Fundamental Theorem of Calculus I help!












2












$begingroup$


If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



b.) The derivative of $F(x)$ does not exist.



c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



I can't seem to find the answer. I found that
$$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
sqrt{5}}{3}$$



and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



so, to me the answer is not provided. Any help would be much appreciated.






real-analysis calculus definite-integrals







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  • 1




    $begingroup$
    Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus.
    $endgroup$
    – Ethan Bolker
    2 hours ago
















2












$begingroup$


If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



b.) The derivative of $F(x)$ does not exist.



c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



I can't seem to find the answer. I found that
$$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
sqrt{5}}{3}$$



and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



so, to me the answer is not provided. Any help would be much appreciated.






real-analysis calculus definite-integrals







share|cite|improve this question







New contributor




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







$endgroup$








  • 1




    $begingroup$
    Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus.
    $endgroup$
    – Ethan Bolker
    2 hours ago














2












2








2


1



$begingroup$


If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



b.) The derivative of $F(x)$ does not exist.



c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



I can't seem to find the answer. I found that
$$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
sqrt{5}}{3}$$



and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



so, to me the answer is not provided. Any help would be much appreciated.






real-analysis calculus definite-integrals







share|cite|improve this question







New contributor




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







$endgroup$




If $F(x)=int_2^{x^3}sqrt{t^2+t^4}dt$



a.) The integral of $F(x)$ is $3x^2sqrt{x^2+x^4}$.



b.) The derivative of $F(x)$ does not exist.



c.) $F'(x)=3x^2sqrt{x^6+x^{12}}$.



I can't seem to find the answer. I found that
$$F(x)=int_2^{x^3}sqrt{t^2+t^4}dt= frac{(x^6+1)^{3/2}}{3}-frac{5
sqrt{5}}{3}$$



and $$frac{d}{dx}F(x)=int_2^{x^3}sqrt{t^2+t^4}dt=3x^5sqrt{x^6+1}$$



so, to me the answer is not provided. Any help would be much appreciated.






real-analysis calculus definite-integrals




real-analysis calculus definite-integrals






share|cite|improve this question







New contributor




Ryan Pennell 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




Ryan Pennell 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






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asked 7 hours ago









Ryan PennellRyan Pennell

172




172




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






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








  • 1




    $begingroup$
    Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus.
    $endgroup$
    – Ethan Bolker
    2 hours ago














  • 1




    $begingroup$
    Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus.
    $endgroup$
    – Ethan Bolker
    2 hours ago








1




1




$begingroup$
Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus.
$endgroup$
– Ethan Bolker
2 hours ago




$begingroup$
Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus.
$endgroup$
– Ethan Bolker
2 hours ago










3 Answers
3






active

oldest

votes


















7












$begingroup$

Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



$F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you Mark!
    $endgroup$
    – Ryan Pennell
    7 hours ago



















4












$begingroup$

Hint:
If $F(x)= int_a^{g(x)} f(t) dt$ , then
$$F’(x) = f(g(x))g’(x)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you Jose!
    $endgroup$
    – Ryan Pennell
    7 hours ago










  • $begingroup$
    Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
    $endgroup$
    – JoseSquare
    6 hours ago





















1












$begingroup$

By the way, aside from the good answers provided by Mark and JoseSquare, note that you actually did compute $F(x)$ and $F'(x)$ correctly! The result you obtained for $F'(x)$ is equivalent to the expression given in choice (c).






share|cite|improve this answer








New contributor




user647194 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













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    3 Answers
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    3 Answers
    3






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    active

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    7












    $begingroup$

    Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



    $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Mark!
      $endgroup$
      – Ryan Pennell
      7 hours ago
















    7












    $begingroup$

    Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



    $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Mark!
      $endgroup$
      – Ryan Pennell
      7 hours ago














    7












    7








    7





    $begingroup$

    Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



    $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$






    share|cite|improve this answer









    $endgroup$



    Let's say $G(x)$ is an antiderivative of $sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:



    $F'(x)=3x^2G'(x^3)=3x^2sqrt{x^6+x^{12}}$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 7 hours ago









    MarkMark

    8,957521




    8,957521












    • $begingroup$
      Thank you Mark!
      $endgroup$
      – Ryan Pennell
      7 hours ago


















    • $begingroup$
      Thank you Mark!
      $endgroup$
      – Ryan Pennell
      7 hours ago
















    $begingroup$
    Thank you Mark!
    $endgroup$
    – Ryan Pennell
    7 hours ago




    $begingroup$
    Thank you Mark!
    $endgroup$
    – Ryan Pennell
    7 hours ago











    4












    $begingroup$

    Hint:
    If $F(x)= int_a^{g(x)} f(t) dt$ , then
    $$F’(x) = f(g(x))g’(x)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Jose!
      $endgroup$
      – Ryan Pennell
      7 hours ago










    • $begingroup$
      Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
      $endgroup$
      – JoseSquare
      6 hours ago


















    4












    $begingroup$

    Hint:
    If $F(x)= int_a^{g(x)} f(t) dt$ , then
    $$F’(x) = f(g(x))g’(x)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you Jose!
      $endgroup$
      – Ryan Pennell
      7 hours ago










    • $begingroup$
      Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
      $endgroup$
      – JoseSquare
      6 hours ago
















    4












    4








    4





    $begingroup$

    Hint:
    If $F(x)= int_a^{g(x)} f(t) dt$ , then
    $$F’(x) = f(g(x))g’(x)$$






    share|cite|improve this answer









    $endgroup$



    Hint:
    If $F(x)= int_a^{g(x)} f(t) dt$ , then
    $$F’(x) = f(g(x))g’(x)$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 7 hours ago









    JoseSquareJoseSquare

    75012




    75012












    • $begingroup$
      Thank you Jose!
      $endgroup$
      – Ryan Pennell
      7 hours ago










    • $begingroup$
      Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
      $endgroup$
      – JoseSquare
      6 hours ago




















    • $begingroup$
      Thank you Jose!
      $endgroup$
      – Ryan Pennell
      7 hours ago










    • $begingroup$
      Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
      $endgroup$
      – JoseSquare
      6 hours ago


















    $begingroup$
    Thank you Jose!
    $endgroup$
    – Ryan Pennell
    7 hours ago




    $begingroup$
    Thank you Jose!
    $endgroup$
    – Ryan Pennell
    7 hours ago












    $begingroup$
    Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
    $endgroup$
    – JoseSquare
    6 hours ago






    $begingroup$
    Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $int e^{-x^2} dx$) @RyanPennell
    $endgroup$
    – JoseSquare
    6 hours ago













    1












    $begingroup$

    By the way, aside from the good answers provided by Mark and JoseSquare, note that you actually did compute $F(x)$ and $F'(x)$ correctly! The result you obtained for $F'(x)$ is equivalent to the expression given in choice (c).






    share|cite|improve this answer








    New contributor




    user647194 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$


















      1












      $begingroup$

      By the way, aside from the good answers provided by Mark and JoseSquare, note that you actually did compute $F(x)$ and $F'(x)$ correctly! The result you obtained for $F'(x)$ is equivalent to the expression given in choice (c).






      share|cite|improve this answer








      New contributor




      user647194 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$
















        1












        1








        1





        $begingroup$

        By the way, aside from the good answers provided by Mark and JoseSquare, note that you actually did compute $F(x)$ and $F'(x)$ correctly! The result you obtained for $F'(x)$ is equivalent to the expression given in choice (c).






        share|cite|improve this answer








        New contributor




        user647194 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$



        By the way, aside from the good answers provided by Mark and JoseSquare, note that you actually did compute $F(x)$ and $F'(x)$ correctly! The result you obtained for $F'(x)$ is equivalent to the expression given in choice (c).







        share|cite|improve this answer








        New contributor




        user647194 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 answer



        share|cite|improve this answer






        New contributor




        user647194 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 2 hours ago









        user647194user647194

        111




        111




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        New contributor





        user647194 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        user647194 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






















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            Tu-95轟炸機

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            body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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