Calculus II Question The Next CEO of Stack OverflowLength of an AstroidUnderstanding this calculus simplificationIntegration problem: $int x^2 -x 4^-x^2 dx$Finding the parametric form of a standard equationApplication of “twice the integral” even if the function is not graphically even?Find the length of the parametric curveFind the exact length of the parametric curve(Not sure what I'm doing wrong)Calculus 2 moments question.The length of a parametric curveParametric curve length - calculus

Why didn't Khan get resurrected in the Genesis Explosion?

How did the Bene Gesserit know how to make a Kwisatz Haderach?

Bold, vivid family

What flight has the highest ratio of time difference to flight time?

In excess I'm lethal

Why don't programming languages automatically manage the synchronous/asynchronous problem?

What was the first Unix version to run on a microcomputer?

Help understanding this unsettling image of Titan, Epimetheus, and Saturn's rings?

How to start emacs in "nothing" mode (`fundamental-mode`)

Is it ever safe to open a suspicious html file (e.g. email attachment)?

If the heap is initialized for security, then why is the stack uninitialized?

Why do remote companies require working in the US?

Is there an analogue of projective spaces for proper schemes?

Does it take more energy to get to Venus or to Mars?

Several mode to write the symbol of a vector

Novel about a guy who is possessed by the divine essence and the world ends?

Are there any unintended negative consequences to allowing PCs to gain multiple levels at once in a short milestone-XP game?

What is "(CFMCC)" on an ILS approach chart?

Elegant way to replace substring in a regex with optional groups in Python?

Non-deterministic sum of floats

Are there any limitations on attacking while grappling?

Inappropriate reference requests from Journal reviewers

Complex fractions

What can we do to stop prior company from asking us questions?



Calculus II Question



The Next CEO of Stack OverflowLength of an AstroidUnderstanding this calculus simplificationIntegration problem: $int x^2 -x 4^-x^2 dx$Finding the parametric form of a standard equationApplication of “twice the integral” even if the function is not graphically even?Find the length of the parametric curveFind the exact length of the parametric curve(Not sure what I'm doing wrong)Calculus 2 moments question.The length of a parametric curveParametric curve length - calculus










2












$begingroup$


Find the length of the following parametric curve.



$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$



I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.



My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$



$$frac23cdot 17^3/2+4-frac23$$










share|cite|improve this question









New contributor




curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
    $endgroup$
    – Ross Millikan
    1 hour ago







  • 1




    $begingroup$
    Isn't there a square root missing in your length formula?
    $endgroup$
    – John Wayland Bales
    1 hour ago






  • 1




    $begingroup$
    We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
    $endgroup$
    – David Peterson
    1 hour ago






  • 1




    $begingroup$
    @curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
    $endgroup$
    – John Omielan
    56 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    55 mins ago















2












$begingroup$


Find the length of the following parametric curve.



$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$



I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.



My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$



$$frac23cdot 17^3/2+4-frac23$$










share|cite|improve this question









New contributor




curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
    $endgroup$
    – Ross Millikan
    1 hour ago







  • 1




    $begingroup$
    Isn't there a square root missing in your length formula?
    $endgroup$
    – John Wayland Bales
    1 hour ago






  • 1




    $begingroup$
    We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
    $endgroup$
    – David Peterson
    1 hour ago






  • 1




    $begingroup$
    @curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
    $endgroup$
    – John Omielan
    56 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    55 mins ago













2












2








2





$begingroup$


Find the length of the following parametric curve.



$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$



I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.



My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$



$$frac23cdot 17^3/2+4-frac23$$










share|cite|improve this question









New contributor




curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Find the length of the following parametric curve.



$$x(t)=5+6t^4 ,quad y(t)=5+4t^6 ,qquad 0  ≤  t  ≤  2.$$



I used the formula
$$int_0^2sqrtleft(fracdxdtright)^2+left(fracdydtright)^2dt$$
And I found
$$frac23cdot 17^3/2+4-frac23$$
However I got it wrong. I don't know where I went wrong. Any help would be apriciated.



My steps:
$$left(fracdxdtright) = 24cdot t^3 $$
$$left(fracdydtright) = 24cdot t^5 $$
$$int_0^2sqrtleft(24cdot t^3right)^2+left(24cdot t^5right)^2dt$$
$$int_0^2sqrtleft(576cdot t^6right)+left(576cdot t^10right)dt$$
$$int_0^2sqrtleft(576cdot t^6right) cdot left(1+t^4right)dt$$
$$24+int_0^2sqrtleft(t^6right) cdot left(1+t^4right)dt$$



$$frac23cdot 17^3/2+4-frac23$$







calculus integration






share|cite|improve this question









New contributor




curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 50 mins ago









rash

585116




585116






New contributor




curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









curiousengcuriouseng

134




134




New contributor




curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






curiouseng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 3




    $begingroup$
    What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
    $endgroup$
    – Ross Millikan
    1 hour ago







  • 1




    $begingroup$
    Isn't there a square root missing in your length formula?
    $endgroup$
    – John Wayland Bales
    1 hour ago






  • 1




    $begingroup$
    We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
    $endgroup$
    – David Peterson
    1 hour ago






  • 1




    $begingroup$
    @curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
    $endgroup$
    – John Omielan
    56 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    55 mins ago












  • 3




    $begingroup$
    What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
    $endgroup$
    – Ross Millikan
    1 hour ago







  • 1




    $begingroup$
    Isn't there a square root missing in your length formula?
    $endgroup$
    – John Wayland Bales
    1 hour ago






  • 1




    $begingroup$
    We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
    $endgroup$
    – David Peterson
    1 hour ago






  • 1




    $begingroup$
    @curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
    $endgroup$
    – John Omielan
    56 mins ago






  • 1




    $begingroup$
    @JohnOmielan that’s exactly what’s wrong
    $endgroup$
    – Shalop
    55 mins ago







3




3




$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago





$begingroup$
What is 6t4? What is 4t6? Without seeing your work we can't see where you went wrong. Answer keys are wrong sometimes. You should have a square root of the sum of the squares in your integral.
$endgroup$
– Ross Millikan
1 hour ago





1




1




$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago




$begingroup$
Isn't there a square root missing in your length formula?
$endgroup$
– John Wayland Bales
1 hour ago




1




1




$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago




$begingroup$
We probably cannot figure out what you did wrong unless you show the work ending with that as an answer..
$endgroup$
– David Peterson
1 hour ago




1




1




$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
56 mins ago




$begingroup$
@curiouseng At the start of your second last line, is "$24 + $" part what you actually used, or is it a typo as you meant it to be $24$ times the integral?
$endgroup$
– John Omielan
56 mins ago




1




1




$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
55 mins ago




$begingroup$
@JohnOmielan that’s exactly what’s wrong
$endgroup$
– Shalop
55 mins ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$



Which gives us:



$$int_0^2 24sqrtt^6+t^10dt$$



Which, when integrated, gives us: $$68sqrt17-4$$



I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
    $endgroup$
    – curiouseng
    58 mins ago










  • $begingroup$
    @curiouseng You are very welcome, regards!
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    57 mins ago


















2












$begingroup$

Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.



Line 5 is correct.



Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.



You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
$$beginalign*
24 int_t=0^2 sqrtt^6(1+t^4) , dt
&= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
&= 6 int_u=1^17 sqrtu , du \
&= 6 left[frac2u^3/23 right]_u=0^17 \
&= 4 (17^3/2 - 1) \
&= 68 sqrt17 - 4.
endalign*$$






share|cite|improve this answer









$endgroup$













    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
    );



    );






    curiouseng is a new contributor. Be nice, and check out our Code of Conduct.









    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167826%2fcalculus-ii-question%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$



    Which gives us:



    $$int_0^2 24sqrtt^6+t^10dt$$



    Which, when integrated, gives us: $$68sqrt17-4$$



    I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
      $endgroup$
      – curiouseng
      58 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      57 mins ago















    2












    $begingroup$

    Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$



    Which gives us:



    $$int_0^2 24sqrtt^6+t^10dt$$



    Which, when integrated, gives us: $$68sqrt17-4$$



    I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
      $endgroup$
      – curiouseng
      58 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      57 mins ago













    2












    2








    2





    $begingroup$

    Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$



    Which gives us:



    $$int_0^2 24sqrtt^6+t^10dt$$



    Which, when integrated, gives us: $$68sqrt17-4$$



    I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.






    share|cite|improve this answer









    $endgroup$



    Okay, start from the beginning $$x'(t)=24t^3; y'(t)=24t^5$$



    Which gives us:



    $$int_0^2 24sqrtt^6+t^10dt$$



    Which, when integrated, gives us: $$68sqrt17-4$$



    I don't, however, know where you went wrong. It could be either a sign error, or a calculation error.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost

    527217




    527217











    • $begingroup$
      Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
      $endgroup$
      – curiouseng
      58 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      57 mins ago
















    • $begingroup$
      Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
      $endgroup$
      – curiouseng
      58 mins ago










    • $begingroup$
      @curiouseng You are very welcome, regards!
      $endgroup$
      – Bertrand Wittgenstein's Ghost
      57 mins ago















    $begingroup$
    Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
    $endgroup$
    – curiouseng
    58 mins ago




    $begingroup$
    Thank you for your help. I used an online integral calculator to see where I went wrong and it was a basic calculation mistake :( Again thank you for your time.
    $endgroup$
    – curiouseng
    58 mins ago












    $begingroup$
    @curiouseng You are very welcome, regards!
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    57 mins ago




    $begingroup$
    @curiouseng You are very welcome, regards!
    $endgroup$
    – Bertrand Wittgenstein's Ghost
    57 mins ago











    2












    $begingroup$

    Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.



    Line 5 is correct.



    Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.



    You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
    $$beginalign*
    24 int_t=0^2 sqrtt^6(1+t^4) , dt
    &= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
    &= 6 int_u=1^17 sqrtu , du \
    &= 6 left[frac2u^3/23 right]_u=0^17 \
    &= 4 (17^3/2 - 1) \
    &= 68 sqrt17 - 4.
    endalign*$$






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.



      Line 5 is correct.



      Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.



      You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
      $$beginalign*
      24 int_t=0^2 sqrtt^6(1+t^4) , dt
      &= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
      &= 6 int_u=1^17 sqrtu , du \
      &= 6 left[frac2u^3/23 right]_u=0^17 \
      &= 4 (17^3/2 - 1) \
      &= 68 sqrt17 - 4.
      endalign*$$






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.



        Line 5 is correct.



        Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.



        You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
        $$beginalign*
        24 int_t=0^2 sqrtt^6(1+t^4) , dt
        &= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
        &= 6 int_u=1^17 sqrtu , du \
        &= 6 left[frac2u^3/23 right]_u=0^17 \
        &= 4 (17^3/2 - 1) \
        &= 68 sqrt17 - 4.
        endalign*$$






        share|cite|improve this answer









        $endgroup$



        Line 4 should read $$int_t=0^2 sqrt576 t^6 + 576 t^10 , dt.$$ This is a typesetting error.



        Line 5 is correct.



        Line 6 should read $$24 int_t=0^2 sqrtt^6 (1+t^4) , dt.$$ The use of the addition sign is incorrect because $24$ is a factor in the integrand, not a term.



        You do not demonstrate how to proceed from Line 6 to Line 7. I would complete the computation as follows:
        $$beginalign*
        24 int_t=0^2 sqrtt^6(1+t^4) , dt
        &= 24 int_t=0^2 t^3 sqrt1+t^4 , dt qquad (u = 1+t^4; ; du = 4t^3 , dt) \
        &= 6 int_u=1^17 sqrtu , du \
        &= 6 left[frac2u^3/23 right]_u=0^17 \
        &= 4 (17^3/2 - 1) \
        &= 68 sqrt17 - 4.
        endalign*$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 29 mins ago









        heropupheropup

        64.8k764103




        64.8k764103




















            curiouseng is a new contributor. Be nice, and check out our Code of Conduct.









            draft saved

            draft discarded


















            curiouseng is a new contributor. Be nice, and check out our Code of Conduct.












            curiouseng is a new contributor. Be nice, and check out our Code of Conduct.











            curiouseng 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.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167826%2fcalculus-ii-question%23new-answer', 'question_page');

            );

            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







            Popular posts from this blog

            Can not update quote_id field of “quote_item” table magento 2Magento 2.1 - We can't remove the item. (Shopping Cart doesnt allow us to remove items before becomes empty)Add value for custom quote item attribute using REST apiREST API endpoint v1/carts/cartId/items always returns error messageCorrect way to save entries to databaseHow to remove all associated quote objects of a customer completelyMagento 2 - Save value from custom input field to quote_itemGet quote_item data using quote id and product id filter in Magento 2How to set additional data to quote_item table from controller in Magento 2?What is the purpose of additional_data column in quote_item table in magento2Set Custom Price to Quote item magento2 from controller

            How to solve knockout JS error in Magento 2 Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?(Magento2) knockout.js:3012 Uncaught ReferenceError: Unable to process bindingUnable to process binding Knockout.js magento 2Cannot read property `scopeLabel` of undefined on Product Detail PageCan't get Customer Data on frontend in Magento 2Magento2 Order Summary - unable to process bindingKO templates are not loading in Magento 2.1 applicationgetting knockout js error magento 2Product grid not load -— Unable to process binding Knockout.js magento 2Product form not loaded in magento2Uncaught ReferenceError: Unable to process binding “if: function()return (isShowLegend()) ” magento 2

            Nissan Patrol Зміст Перше покоління — 4W60 (1951-1960) | Друге покоління — 60 series (1960-1980) | Третє покоління (1980–2002) | Четверте покоління — Y60 (1987–1998) | П'яте покоління — Y61 (1997–2013) | Шосте покоління — Y62 (2010- ) | Посилання | Зноски | Навігаційне менюОфіційний український сайтТест-драйв Nissan Patrol 2010 7-го поколінняNissan PatrolКак мы тестировали Nissan Patrol 2016рвиправивши або дописавши її