How to integrate the curve












2












$begingroup$


We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04
















2












$begingroup$


We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04














2












2








2





$begingroup$


We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields







share|cite|improve this question











$endgroup$




We have a Curve $C:vec{x}left(tright)=begin{pmatrix}1-2t^2\ tend{pmatrix}$ Now you have to calculate $int _Cvec{F}left(vec{x}right)dvec{x}$ for




  • $vec{F}left(vec{x}right)=begin{pmatrix}1\ 0end{pmatrix}$


  • $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$



There are many more practice problems just like this and I'd like to try and solve them but I don't know how to start.



Do I make $1-2t^2 =1$ or $t=0$?



What does $vec{F}left(vec{x}right)=begin{pmatrix}x_2\ x_1end{pmatrix}$ even mean?



Question Pic






calculus integration definite-integrals vectors vector-fields




calculus integration definite-integrals vectors vector-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 4:33







Naochi

















asked Dec 5 '18 at 2:50









NaochiNaochi

626




626








  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04














  • 1




    $begingroup$
    there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:58








  • 2




    $begingroup$
    seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
    $endgroup$
    – Masacroso
    Dec 5 '18 at 3:04








1




1




$begingroup$
there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
$endgroup$
– Masacroso
Dec 5 '18 at 2:58






$begingroup$
there $x:=(x_1,x_2,ldots,x_n)$, assuming that $F:Bbb R^ntoBbb R^m$, so $F(x)=F(x_1,x_2,ldots,x_n)=(x_2,x_1)$ is a function from $Bbb R^ntoBbb R^2$. By the other hand $F(x)=(1,0)$ also goes from $Bbb R^n$ to $Bbb R^2$ (it is a constant function, it maps every $xinBbb R^n$ to the vector $(1,0)inBbb R^2$)
$endgroup$
– Masacroso
Dec 5 '18 at 2:58






2




2




$begingroup$
seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
$endgroup$
– Masacroso
Dec 5 '18 at 3:04




$begingroup$
seeing the definition of $x(t)$ I guess that $n=2$ for this exercise, otherwise the composition $(Fcirc x)(t)=F(x(t))$ would not be defined
$endgroup$
– Masacroso
Dec 5 '18 at 3:04










1 Answer
1






active

oldest

votes


















2












$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026530%2fhow-to-integrate-the-curve%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04


















2












$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04
















2












2








2





$begingroup$

Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?






share|cite|improve this answer











$endgroup$



Note: if you dont know the notation $fcirc g$ just mean the composition of functions $f$ and $g$, that is, $(fcirc g)(s)=f(g(s))$.



There is something that could confuse you: the symbol $x$ is used for two different purposes in the exercise




  1. to denote a vector on the domain of $F$, and


  2. to denote a function from $Bbb RtoBbb R^2$.



Just substituting we have that



$$(Fcirc x)(t)=F(1-2t^2,1)$$



where we assumed here that the domain of $F$ is some subset of $Bbb R^2$ (probably $Bbb R^2$ itself) because the image of the function $x$ is in $Bbb R^2$.



Now, for the first case we have that $(Fcirc x)(t)=(1,0)$, that is, every $tinBbb R$ defines some $x(t)inBbb R^2$, thus $F(x(t))=(1,0)$ because $F$ is already constant.



For the second case we have that $(Fcirc x)(t)=(t,1-2t^2)$, assuming that the notation $x_k$ refer to the coordinates of the vector $x$.



Can you evaluate $int_C F(x)cdot dx$ in each case now?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 3:29

























answered Dec 5 '18 at 3:26









MasacrosoMasacroso

13k41746




13k41746












  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04




















  • $begingroup$
    Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:29










  • $begingroup$
    Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:37










  • $begingroup$
    $int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 3:43










  • $begingroup$
    Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
    $endgroup$
    – Naochi
    Dec 5 '18 at 4:01










  • $begingroup$
    @Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
    $endgroup$
    – Masacroso
    Dec 5 '18 at 4:04


















$begingroup$
Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
$endgroup$
– Naochi
Dec 5 '18 at 3:29




$begingroup$
Ok I will try to do it now. I can evaluate it in the first case. But I am still confused about the second one.
$endgroup$
– Naochi
Dec 5 '18 at 3:29












$begingroup$
Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
$endgroup$
– Naochi
Dec 5 '18 at 3:37




$begingroup$
Ok I still do not understand where am I supposed to use the (1, 0). Before the integration? Will it be used to find t?
$endgroup$
– Naochi
Dec 5 '18 at 3:37












$begingroup$
$int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
$endgroup$
– Naochi
Dec 5 '18 at 3:43




$begingroup$
$int _0^1-4tdt=-2$ This is what I understood from your explanation, I multiply F(x) with the derivative of x(t). Is that correct? @Masacroso
$endgroup$
– Naochi
Dec 5 '18 at 3:43












$begingroup$
Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
$endgroup$
– Naochi
Dec 5 '18 at 4:01




$begingroup$
Ok nvm I could do it :) Thanks for the explanation. So for the second case it would be $int _0^1begin{pmatrix}t\ 1-2t^2end{pmatrix}cdot begin{pmatrix}-4t\ 1end{pmatrix}$ which would then be $int _0^1-4t^2+1-2t^2dt=-1$ No?
$endgroup$
– Naochi
Dec 5 '18 at 4:01












$begingroup$
@Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
$endgroup$
– Masacroso
Dec 5 '18 at 4:04






$begingroup$
@Naochi we have that $$int_C F(x)cdot dx=int_a^b (Fcirc x)(t)cdot x'(t), dt$$ assuming that $[a,b]$ is the domain of the function $x$ that parametrizes the curve $C$. Yes, what you did is right if $[0,1]$ is the domain of $x$
$endgroup$
– Masacroso
Dec 5 '18 at 4:04




















draft saved

draft discarded




















































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%2f3026530%2fhow-to-integrate-the-curve%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

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more