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I have a line integral (I hope it is) question as:

question

And the solution to this is :

Solution

Now when I try to solve the question using parametric method taking:

$x(t)=t$, $y(t)=2t$

and using the below formula.

$\displaystyle \int _{\mathcal {C}}f(\mathbf {r} )\,ds=\int _{a}^{b}f\left(\mathbf {r} (t)\right)|\mathbf {r} '(t)|\,dt$

I get an answer $33\sqrt{5}$

Can please anyone tell if I am somehow wrong or the solution above contains mistake(s)?

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  • $\begingroup$ +1, they didn't count the factor $|\mathbf{r}'(t)|$ $\endgroup$
    – tortue
    Aug 8, 2018 at 10:15
  • $\begingroup$ Corrected comment: I think you are right. I cannot see that they are taking into account that this line segment is $\sqrt 5$ long rather than $1$. But I haven't actually checked the calculations, so this is not an answer. $\endgroup$
    – Arthur
    Aug 8, 2018 at 10:20
  • $\begingroup$ If you can just verify it and put it as an answer it would be great. $\endgroup$
    – paulplusx
    Aug 8, 2018 at 10:26
  • $\begingroup$ In the future, please take the time to enter important parts of your question—in this case, basically the question itself—as text instead of as pictures. Images are neither searchable nor accessible to screen readers. $\endgroup$
    – amd
    Aug 8, 2018 at 20:11
  • $\begingroup$ @amd Yes, will do that. Thank you for the advice. $\endgroup$
    – paulplusx
    Aug 9, 2018 at 2:53

1 Answer 1

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Taking $\mathbf{r}(t)$ as mentioned in your question one has $$\displaystyle \int _{\mathcal {C}}f(\mathbf {r} )\,ds=\int _{a}^{b}f\left(\mathbf {r} (t)\right)|\mathbf {r} '(t)|\,dt = \int_{0}^{1}(4t^3 + 160t^4 ) \cdot \sqrt{5} ~ dt = 33\sqrt{5},$$

where the last equation is due to the solution from your question or simple tedious calculations.

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  • $\begingroup$ Thank you for your verification. $\endgroup$
    – paulplusx
    Aug 8, 2018 at 11:54

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