Integral equation with a square root seems to go on












2












$begingroup$


$$ int_0^1 sqrt{1+y^2} (a_0 + a_1y) dy
$$



I have tried to solve this equation by using integration by parts. I am using the square root as the part which I am differentiating. By using this, the method seems like it is ongoing and does not have a final answer. Any guidance will be much appreciated



I am still trying to figure out the formatting so please forgive me






calculus integration







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$endgroup$












  • $begingroup$
    MathJax hint: To group things under a square root, enclose them in braces, so sqrt{1+2} gives $sqrt {1+2}$. It works everywhere, like superscripts and subscripts. Is the last term just $a_1y$ or is $a_1$ a function of $y$?
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 16:33












  • $begingroup$
    it is a1y @RossMillikan. So there should be no brackets around y
    $endgroup$
    – p s
    Dec 1 '18 at 16:38








  • 1




    $begingroup$
    Break it into two integrals. For the first, try a trig substitution. For the second, do a change of variables.
    $endgroup$
    – Cameron Williams
    Dec 1 '18 at 16:40










  • $begingroup$
    @CameronWilliams. For the second integral, is that the same as integration by substitution.
    $endgroup$
    – p s
    Dec 1 '18 at 16:48
















2












$begingroup$


$$ int_0^1 sqrt{1+y^2} (a_0 + a_1y) dy
$$



I have tried to solve this equation by using integration by parts. I am using the square root as the part which I am differentiating. By using this, the method seems like it is ongoing and does not have a final answer. Any guidance will be much appreciated



I am still trying to figure out the formatting so please forgive me






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    MathJax hint: To group things under a square root, enclose them in braces, so sqrt{1+2} gives $sqrt {1+2}$. It works everywhere, like superscripts and subscripts. Is the last term just $a_1y$ or is $a_1$ a function of $y$?
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 16:33












  • $begingroup$
    it is a1y @RossMillikan. So there should be no brackets around y
    $endgroup$
    – p s
    Dec 1 '18 at 16:38








  • 1




    $begingroup$
    Break it into two integrals. For the first, try a trig substitution. For the second, do a change of variables.
    $endgroup$
    – Cameron Williams
    Dec 1 '18 at 16:40










  • $begingroup$
    @CameronWilliams. For the second integral, is that the same as integration by substitution.
    $endgroup$
    – p s
    Dec 1 '18 at 16:48














2












2








2





$begingroup$


$$ int_0^1 sqrt{1+y^2} (a_0 + a_1y) dy
$$



I have tried to solve this equation by using integration by parts. I am using the square root as the part which I am differentiating. By using this, the method seems like it is ongoing and does not have a final answer. Any guidance will be much appreciated



I am still trying to figure out the formatting so please forgive me






calculus integration







share|cite|improve this question











$endgroup$




$$ int_0^1 sqrt{1+y^2} (a_0 + a_1y) dy
$$



I have tried to solve this equation by using integration by parts. I am using the square root as the part which I am differentiating. By using this, the method seems like it is ongoing and does not have a final answer. Any guidance will be much appreciated



I am still trying to figure out the formatting so please forgive me






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 16:38







p s

















asked Dec 1 '18 at 16:27









p sp s

357




357












  • $begingroup$
    MathJax hint: To group things under a square root, enclose them in braces, so sqrt{1+2} gives $sqrt {1+2}$. It works everywhere, like superscripts and subscripts. Is the last term just $a_1y$ or is $a_1$ a function of $y$?
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 16:33












  • $begingroup$
    it is a1y @RossMillikan. So there should be no brackets around y
    $endgroup$
    – p s
    Dec 1 '18 at 16:38








  • 1




    $begingroup$
    Break it into two integrals. For the first, try a trig substitution. For the second, do a change of variables.
    $endgroup$
    – Cameron Williams
    Dec 1 '18 at 16:40










  • $begingroup$
    @CameronWilliams. For the second integral, is that the same as integration by substitution.
    $endgroup$
    – p s
    Dec 1 '18 at 16:48


















  • $begingroup$
    MathJax hint: To group things under a square root, enclose them in braces, so sqrt{1+2} gives $sqrt {1+2}$. It works everywhere, like superscripts and subscripts. Is the last term just $a_1y$ or is $a_1$ a function of $y$?
    $endgroup$
    – Ross Millikan
    Dec 1 '18 at 16:33












  • $begingroup$
    it is a1y @RossMillikan. So there should be no brackets around y
    $endgroup$
    – p s
    Dec 1 '18 at 16:38








  • 1




    $begingroup$
    Break it into two integrals. For the first, try a trig substitution. For the second, do a change of variables.
    $endgroup$
    – Cameron Williams
    Dec 1 '18 at 16:40










  • $begingroup$
    @CameronWilliams. For the second integral, is that the same as integration by substitution.
    $endgroup$
    – p s
    Dec 1 '18 at 16:48
















$begingroup$
MathJax hint: To group things under a square root, enclose them in braces, so sqrt{1+2} gives $sqrt {1+2}$. It works everywhere, like superscripts and subscripts. Is the last term just $a_1y$ or is $a_1$ a function of $y$?
$endgroup$
– Ross Millikan
Dec 1 '18 at 16:33






$begingroup$
MathJax hint: To group things under a square root, enclose them in braces, so sqrt{1+2} gives $sqrt {1+2}$. It works everywhere, like superscripts and subscripts. Is the last term just $a_1y$ or is $a_1$ a function of $y$?
$endgroup$
– Ross Millikan
Dec 1 '18 at 16:33














$begingroup$
it is a1y @RossMillikan. So there should be no brackets around y
$endgroup$
– p s
Dec 1 '18 at 16:38






$begingroup$
it is a1y @RossMillikan. So there should be no brackets around y
$endgroup$
– p s
Dec 1 '18 at 16:38






1




1




$begingroup$
Break it into two integrals. For the first, try a trig substitution. For the second, do a change of variables.
$endgroup$
– Cameron Williams
Dec 1 '18 at 16:40




$begingroup$
Break it into two integrals. For the first, try a trig substitution. For the second, do a change of variables.
$endgroup$
– Cameron Williams
Dec 1 '18 at 16:40












$begingroup$
@CameronWilliams. For the second integral, is that the same as integration by substitution.
$endgroup$
– p s
Dec 1 '18 at 16:48




$begingroup$
@CameronWilliams. For the second integral, is that the same as integration by substitution.
$endgroup$
– p s
Dec 1 '18 at 16:48










1 Answer
1






active

oldest

votes


















0












$begingroup$

$$I=int_0^1(a_0+a_1x)sqrt{1+x^2} mathrm dx$$
$$I=a_0int_0^1sqrt{1+x^2} mathrm dx+a_1int_0^1xsqrt{1+x^2} mathrm dx$$
$$I=a_0I_0+a_1I_1$$




$$I_0=int_0^1sqrt{1+x^2} mathrm dx$$
For this we will use a trigonometric substitution:
$$x=tan uRightarrow mathrm dx=sec^2u mathrm du$$
The change of bounds:
$$int_0^1mapsto int_0^{pi/4}$$
Hence
$$I_0=int_0^{pi/4}sqrt{1+tan^2u}sec^2u mathrm du$$
$$I_0=int_0^{pi/4}sqrt{sec^2u}sec^2u mathrm du$$
$$I_0=int_0^{pi/4}sec usec^2u mathrm du=int_0^{pi/4}sec^3u mathrm du$$
We then integrate by parts:
$$mathrm dV=sec^2u mathrm duRightarrow V=tan u$$
$$U=sec uRightarrow mathrm dU=sec utan u mathrm du$$
Which gives
$$I_0=sec utan ubig|_0^{pi/4}-int_0^{pi/4}tan usec utan u mathrm du$$
$$I_0=sqrt{2}-int_0^{pi/4}sec utan^2u mathrm du$$
$$I_0=sqrt{2}-int_0^{pi/4}sec u(sec^2u-1) mathrm du$$
$$I_0=sqrt{2}-int_0^{pi/4}sec^3u mathrm du+int_0^{pi/4}sec u mathrm du$$
$$I_0=sqrt{2}-I_0+int_0^{pi/4}sec u mathrm du$$
$$2I_0=sqrt{2}+int_0^{pi/4}sec u mathrm du$$
$$2I_0=sqrt{2}+log(sec u+tan u)big|_0^{pi/4}$$
$$2I_0=sqrt{2}+log(1+sqrt2)$$
$$I_0=frac{sqrt{2}+log(1+sqrt2)}2$$




$$I_1=int_0^1xsqrt{1+x^2} mathrm dx$$
Substitution:
$$w^2=x^2+1Rightarrow wmathrm dw=xmathrm dx$$
$$int_0^1mapsto int_1^{sqrt2}$$
This gives
$$I_1=int_1^{sqrt2}sqrt{w^2}w mathrm dw$$
$$I_1=int_1^{sqrt2}w^2 mathrm dw$$
$$I_1=frac{w^3}3bigg|_1^{sqrt2}$$
$$I_1=frac{2^{3/2}-1}3$$




Finally:
$$I=frac{a_0}2bigg(sqrt2+log(1+sqrt2)bigg)+frac{a_1}3big(2^{3/2}-1big)$$






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    1 Answer
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    1 Answer
    1






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    active

    oldest

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    0












    $begingroup$

    $$I=int_0^1(a_0+a_1x)sqrt{1+x^2} mathrm dx$$
    $$I=a_0int_0^1sqrt{1+x^2} mathrm dx+a_1int_0^1xsqrt{1+x^2} mathrm dx$$
    $$I=a_0I_0+a_1I_1$$




    $$I_0=int_0^1sqrt{1+x^2} mathrm dx$$
    For this we will use a trigonometric substitution:
    $$x=tan uRightarrow mathrm dx=sec^2u mathrm du$$
    The change of bounds:
    $$int_0^1mapsto int_0^{pi/4}$$
    Hence
    $$I_0=int_0^{pi/4}sqrt{1+tan^2u}sec^2u mathrm du$$
    $$I_0=int_0^{pi/4}sqrt{sec^2u}sec^2u mathrm du$$
    $$I_0=int_0^{pi/4}sec usec^2u mathrm du=int_0^{pi/4}sec^3u mathrm du$$
    We then integrate by parts:
    $$mathrm dV=sec^2u mathrm duRightarrow V=tan u$$
    $$U=sec uRightarrow mathrm dU=sec utan u mathrm du$$
    Which gives
    $$I_0=sec utan ubig|_0^{pi/4}-int_0^{pi/4}tan usec utan u mathrm du$$
    $$I_0=sqrt{2}-int_0^{pi/4}sec utan^2u mathrm du$$
    $$I_0=sqrt{2}-int_0^{pi/4}sec u(sec^2u-1) mathrm du$$
    $$I_0=sqrt{2}-int_0^{pi/4}sec^3u mathrm du+int_0^{pi/4}sec u mathrm du$$
    $$I_0=sqrt{2}-I_0+int_0^{pi/4}sec u mathrm du$$
    $$2I_0=sqrt{2}+int_0^{pi/4}sec u mathrm du$$
    $$2I_0=sqrt{2}+log(sec u+tan u)big|_0^{pi/4}$$
    $$2I_0=sqrt{2}+log(1+sqrt2)$$
    $$I_0=frac{sqrt{2}+log(1+sqrt2)}2$$




    $$I_1=int_0^1xsqrt{1+x^2} mathrm dx$$
    Substitution:
    $$w^2=x^2+1Rightarrow wmathrm dw=xmathrm dx$$
    $$int_0^1mapsto int_1^{sqrt2}$$
    This gives
    $$I_1=int_1^{sqrt2}sqrt{w^2}w mathrm dw$$
    $$I_1=int_1^{sqrt2}w^2 mathrm dw$$
    $$I_1=frac{w^3}3bigg|_1^{sqrt2}$$
    $$I_1=frac{2^{3/2}-1}3$$




    Finally:
    $$I=frac{a_0}2bigg(sqrt2+log(1+sqrt2)bigg)+frac{a_1}3big(2^{3/2}-1big)$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $$I=int_0^1(a_0+a_1x)sqrt{1+x^2} mathrm dx$$
      $$I=a_0int_0^1sqrt{1+x^2} mathrm dx+a_1int_0^1xsqrt{1+x^2} mathrm dx$$
      $$I=a_0I_0+a_1I_1$$




      $$I_0=int_0^1sqrt{1+x^2} mathrm dx$$
      For this we will use a trigonometric substitution:
      $$x=tan uRightarrow mathrm dx=sec^2u mathrm du$$
      The change of bounds:
      $$int_0^1mapsto int_0^{pi/4}$$
      Hence
      $$I_0=int_0^{pi/4}sqrt{1+tan^2u}sec^2u mathrm du$$
      $$I_0=int_0^{pi/4}sqrt{sec^2u}sec^2u mathrm du$$
      $$I_0=int_0^{pi/4}sec usec^2u mathrm du=int_0^{pi/4}sec^3u mathrm du$$
      We then integrate by parts:
      $$mathrm dV=sec^2u mathrm duRightarrow V=tan u$$
      $$U=sec uRightarrow mathrm dU=sec utan u mathrm du$$
      Which gives
      $$I_0=sec utan ubig|_0^{pi/4}-int_0^{pi/4}tan usec utan u mathrm du$$
      $$I_0=sqrt{2}-int_0^{pi/4}sec utan^2u mathrm du$$
      $$I_0=sqrt{2}-int_0^{pi/4}sec u(sec^2u-1) mathrm du$$
      $$I_0=sqrt{2}-int_0^{pi/4}sec^3u mathrm du+int_0^{pi/4}sec u mathrm du$$
      $$I_0=sqrt{2}-I_0+int_0^{pi/4}sec u mathrm du$$
      $$2I_0=sqrt{2}+int_0^{pi/4}sec u mathrm du$$
      $$2I_0=sqrt{2}+log(sec u+tan u)big|_0^{pi/4}$$
      $$2I_0=sqrt{2}+log(1+sqrt2)$$
      $$I_0=frac{sqrt{2}+log(1+sqrt2)}2$$




      $$I_1=int_0^1xsqrt{1+x^2} mathrm dx$$
      Substitution:
      $$w^2=x^2+1Rightarrow wmathrm dw=xmathrm dx$$
      $$int_0^1mapsto int_1^{sqrt2}$$
      This gives
      $$I_1=int_1^{sqrt2}sqrt{w^2}w mathrm dw$$
      $$I_1=int_1^{sqrt2}w^2 mathrm dw$$
      $$I_1=frac{w^3}3bigg|_1^{sqrt2}$$
      $$I_1=frac{2^{3/2}-1}3$$




      Finally:
      $$I=frac{a_0}2bigg(sqrt2+log(1+sqrt2)bigg)+frac{a_1}3big(2^{3/2}-1big)$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$I=int_0^1(a_0+a_1x)sqrt{1+x^2} mathrm dx$$
        $$I=a_0int_0^1sqrt{1+x^2} mathrm dx+a_1int_0^1xsqrt{1+x^2} mathrm dx$$
        $$I=a_0I_0+a_1I_1$$




        $$I_0=int_0^1sqrt{1+x^2} mathrm dx$$
        For this we will use a trigonometric substitution:
        $$x=tan uRightarrow mathrm dx=sec^2u mathrm du$$
        The change of bounds:
        $$int_0^1mapsto int_0^{pi/4}$$
        Hence
        $$I_0=int_0^{pi/4}sqrt{1+tan^2u}sec^2u mathrm du$$
        $$I_0=int_0^{pi/4}sqrt{sec^2u}sec^2u mathrm du$$
        $$I_0=int_0^{pi/4}sec usec^2u mathrm du=int_0^{pi/4}sec^3u mathrm du$$
        We then integrate by parts:
        $$mathrm dV=sec^2u mathrm duRightarrow V=tan u$$
        $$U=sec uRightarrow mathrm dU=sec utan u mathrm du$$
        Which gives
        $$I_0=sec utan ubig|_0^{pi/4}-int_0^{pi/4}tan usec utan u mathrm du$$
        $$I_0=sqrt{2}-int_0^{pi/4}sec utan^2u mathrm du$$
        $$I_0=sqrt{2}-int_0^{pi/4}sec u(sec^2u-1) mathrm du$$
        $$I_0=sqrt{2}-int_0^{pi/4}sec^3u mathrm du+int_0^{pi/4}sec u mathrm du$$
        $$I_0=sqrt{2}-I_0+int_0^{pi/4}sec u mathrm du$$
        $$2I_0=sqrt{2}+int_0^{pi/4}sec u mathrm du$$
        $$2I_0=sqrt{2}+log(sec u+tan u)big|_0^{pi/4}$$
        $$2I_0=sqrt{2}+log(1+sqrt2)$$
        $$I_0=frac{sqrt{2}+log(1+sqrt2)}2$$




        $$I_1=int_0^1xsqrt{1+x^2} mathrm dx$$
        Substitution:
        $$w^2=x^2+1Rightarrow wmathrm dw=xmathrm dx$$
        $$int_0^1mapsto int_1^{sqrt2}$$
        This gives
        $$I_1=int_1^{sqrt2}sqrt{w^2}w mathrm dw$$
        $$I_1=int_1^{sqrt2}w^2 mathrm dw$$
        $$I_1=frac{w^3}3bigg|_1^{sqrt2}$$
        $$I_1=frac{2^{3/2}-1}3$$




        Finally:
        $$I=frac{a_0}2bigg(sqrt2+log(1+sqrt2)bigg)+frac{a_1}3big(2^{3/2}-1big)$$






        share|cite|improve this answer









        $endgroup$



        $$I=int_0^1(a_0+a_1x)sqrt{1+x^2} mathrm dx$$
        $$I=a_0int_0^1sqrt{1+x^2} mathrm dx+a_1int_0^1xsqrt{1+x^2} mathrm dx$$
        $$I=a_0I_0+a_1I_1$$




        $$I_0=int_0^1sqrt{1+x^2} mathrm dx$$
        For this we will use a trigonometric substitution:
        $$x=tan uRightarrow mathrm dx=sec^2u mathrm du$$
        The change of bounds:
        $$int_0^1mapsto int_0^{pi/4}$$
        Hence
        $$I_0=int_0^{pi/4}sqrt{1+tan^2u}sec^2u mathrm du$$
        $$I_0=int_0^{pi/4}sqrt{sec^2u}sec^2u mathrm du$$
        $$I_0=int_0^{pi/4}sec usec^2u mathrm du=int_0^{pi/4}sec^3u mathrm du$$
        We then integrate by parts:
        $$mathrm dV=sec^2u mathrm duRightarrow V=tan u$$
        $$U=sec uRightarrow mathrm dU=sec utan u mathrm du$$
        Which gives
        $$I_0=sec utan ubig|_0^{pi/4}-int_0^{pi/4}tan usec utan u mathrm du$$
        $$I_0=sqrt{2}-int_0^{pi/4}sec utan^2u mathrm du$$
        $$I_0=sqrt{2}-int_0^{pi/4}sec u(sec^2u-1) mathrm du$$
        $$I_0=sqrt{2}-int_0^{pi/4}sec^3u mathrm du+int_0^{pi/4}sec u mathrm du$$
        $$I_0=sqrt{2}-I_0+int_0^{pi/4}sec u mathrm du$$
        $$2I_0=sqrt{2}+int_0^{pi/4}sec u mathrm du$$
        $$2I_0=sqrt{2}+log(sec u+tan u)big|_0^{pi/4}$$
        $$2I_0=sqrt{2}+log(1+sqrt2)$$
        $$I_0=frac{sqrt{2}+log(1+sqrt2)}2$$




        $$I_1=int_0^1xsqrt{1+x^2} mathrm dx$$
        Substitution:
        $$w^2=x^2+1Rightarrow wmathrm dw=xmathrm dx$$
        $$int_0^1mapsto int_1^{sqrt2}$$
        This gives
        $$I_1=int_1^{sqrt2}sqrt{w^2}w mathrm dw$$
        $$I_1=int_1^{sqrt2}w^2 mathrm dw$$
        $$I_1=frac{w^3}3bigg|_1^{sqrt2}$$
        $$I_1=frac{2^{3/2}-1}3$$




        Finally:
        $$I=frac{a_0}2bigg(sqrt2+log(1+sqrt2)bigg)+frac{a_1}3big(2^{3/2}-1big)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 0:11









        clathratusclathratus

        3,499331




        3,499331






























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            Aardman Animations

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            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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            Are they similar matrix

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            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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