Calculus II Professor will not accept my correct integral evaluation that uses a different method, should I...
$begingroup$
I am a freshman enrolled at an American University. Recently, I took an examination in which the following problem appeared:
Evaluate the following integral:
$int_0^4sqrt{16-x^2}dx$
My answer: 4$pi$, was correct. However, I received reduced credit for this answer because I did not solve it correctly (according to the professor). The exams are time-limited and have a fair amount of content, so when I saw this problem, I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
The context of the test was surrounding our unit on inverse trigonometry and integration by parts. This section of the test did not list any other instructions besides evaluating the definite integrals. I've talked to my professor about it and his only response was that I solved it wrong:
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Is my interpretation of evaluating the integral different? Does the instruction "Find the antiderivative and then evaluate" not need to exist for that to be required?
Thank you.
calculus integration
asked 4 hours ago
user146073user146073
342
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
I am a freshman enrolled at an American University. Recently, I took an examination in which the following problem appeared:
Evaluate the following integral:
$int_0^4sqrt{16-x^2}dx$
My answer: 4$pi$, was correct. However, I received reduced credit for this answer because I did not solve it correctly (according to the professor). The exams are time-limited and have a fair amount of content, so when I saw this problem, I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
The context of the test was surrounding our unit on inverse trigonometry and integration by parts. This section of the test did not list any other instructions besides evaluating the definite integrals. I've talked to my professor about it and his only response was that I solved it wrong:
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Is my interpretation of evaluating the integral different? Does the instruction "Find the antiderivative and then evaluate" not need to exist for that to be required?
Thank you.
calculus integration
asked 4 hours ago
user146073user146073
342
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
you profesor in mixed up
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
can you post the exact problem question?
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
@MikeySpivak The question sheet is no longer in my possession, but I remember exactly what it said "Evaluate the following integrals:" (there were several). I wrote the integral exactly as it appeared.
$endgroup$
– user146073
4 hours ago
$begingroup$
He was obviously nitpicking, since you were able to bypass a direct (tedious) evaluation, which is what he wanted.
$endgroup$
– herb steinberg
4 hours ago
1
$begingroup$
Did you obtain the correct answer via a logically sound argument? If so (as you did here), then you solved the problem. Of course, the professor can use whatever arbitrary grading metric he wants (and if he meant "evaluate the integral by computing an antiderivative for some reason, then he probably should have said so); but as far as the underlying math goes, recognizing that an integral corresponds to some other quantity that's easier to compute via a different method is completely legit.
$endgroup$
– anomaly
3 hours ago
|
show 3 more comments
$begingroup$
I am a freshman enrolled at an American University. Recently, I took an examination in which the following problem appeared:
Evaluate the following integral:
$int_0^4sqrt{16-x^2}dx$
My answer: 4$pi$, was correct. However, I received reduced credit for this answer because I did not solve it correctly (according to the professor). The exams are time-limited and have a fair amount of content, so when I saw this problem, I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
The context of the test was surrounding our unit on inverse trigonometry and integration by parts. This section of the test did not list any other instructions besides evaluating the definite integrals. I've talked to my professor about it and his only response was that I solved it wrong:
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Is my interpretation of evaluating the integral different? Does the instruction "Find the antiderivative and then evaluate" not need to exist for that to be required?
Thank you.
calculus integration
asked 4 hours ago
user146073user146073
342
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am a freshman enrolled at an American University. Recently, I took an examination in which the following problem appeared:
Evaluate the following integral:
$int_0^4sqrt{16-x^2}dx$
My answer: 4$pi$, was correct. However, I received reduced credit for this answer because I did not solve it correctly (according to the professor). The exams are time-limited and have a fair amount of content, so when I saw this problem, I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
The context of the test was surrounding our unit on inverse trigonometry and integration by parts. This section of the test did not list any other instructions besides evaluating the definite integrals. I've talked to my professor about it and his only response was that I solved it wrong:
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Is my interpretation of evaluating the integral different? Does the instruction "Find the antiderivative and then evaluate" not need to exist for that to be required?
Thank you.
calculus integration
calculus integration
asked 4 hours ago
user146073user146073
342
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user146073user146073
342
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user146073user146073
342
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user146073user146073
342
asked 4 hours ago
user146073user146073
342
342
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user146073 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
you profesor in mixed up
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
can you post the exact problem question?
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
@MikeySpivak The question sheet is no longer in my possession, but I remember exactly what it said "Evaluate the following integrals:" (there were several). I wrote the integral exactly as it appeared.
$endgroup$
– user146073
4 hours ago
$begingroup$
He was obviously nitpicking, since you were able to bypass a direct (tedious) evaluation, which is what he wanted.
$endgroup$
– herb steinberg
4 hours ago
1
$begingroup$
Did you obtain the correct answer via a logically sound argument? If so (as you did here), then you solved the problem. Of course, the professor can use whatever arbitrary grading metric he wants (and if he meant "evaluate the integral by computing an antiderivative for some reason, then he probably should have said so); but as far as the underlying math goes, recognizing that an integral corresponds to some other quantity that's easier to compute via a different method is completely legit.
$endgroup$
– anomaly
3 hours ago
|
show 3 more comments
$begingroup$
you profesor in mixed up
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
can you post the exact problem question?
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
@MikeySpivak The question sheet is no longer in my possession, but I remember exactly what it said "Evaluate the following integrals:" (there were several). I wrote the integral exactly as it appeared.
$endgroup$
– user146073
4 hours ago
$begingroup$
He was obviously nitpicking, since you were able to bypass a direct (tedious) evaluation, which is what he wanted.
$endgroup$
– herb steinberg
4 hours ago
1
$begingroup$
Did you obtain the correct answer via a logically sound argument? If so (as you did here), then you solved the problem. Of course, the professor can use whatever arbitrary grading metric he wants (and if he meant "evaluate the integral by computing an antiderivative for some reason, then he probably should have said so); but as far as the underlying math goes, recognizing that an integral corresponds to some other quantity that's easier to compute via a different method is completely legit.
$endgroup$
– anomaly
3 hours ago
$begingroup$
you profesor in mixed up
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
you profesor in mixed up
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
can you post the exact problem question?
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
can you post the exact problem question?
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
@MikeySpivak The question sheet is no longer in my possession, but I remember exactly what it said "Evaluate the following integrals:" (there were several). I wrote the integral exactly as it appeared.
$endgroup$
– user146073
4 hours ago
$begingroup$
@MikeySpivak The question sheet is no longer in my possession, but I remember exactly what it said "Evaluate the following integrals:" (there were several). I wrote the integral exactly as it appeared.
$endgroup$
– user146073
4 hours ago
$begingroup$
He was obviously nitpicking, since you were able to bypass a direct (tedious) evaluation, which is what he wanted.
$endgroup$
– herb steinberg
4 hours ago
$begingroup$
He was obviously nitpicking, since you were able to bypass a direct (tedious) evaluation, which is what he wanted.
$endgroup$
– herb steinberg
4 hours ago
1
1
$begingroup$
Did you obtain the correct answer via a logically sound argument? If so (as you did here), then you solved the problem. Of course, the professor can use whatever arbitrary grading metric he wants (and if he meant "evaluate the integral by computing an antiderivative for some reason, then he probably should have said so); but as far as the underlying math goes, recognizing that an integral corresponds to some other quantity that's easier to compute via a different method is completely legit.
$endgroup$
– anomaly
3 hours ago
$begingroup$
Did you obtain the correct answer via a logically sound argument? If so (as you did here), then you solved the problem. Of course, the professor can use whatever arbitrary grading metric he wants (and if he meant "evaluate the integral by computing an antiderivative for some reason, then he probably should have said so); but as far as the underlying math goes, recognizing that an integral corresponds to some other quantity that's easier to compute via a different method is completely legit.
$endgroup$
– anomaly
3 hours ago
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.
You could argue your case. Your professor could argue back that solving the integral by trig substitution does not require the formula $A = pi r^2$, and he did not permit the use of that formula. He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $int_{-r}^r sqrt{r^2-x^2}dx$).
It could go either way for you. But I think it would be a waste of your and your professor's time.
answered 3 hours ago
D_SD_S
13.9k61553
$endgroup$
12
$begingroup$
+1 for "circular reasoning".
$endgroup$
– JonathanZ
3 hours ago
$begingroup$
Got eem $space$
$endgroup$
– D_S
2 hours ago
$begingroup$
Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense.
$endgroup$
– user146073
2 hours ago
1
$begingroup$
Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution
$endgroup$
– D_S
2 hours ago
add a comment |
$begingroup$
An argument could be made that you should include a proof that the integral evaluates the area of a half-disk, rather than just asserting the answer.
Whether you “should” have gotten full points is more a matter of pedagogy than of mathematics, but as a practical tip: using (correct) method Y to solve a problem with instructions to use method X (especially in an intro class and when you are not familiar with the instructor and their teaching philosophy) is always a gamble.
answered 4 hours ago
user7530user7530
35k761113
$endgroup$
add a comment |
$begingroup$
As a student who has had a similar thing happen and heard of it happening to others, my personal recommendation would be to not bring it up again, it will probably be a losing battle. :(
But I would not consider your method "wrong" or "incorrect." There are many ways to solve a problem, you simply just made use of one of them, that wasn't the desired one.
As long as you explained how you came to your answer, the reason why your professor $textit{probably}$ marked you down is that based on the class and section the test covered, the question was designed so that you would display and make use of your knowledge of trig substitution to solve the problem.
I would hope your professor didn't take too many points off (since the method does work), but in these classes you will usually be expected to give the professor a specific method that they are looking for.
answered 4 hours ago
HotdogHotdog
627
$endgroup$
add a comment |
$begingroup$
I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
Did you write this clearly in your test (as you did here)? If not, it is fair to give reduced points. One should always explain where answers come from. If yes, proceed reading.
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Do the instructions clearly disallow your solution? If not, it was not fair to you, and you should insist on it. If yes, read further.
Were these instructions available a priori, or they were included in the test itself? If available a priori, you should have complained about them before the test. If not, the instructions are unfair, and you should try to insist about it as well.
Also, recall that most of this is up to the professor, so you might be with bad luck, sadly.
answered 3 hours ago
Pedro APedro A
2,0461827
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
user146073 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148705%2fcalculus-ii-professor-will-not-accept-my-correct-integral-evaluation-that-uses-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.
You could argue your case. Your professor could argue back that solving the integral by trig substitution does not require the formula $A = pi r^2$, and he did not permit the use of that formula. He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $int_{-r}^r sqrt{r^2-x^2}dx$).
It could go either way for you. But I think it would be a waste of your and your professor's time.
answered 3 hours ago
D_SD_S
13.9k61553
$endgroup$
12
$begingroup$
+1 for "circular reasoning".
$endgroup$
– JonathanZ
3 hours ago
$begingroup$
Got eem $space$
$endgroup$
– D_S
2 hours ago
$begingroup$
Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense.
$endgroup$
– user146073
2 hours ago
1
$begingroup$
Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution
$endgroup$
– D_S
2 hours ago
add a comment |
$begingroup$
I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.
You could argue your case. Your professor could argue back that solving the integral by trig substitution does not require the formula $A = pi r^2$, and he did not permit the use of that formula. He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $int_{-r}^r sqrt{r^2-x^2}dx$).
It could go either way for you. But I think it would be a waste of your and your professor's time.
answered 3 hours ago
D_SD_S
13.9k61553
$endgroup$
12
$begingroup$
+1 for "circular reasoning".
$endgroup$
– JonathanZ
3 hours ago
$begingroup$
Got eem $space$
$endgroup$
– D_S
2 hours ago
$begingroup$
Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense.
$endgroup$
– user146073
2 hours ago
1
$begingroup$
Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution
$endgroup$
– D_S
2 hours ago
add a comment |
$begingroup$
I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.
You could argue your case. Your professor could argue back that solving the integral by trig substitution does not require the formula $A = pi r^2$, and he did not permit the use of that formula. He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $int_{-r}^r sqrt{r^2-x^2}dx$).
It could go either way for you. But I think it would be a waste of your and your professor's time.
answered 3 hours ago
D_SD_S
13.9k61553
$endgroup$
I've done some undergraduate teaching and my policy is always if you get the correct answer by any means then you get full credit, but others have different policies and it's really up to them.
You could argue your case. Your professor could argue back that solving the integral by trig substitution does not require the formula $A = pi r^2$, and he did not permit the use of that formula. He could argue that using that formula entails circular reasoning (the formula for the area of a circle has to be gotten by some limiting or integration method equivalent to evaluating $int_{-r}^r sqrt{r^2-x^2}dx$).
It could go either way for you. But I think it would be a waste of your and your professor's time.
answered 3 hours ago
D_SD_S
13.9k61553
answered 3 hours ago
D_SD_S
13.9k61553
answered 3 hours ago
D_SD_S
13.9k61553
answered 3 hours ago
D_SD_S
13.9k61553
13.9k61553
12
$begingroup$
+1 for "circular reasoning".
$endgroup$
– JonathanZ
3 hours ago
$begingroup$
Got eem $space$
$endgroup$
– D_S
2 hours ago
$begingroup$
Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense.
$endgroup$
– user146073
2 hours ago
1
$begingroup$
Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution
$endgroup$
– D_S
2 hours ago
add a comment |
12
$begingroup$
+1 for "circular reasoning".
$endgroup$
– JonathanZ
3 hours ago
$begingroup$
Got eem $space$
$endgroup$
– D_S
2 hours ago
$begingroup$
Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense.
$endgroup$
– user146073
2 hours ago
1
$begingroup$
Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution
$endgroup$
– D_S
2 hours ago
12
12
$begingroup$
+1 for "circular reasoning".
$endgroup$
– JonathanZ
3 hours ago
$begingroup$
+1 for "circular reasoning".
$endgroup$
– JonathanZ
3 hours ago
$begingroup$
Got eem $space$
$endgroup$
– D_S
2 hours ago
$begingroup$
Got eem $space$
$endgroup$
– D_S
2 hours ago
$begingroup$
Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense.
$endgroup$
– user146073
2 hours ago
$begingroup$
Thank you for the response, fair enough, I don't think I'll pursue further. This made a lot of sense.
$endgroup$
– user146073
2 hours ago
1
1
$begingroup$
Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution
$endgroup$
– D_S
2 hours ago
$begingroup$
Even if you lost points, good on you for noticing a slick way to solve the problem instead of mindlessly grinding out a trig substitution
$endgroup$
– D_S
2 hours ago
add a comment |
$begingroup$
An argument could be made that you should include a proof that the integral evaluates the area of a half-disk, rather than just asserting the answer.
Whether you “should” have gotten full points is more a matter of pedagogy than of mathematics, but as a practical tip: using (correct) method Y to solve a problem with instructions to use method X (especially in an intro class and when you are not familiar with the instructor and their teaching philosophy) is always a gamble.
answered 4 hours ago
user7530user7530
35k761113
$endgroup$
add a comment |
$begingroup$
An argument could be made that you should include a proof that the integral evaluates the area of a half-disk, rather than just asserting the answer.
Whether you “should” have gotten full points is more a matter of pedagogy than of mathematics, but as a practical tip: using (correct) method Y to solve a problem with instructions to use method X (especially in an intro class and when you are not familiar with the instructor and their teaching philosophy) is always a gamble.
answered 4 hours ago
user7530user7530
35k761113
$endgroup$
add a comment |
$begingroup$
An argument could be made that you should include a proof that the integral evaluates the area of a half-disk, rather than just asserting the answer.
Whether you “should” have gotten full points is more a matter of pedagogy than of mathematics, but as a practical tip: using (correct) method Y to solve a problem with instructions to use method X (especially in an intro class and when you are not familiar with the instructor and their teaching philosophy) is always a gamble.
answered 4 hours ago
user7530user7530
35k761113
$endgroup$
An argument could be made that you should include a proof that the integral evaluates the area of a half-disk, rather than just asserting the answer.
Whether you “should” have gotten full points is more a matter of pedagogy than of mathematics, but as a practical tip: using (correct) method Y to solve a problem with instructions to use method X (especially in an intro class and when you are not familiar with the instructor and their teaching philosophy) is always a gamble.
answered 4 hours ago
user7530user7530
35k761113
answered 4 hours ago
user7530user7530
35k761113
answered 4 hours ago
user7530user7530
35k761113
answered 4 hours ago
user7530user7530
35k761113
35k761113
add a comment |
add a comment |
$begingroup$
As a student who has had a similar thing happen and heard of it happening to others, my personal recommendation would be to not bring it up again, it will probably be a losing battle. :(
But I would not consider your method "wrong" or "incorrect." There are many ways to solve a problem, you simply just made use of one of them, that wasn't the desired one.
As long as you explained how you came to your answer, the reason why your professor $textit{probably}$ marked you down is that based on the class and section the test covered, the question was designed so that you would display and make use of your knowledge of trig substitution to solve the problem.
I would hope your professor didn't take too many points off (since the method does work), but in these classes you will usually be expected to give the professor a specific method that they are looking for.
answered 4 hours ago
HotdogHotdog
627
$endgroup$
add a comment |
$begingroup$
As a student who has had a similar thing happen and heard of it happening to others, my personal recommendation would be to not bring it up again, it will probably be a losing battle. :(
But I would not consider your method "wrong" or "incorrect." There are many ways to solve a problem, you simply just made use of one of them, that wasn't the desired one.
As long as you explained how you came to your answer, the reason why your professor $textit{probably}$ marked you down is that based on the class and section the test covered, the question was designed so that you would display and make use of your knowledge of trig substitution to solve the problem.
I would hope your professor didn't take too many points off (since the method does work), but in these classes you will usually be expected to give the professor a specific method that they are looking for.
answered 4 hours ago
HotdogHotdog
627
$endgroup$
add a comment |
$begingroup$
As a student who has had a similar thing happen and heard of it happening to others, my personal recommendation would be to not bring it up again, it will probably be a losing battle. :(
But I would not consider your method "wrong" or "incorrect." There are many ways to solve a problem, you simply just made use of one of them, that wasn't the desired one.
As long as you explained how you came to your answer, the reason why your professor $textit{probably}$ marked you down is that based on the class and section the test covered, the question was designed so that you would display and make use of your knowledge of trig substitution to solve the problem.
I would hope your professor didn't take too many points off (since the method does work), but in these classes you will usually be expected to give the professor a specific method that they are looking for.
answered 4 hours ago
HotdogHotdog
627
$endgroup$
As a student who has had a similar thing happen and heard of it happening to others, my personal recommendation would be to not bring it up again, it will probably be a losing battle. :(
But I would not consider your method "wrong" or "incorrect." There are many ways to solve a problem, you simply just made use of one of them, that wasn't the desired one.
As long as you explained how you came to your answer, the reason why your professor $textit{probably}$ marked you down is that based on the class and section the test covered, the question was designed so that you would display and make use of your knowledge of trig substitution to solve the problem.
I would hope your professor didn't take too many points off (since the method does work), but in these classes you will usually be expected to give the professor a specific method that they are looking for.
answered 4 hours ago
HotdogHotdog
627
answered 4 hours ago
HotdogHotdog
627
answered 4 hours ago
HotdogHotdog
627
answered 4 hours ago
HotdogHotdog
627
627
add a comment |
add a comment |
$begingroup$
I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
Did you write this clearly in your test (as you did here)? If not, it is fair to give reduced points. One should always explain where answers come from. If yes, proceed reading.
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Do the instructions clearly disallow your solution? If not, it was not fair to you, and you should insist on it. If yes, read further.
Were these instructions available a priori, or they were included in the test itself? If available a priori, you should have complained about them before the test. If not, the instructions are unfair, and you should try to insist about it as well.
Also, recall that most of this is up to the professor, so you might be with bad luck, sadly.
answered 3 hours ago
Pedro APedro A
2,0461827
$endgroup$
add a comment |
$begingroup$
I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
Did you write this clearly in your test (as you did here)? If not, it is fair to give reduced points. One should always explain where answers come from. If yes, proceed reading.
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Do the instructions clearly disallow your solution? If not, it was not fair to you, and you should insist on it. If yes, read further.
Were these instructions available a priori, or they were included in the test itself? If available a priori, you should have complained about them before the test. If not, the instructions are unfair, and you should try to insist about it as well.
Also, recall that most of this is up to the professor, so you might be with bad luck, sadly.
answered 3 hours ago
Pedro APedro A
2,0461827
$endgroup$
add a comment |
$begingroup$
I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
Did you write this clearly in your test (as you did here)? If not, it is fair to give reduced points. One should always explain where answers come from. If yes, proceed reading.
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Do the instructions clearly disallow your solution? If not, it was not fair to you, and you should insist on it. If yes, read further.
Were these instructions available a priori, or they were included in the test itself? If available a priori, you should have complained about them before the test. If not, the instructions are unfair, and you should try to insist about it as well.
Also, recall that most of this is up to the professor, so you might be with bad luck, sadly.
answered 3 hours ago
Pedro APedro A
2,0461827
$endgroup$
I noticed it was the equation of the top half of a circle centered at (0, 0) and with radius 4. Knowing this, and my knowledge of the integral indicating the signed area under a curve, I merely took the area of a quarter-circle of radius 4, $frac{1}{4}$$pi$$r^2$ and wrote my answer of 4$pi$.
Did you write this clearly in your test (as you did here)? If not, it is fair to give reduced points. One should always explain where answers come from. If yes, proceed reading.
To receive full credit, you would have had to evaluate an integral, as the instructions indicated.
Do the instructions clearly disallow your solution? If not, it was not fair to you, and you should insist on it. If yes, read further.
Were these instructions available a priori, or they were included in the test itself? If available a priori, you should have complained about them before the test. If not, the instructions are unfair, and you should try to insist about it as well.
Also, recall that most of this is up to the professor, so you might be with bad luck, sadly.
answered 3 hours ago
Pedro APedro A
2,0461827
answered 3 hours ago
Pedro APedro A
2,0461827
answered 3 hours ago
Pedro APedro A
2,0461827
answered 3 hours ago
Pedro APedro A
2,0461827
2,0461827
add a comment |
add a comment |
user146073 is a new contributor. Be nice, and check out our Code of Conduct.
user146073 is a new contributor. Be nice, and check out our Code of Conduct.
user146073 is a new contributor. Be nice, and check out our Code of Conduct.
user146073 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3148705%2fcalculus-ii-professor-will-not-accept-my-correct-integral-evaluation-that-uses-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
you profesor in mixed up
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
can you post the exact problem question?
$endgroup$
– Mikey Spivak
4 hours ago
$begingroup$
@MikeySpivak The question sheet is no longer in my possession, but I remember exactly what it said "Evaluate the following integrals:" (there were several). I wrote the integral exactly as it appeared.
$endgroup$
– user146073
4 hours ago
$begingroup$
He was obviously nitpicking, since you were able to bypass a direct (tedious) evaluation, which is what he wanted.
$endgroup$
– herb steinberg
4 hours ago
1
$begingroup$
Did you obtain the correct answer via a logically sound argument? If so (as you did here), then you solved the problem. Of course, the professor can use whatever arbitrary grading metric he wants (and if he meant "evaluate the integral by computing an antiderivative for some reason, then he probably should have said so); but as far as the underlying math goes, recognizing that an integral corresponds to some other quantity that's easier to compute via a different method is completely legit.
$endgroup$
– anomaly
3 hours ago