If $F(x)=frac{x^4-3}{x^4+1}$ is a primitive of $f(x)$ find $int_{0}^{1} xf(x) dx$












2














Let $f:mathbb{R}tomathbb{R}$ be a differentiable function.



If $F(x)=frac{x^4-3}{x^4+1}$ is a primitive of $f(x)$ find $int_{0}^{1} xf(x) dx$



I literally have no idea how to integrate this.



I tried integrating by parts (and finding the derivative of $F(x)$) but I end up getting a even worse integral...



The correct answer apparently is $-3$.






calculus real-analysis integration derivatives definite-integrals







share|cite|improve this question






















  • Please check the statement of your exercise. The correct result is not $-3$!
    – Robert Z
    Nov 26 at 15:44






  • 1




    Since $f(x)=F'(x)=frac{16x^3}{x^4+1}$, there's no way that the integral is $-3$, because you're doing $int_0^1frac{16x^4}{x^4+1},dx$.
    – egreg
    Nov 26 at 15:49










  • @parishilton egreg is right! Something is wrong here. Please check!!
    – Robert Z
    Nov 26 at 15:50










  • Yep! Clearly it's not $-3$ but my solution says it's... I'm getting something very far away from -3. I'll try to ask my teacher about it. Thank you!
    – parishilton
    Nov 26 at 15:53


















2














Let $f:mathbb{R}tomathbb{R}$ be a differentiable function.



If $F(x)=frac{x^4-3}{x^4+1}$ is a primitive of $f(x)$ find $int_{0}^{1} xf(x) dx$



I literally have no idea how to integrate this.



I tried integrating by parts (and finding the derivative of $F(x)$) but I end up getting a even worse integral...



The correct answer apparently is $-3$.






calculus real-analysis integration derivatives definite-integrals







share|cite|improve this question






















  • Please check the statement of your exercise. The correct result is not $-3$!
    – Robert Z
    Nov 26 at 15:44






  • 1




    Since $f(x)=F'(x)=frac{16x^3}{x^4+1}$, there's no way that the integral is $-3$, because you're doing $int_0^1frac{16x^4}{x^4+1},dx$.
    – egreg
    Nov 26 at 15:49










  • @parishilton egreg is right! Something is wrong here. Please check!!
    – Robert Z
    Nov 26 at 15:50










  • Yep! Clearly it's not $-3$ but my solution says it's... I'm getting something very far away from -3. I'll try to ask my teacher about it. Thank you!
    – parishilton
    Nov 26 at 15:53
















2












2








2







Let $f:mathbb{R}tomathbb{R}$ be a differentiable function.



If $F(x)=frac{x^4-3}{x^4+1}$ is a primitive of $f(x)$ find $int_{0}^{1} xf(x) dx$



I literally have no idea how to integrate this.



I tried integrating by parts (and finding the derivative of $F(x)$) but I end up getting a even worse integral...



The correct answer apparently is $-3$.






calculus real-analysis integration derivatives definite-integrals







share|cite|improve this question













Let $f:mathbb{R}tomathbb{R}$ be a differentiable function.



If $F(x)=frac{x^4-3}{x^4+1}$ is a primitive of $f(x)$ find $int_{0}^{1} xf(x) dx$



I literally have no idea how to integrate this.



I tried integrating by parts (and finding the derivative of $F(x)$) but I end up getting a even worse integral...



The correct answer apparently is $-3$.






calculus real-analysis integration derivatives definite-integrals




calculus real-analysis integration derivatives definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 15:16









parishilton

18110




18110












  • Please check the statement of your exercise. The correct result is not $-3$!
    – Robert Z
    Nov 26 at 15:44






  • 1




    Since $f(x)=F'(x)=frac{16x^3}{x^4+1}$, there's no way that the integral is $-3$, because you're doing $int_0^1frac{16x^4}{x^4+1},dx$.
    – egreg
    Nov 26 at 15:49










  • @parishilton egreg is right! Something is wrong here. Please check!!
    – Robert Z
    Nov 26 at 15:50










  • Yep! Clearly it's not $-3$ but my solution says it's... I'm getting something very far away from -3. I'll try to ask my teacher about it. Thank you!
    – parishilton
    Nov 26 at 15:53




















  • Please check the statement of your exercise. The correct result is not $-3$!
    – Robert Z
    Nov 26 at 15:44






  • 1




    Since $f(x)=F'(x)=frac{16x^3}{x^4+1}$, there's no way that the integral is $-3$, because you're doing $int_0^1frac{16x^4}{x^4+1},dx$.
    – egreg
    Nov 26 at 15:49










  • @parishilton egreg is right! Something is wrong here. Please check!!
    – Robert Z
    Nov 26 at 15:50










  • Yep! Clearly it's not $-3$ but my solution says it's... I'm getting something very far away from -3. I'll try to ask my teacher about it. Thank you!
    – parishilton
    Nov 26 at 15:53


















Please check the statement of your exercise. The correct result is not $-3$!
– Robert Z
Nov 26 at 15:44




Please check the statement of your exercise. The correct result is not $-3$!
– Robert Z
Nov 26 at 15:44




1




1




Since $f(x)=F'(x)=frac{16x^3}{x^4+1}$, there's no way that the integral is $-3$, because you're doing $int_0^1frac{16x^4}{x^4+1},dx$.
– egreg
Nov 26 at 15:49




Since $f(x)=F'(x)=frac{16x^3}{x^4+1}$, there's no way that the integral is $-3$, because you're doing $int_0^1frac{16x^4}{x^4+1},dx$.
– egreg
Nov 26 at 15:49












@parishilton egreg is right! Something is wrong here. Please check!!
– Robert Z
Nov 26 at 15:50




@parishilton egreg is right! Something is wrong here. Please check!!
– Robert Z
Nov 26 at 15:50












Yep! Clearly it's not $-3$ but my solution says it's... I'm getting something very far away from -3. I'll try to ask my teacher about it. Thank you!
– parishilton
Nov 26 at 15:53






Yep! Clearly it's not $-3$ but my solution says it's... I'm getting something very far away from -3. I'll try to ask my teacher about it. Thank you!
– parishilton
Nov 26 at 15:53












1 Answer
1






active

oldest

votes


















5














HINT



Integrating by parts we have



$$int_{0}^{1} xf(x) dx=[xF(x)]_{0}^{1}-int_{0}^{1} F(x) dx$$






share|cite|improve this answer



















  • 3




    It should be $int xf(x) dx=[xF(x)]_0^1-int_{0}^{1} F(x) dx$.
    – Robert Z
    Nov 26 at 15:29










  • @egreg Ah ok that's indeed what I wrote up there but not in the answer! I fix that :)
    – gimusi
    Nov 26 at 15:37










  • Thank you to everyone!!
    – parishilton
    Nov 26 at 15:38










  • @RobertZ Thanks a lot! I was sleeping at all :)
    – gimusi
    Nov 26 at 15:39










  • @parishilton You are welcome and sorry for the typo in the last part! I had in mind something but I wrote something different! Bye
    – gimusi
    Nov 26 at 15:40











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














HINT



Integrating by parts we have



$$int_{0}^{1} xf(x) dx=[xF(x)]_{0}^{1}-int_{0}^{1} F(x) dx$$






share|cite|improve this answer



















  • 3




    It should be $int xf(x) dx=[xF(x)]_0^1-int_{0}^{1} F(x) dx$.
    – Robert Z
    Nov 26 at 15:29










  • @egreg Ah ok that's indeed what I wrote up there but not in the answer! I fix that :)
    – gimusi
    Nov 26 at 15:37










  • Thank you to everyone!!
    – parishilton
    Nov 26 at 15:38










  • @RobertZ Thanks a lot! I was sleeping at all :)
    – gimusi
    Nov 26 at 15:39










  • @parishilton You are welcome and sorry for the typo in the last part! I had in mind something but I wrote something different! Bye
    – gimusi
    Nov 26 at 15:40
















5














HINT



Integrating by parts we have



$$int_{0}^{1} xf(x) dx=[xF(x)]_{0}^{1}-int_{0}^{1} F(x) dx$$






share|cite|improve this answer



















  • 3




    It should be $int xf(x) dx=[xF(x)]_0^1-int_{0}^{1} F(x) dx$.
    – Robert Z
    Nov 26 at 15:29










  • @egreg Ah ok that's indeed what I wrote up there but not in the answer! I fix that :)
    – gimusi
    Nov 26 at 15:37










  • Thank you to everyone!!
    – parishilton
    Nov 26 at 15:38










  • @RobertZ Thanks a lot! I was sleeping at all :)
    – gimusi
    Nov 26 at 15:39










  • @parishilton You are welcome and sorry for the typo in the last part! I had in mind something but I wrote something different! Bye
    – gimusi
    Nov 26 at 15:40














5












5








5






HINT



Integrating by parts we have



$$int_{0}^{1} xf(x) dx=[xF(x)]_{0}^{1}-int_{0}^{1} F(x) dx$$






share|cite|improve this answer














HINT



Integrating by parts we have



$$int_{0}^{1} xf(x) dx=[xF(x)]_{0}^{1}-int_{0}^{1} F(x) dx$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 15:37

























answered Nov 26 at 15:21









gimusi

1




1








  • 3




    It should be $int xf(x) dx=[xF(x)]_0^1-int_{0}^{1} F(x) dx$.
    – Robert Z
    Nov 26 at 15:29










  • @egreg Ah ok that's indeed what I wrote up there but not in the answer! I fix that :)
    – gimusi
    Nov 26 at 15:37










  • Thank you to everyone!!
    – parishilton
    Nov 26 at 15:38










  • @RobertZ Thanks a lot! I was sleeping at all :)
    – gimusi
    Nov 26 at 15:39










  • @parishilton You are welcome and sorry for the typo in the last part! I had in mind something but I wrote something different! Bye
    – gimusi
    Nov 26 at 15:40














  • 3




    It should be $int xf(x) dx=[xF(x)]_0^1-int_{0}^{1} F(x) dx$.
    – Robert Z
    Nov 26 at 15:29










  • @egreg Ah ok that's indeed what I wrote up there but not in the answer! I fix that :)
    – gimusi
    Nov 26 at 15:37










  • Thank you to everyone!!
    – parishilton
    Nov 26 at 15:38










  • @RobertZ Thanks a lot! I was sleeping at all :)
    – gimusi
    Nov 26 at 15:39










  • @parishilton You are welcome and sorry for the typo in the last part! I had in mind something but I wrote something different! Bye
    – gimusi
    Nov 26 at 15:40








3




3




It should be $int xf(x) dx=[xF(x)]_0^1-int_{0}^{1} F(x) dx$.
– Robert Z
Nov 26 at 15:29




It should be $int xf(x) dx=[xF(x)]_0^1-int_{0}^{1} F(x) dx$.
– Robert Z
Nov 26 at 15:29












@egreg Ah ok that's indeed what I wrote up there but not in the answer! I fix that :)
– gimusi
Nov 26 at 15:37




@egreg Ah ok that's indeed what I wrote up there but not in the answer! I fix that :)
– gimusi
Nov 26 at 15:37












Thank you to everyone!!
– parishilton
Nov 26 at 15:38




Thank you to everyone!!
– parishilton
Nov 26 at 15:38












@RobertZ Thanks a lot! I was sleeping at all :)
– gimusi
Nov 26 at 15:39




@RobertZ Thanks a lot! I was sleeping at all :)
– gimusi
Nov 26 at 15:39












@parishilton You are welcome and sorry for the typo in the last part! I had in mind something but I wrote something different! Bye
– gimusi
Nov 26 at 15:40




@parishilton You are welcome and sorry for the typo in the last part! I had in mind something but I wrote something different! Bye
– gimusi
Nov 26 at 15:40


















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