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.
Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
$$lim_xtopminfty fracxsqrtx^2+1 = pm 1$$
The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:
$$lim_xtopminfty fracxcdot 1/xsqrt(x^2+1)cdot 1/x^2 = lim_xtopminfty frac1sqrt1+1/x^2=1$$
But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.
In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)
So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac1xsqrtx^2-1$ is definitely negative there.
Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.
Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)
$endgroup$
add a comment |
$begingroup$
This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.
Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
$$lim_xtopminfty fracxsqrtx^2+1 = pm 1$$
The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:
$$lim_xtopminfty fracxcdot 1/xsqrt(x^2+1)cdot 1/x^2 = lim_xtopminfty frac1sqrt1+1/x^2=1$$
But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.
In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)
So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac1xsqrtx^2-1$ is definitely negative there.
Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.
Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)
$endgroup$
add a comment |
$begingroup$
This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.
Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
$$lim_xtopminfty fracxsqrtx^2+1 = pm 1$$
The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:
$$lim_xtopminfty fracxcdot 1/xsqrt(x^2+1)cdot 1/x^2 = lim_xtopminfty frac1sqrt1+1/x^2=1$$
But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.
In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)
So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac1xsqrtx^2-1$ is definitely negative there.
Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.
Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)
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This is a hard question, because students are so used to manipulation of this kind. I have found you are right that absolute values can cause the worst of these examples.
Here is an example I ran into recently, which I hope will help your thinking. Observe that there are two different limits here:
$$lim_xtopminfty fracxsqrtx^2+1 = pm 1$$
The "usual" way to proceed with these (informally, in many texts nowadays) is to divide numerator and denominator by the highest power, so:
$$lim_xtopminfty fracxcdot 1/xsqrt(x^2+1)cdot 1/x^2 = lim_xtopminfty frac1sqrt1+1/x^2=1$$
But of course bringing the $1/x$ inside the root like that is the same invalid manipulation you are mentioning.
In this case, we actually talked through it at an even more naive level, not more rigorous! Namely, as $xto -infty$, the numerator is negative and the denominator is positive. So the overall answer must be negative, no matter what the manipulation says. (You can graph it for them too.)
So in your case, I would go more naive as well. Do a very rough sketch of $arcsin(1/x)$ (you can basically do this by drawing $-arcsin(x)$ and then "flipping over $x=1$ to infinity"), and then ask them whether this function is increasing or decreasing. When $x<-1$ it should be increasing (in fact, it should be increasing on the whole domain), so its derivative should be positive (by whichever numbering of the fundamental theorems of calculus you like). But $frac1xsqrtx^2-1$ is definitely negative there.
Now you can explain why you are picky, instead of just being picky because of some "dumb" $|x|$ thing students may find to be a little too abstract.
Another answer brings up the question of whether this is a good question at all. But I think it is reasonable. What you may want to do, though, is find a way to discuss this "naturally", i.e. using the disconnect between what people write and then if they see what seems to be a "wrong" answer in the back of a book or something. Taking it as a first example in class probably will not register with them unless they are quite good at the concepts of calculus (not just mechanics), whereas pointing out why it is wrong/right should be better. (On another note, presumably there are branch cut issues here as well but presumably your class isn't ready for that!)
answered yesterday
kcrismankcrisman
3,633732
3,633732
add a comment |
add a comment |
$begingroup$
You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.
In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!
The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.
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add a comment |
$begingroup$
You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.
In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!
The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.
$endgroup$
add a comment |
$begingroup$
You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.
In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!
The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.
$endgroup$
You are right to be concerned that the students are "missing something", but IMO the real problem here is that the question is completely artificial.
In any application of this type of integral, most likely $x$ will be known to be either positive or negative, but not both, and only one part of the "either-or" answer would apply. And there had better be a good reason why the rest of the problem needs $x$ to be negative, when it could have been replaced by $-x$ right from the start!
The same is true for the more common case of the indefinite integral of $1/x$ when $x < 0$, of course.
answered yesterday
alephzeroalephzero
31113
31113
add a comment |
add a comment |
$begingroup$
To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.
I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.
As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.
New contributor
$endgroup$
add a comment |
$begingroup$
To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.
I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.
As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.
New contributor
$endgroup$
add a comment |
$begingroup$
To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.
I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.
As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.
New contributor
$endgroup$
To me, the key point here is that the integral runs over a singularity. If you naively calculates a definite form that runs over the singularity you get the wrong answer. This is something I have done enough so that I have taught myself to be careful in this case.
I am more a physicist than a mathematician, so what I care about is the connection to a practical situation, rather than the formal manipulation of symbols. If you or the students are of a similar inclination, then the presence of the singularity is what tips you off.
As you know, different students respond well to different approaches, so I mention this one so that you might add it to your arsenal.
New contributor
New contributor
answered yesterday
AndrewAndrew
511
511
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
It is interesting to note that if we instead write the antiderivative as $I=-arctanleft(frac1sqrtx^2-1right)$, then this form is valid for both $x<1$ and $x>1$. In other words, using this arctan representation, we avoid a need for a piecewise representation for the antiderivative.
This problem is well suited for formative assessment. I would give full credit for an answer of the form $I=-arcsinleft(frac1xright)$, and then use this as a launching point for a discussion of these more involved issues that you have uncovered, depending on the mathematical maturity of the student.
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add a comment |
$begingroup$
It is interesting to note that if we instead write the antiderivative as $I=-arctanleft(frac1sqrtx^2-1right)$, then this form is valid for both $x<1$ and $x>1$. In other words, using this arctan representation, we avoid a need for a piecewise representation for the antiderivative.
This problem is well suited for formative assessment. I would give full credit for an answer of the form $I=-arcsinleft(frac1xright)$, and then use this as a launching point for a discussion of these more involved issues that you have uncovered, depending on the mathematical maturity of the student.
$endgroup$
add a comment |
$begingroup$
It is interesting to note that if we instead write the antiderivative as $I=-arctanleft(frac1sqrtx^2-1right)$, then this form is valid for both $x<1$ and $x>1$. In other words, using this arctan representation, we avoid a need for a piecewise representation for the antiderivative.
This problem is well suited for formative assessment. I would give full credit for an answer of the form $I=-arcsinleft(frac1xright)$, and then use this as a launching point for a discussion of these more involved issues that you have uncovered, depending on the mathematical maturity of the student.
$endgroup$
It is interesting to note that if we instead write the antiderivative as $I=-arctanleft(frac1sqrtx^2-1right)$, then this form is valid for both $x<1$ and $x>1$. In other words, using this arctan representation, we avoid a need for a piecewise representation for the antiderivative.
This problem is well suited for formative assessment. I would give full credit for an answer of the form $I=-arcsinleft(frac1xright)$, and then use this as a launching point for a discussion of these more involved issues that you have uncovered, depending on the mathematical maturity of the student.
answered 11 hours ago
user52817user52817
3,359618
3,359618
add a comment |
add a comment |
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-teacher-preparation, teaching
2
$begingroup$
Seems to me this is more of a general simplification/substitution issue, and the fact that the simplification appears within an integral is a side issue.
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– Acccumulation
yesterday
7
$begingroup$
The correct answer should in fact have two different arbitrary constants, one for each connected component of the domain.
$endgroup$
– Javier
yesterday
2
$begingroup$
By the way, there are so many equiv ways ... Wolfram Alpha gives a form of the arctangent, Sage/Maxima says your answer but with absolute value of 1/x, and Sympy gives a nice cases result including I*arccosh(1/x). And it looks like arc secant :)
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– kcrisman
yesterday
$begingroup$
Very nice question. Inattention to this detail is indeed a shortcoming of many texts and many of my own lectures. Probably adding a condition ($x>1$ or $x<-1$) to focus attention on one case is the smart solution to not overwhelm students and yet be true to detail. Some texts are more careful than others...
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– James S. Cook
2 hours ago