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Math Help - U - Substitution for integrals

  1. #1
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    U - Substitution for integrals

    lower boundary = 0

    upper boundary = 4

    x / (sqrt 1 + 2x)

    I've let u = 1 + 2x

    du = xdx

    new boundaries = 1 to 9

    u^-1/2 du. If I integrate it, it will 2u^1/2

    2(9)^1/2 - 2(1)^1/2 = 4

    This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem . The answer should be 10/3
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  2. #2
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    Quote Originally Posted by JonathanEyoon View Post
    lower boundary = 0

    upper boundary = 4

    x / (sqrt 1 + 2x)

    I've let u = 1 + 2x

    du = xdx

    new boundaries = 1 to 9

    u^-1/2 du. If I integrate it, it will 2u^1/2

    2(9)^1/2 - 2(1)^1/2 = 4

    This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem . The answer should be 10/3
    u=1+2x \to du=2dx it does not equal xdx
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    lower boundary = 0

    upper boundary = 4

    x / (sqrt 1 + 2x)

    I've let u = 1 + 2x

    du = xdx

    new boundaries = 1 to 9

    u^-1/2 du. If I integrate it, it will 2u^1/2

    2(9)^1/2 - 2(1)^1/2 = 4

    This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem . The answer should be 10/3
    \int_0^{4}\frac{\varphi}{1+2\varphi}d\varphi

    Let \lambda=\sqrt{1+2\varphi}\Rightarrow\varphi=\frac{  \lambda^2-1}{2}

    So d\varphi=\lambda{d\lambda}

    And seeing that at \lambda(0)=\sqrt{1+2(0)}=1

    and \lambda(4)=\sqrt{1+2(4)}=3

    We have

    \frac{1}{2}\int_1^{3}\frac{(\lambda^2-1)\lambda}{\lambda}d\lambda=\frac{1}{2}\bigg[\frac{\lambda^3}{3}-\lambda\bigg]\bigg|_{1}^{3}=\frac{10}{3}
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    u=1+2x \to du=2dx it does not equal xdx

    forgive me, i've been studying all day and it's late now. I should have definitely caught that. I'll rework the problem and try again.
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  5. #5
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    This is a fun different sub try

    t^2=1+2x \implies 2tdt=2dx \iff tdt=dx

    Solving the first for x we get t^2=1+2x \iff x=\frac{t^2-1}{2}

    \int_{0}^{4}\frac{x}{\sqrt{1+2x}}dx=\int_{1}^{3}\f  rac{\frac{t^2-1}{2}}{\sqrt{t^2}}tdt=\frac{1}{2}\int_{1}^{3}t^2-1dt=\frac{1}{2}\left( \frac{1}{3}t^3-t\right)\bigg|_{1}^{3}=\frac{10}{3}
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \int_0^{4}\frac{\varphi}{1+2\varphi}d\varphi

    Let \lambda=\sqrt{1+2\varphi}\Rightarrow\varphi=\frac{  \lambda^2-1}{2}

    So d\varphi=\lambda{d\lambda}

    And seeing that at \lambda(0)=\sqrt{1+2(0)}=1

    and \lambda(4)=\sqrt{1+2(4)}=3

    We have

    \frac{1}{2}\int_1^{3}\frac{(\lambda^2-1)\lambda}{\lambda}d\lambda=\frac{1}{2}\bigg[\frac{\lambda^3}{3}-\lambda\bigg]\bigg|_{1}^{3}=\frac{10}{3}
    Quote Originally Posted by TheEmptySet View Post
    This is a fun different sub try

    t^2=1+2x \implies 2tdt=2dx \iff tdt=dx

    Solving the first for x we get t^2=1+2x \iff x=\frac{t^2-1}{2}

    \int_{0}^{4}\frac{x}{\sqrt{1+2x}}dx=\int_{1}^{3}\f  rac{\frac{t^2-1}{2}}{\sqrt{t^2}}tdt=\frac{1}{2}\int_{1}^{3}t^2-1dt=\frac{1}{2}\left( \frac{1}{3}t^3-t\right)\bigg|_{1}^{3}=\frac{10}{3}
    I think I have seen this method done before somewhere...hmm
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  7. #7
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    Quote Originally Posted by JonathanEyoon View Post
    forgive me, i've been studying all day and it's late now. I should have definitely caught that. I'll rework the problem and try again.
    I know you were on this morning before I went to work and you are still on now that I'm home.
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  8. #8
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    question :

    Could I just substitute how I did originally?

    u = 1 + 2x

    du / 2 = dx


    As for the last substitution for the x in the numerator, can I just let x = (u - 1) / 2?
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  9. #9
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    Quote Originally Posted by JonathanEyoon View Post
    question :

    Could I just substitute how I did originally?

    u = 1 + 2x

    du / 2 = dx


    As for the last substitution for the x in the numerator, can I just let x = (u - 1) / 2?

    Yes that is what you need to do!

    Nice
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  10. #10
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    lower boundary = 0

    upper boundary = 4

    x / (sqrt 1 + 2x)

    I've let u = 1 + 2x

    du = xdx

    new boundaries = 1 to 9

    u^-1/2 du. If I integrate it, it will 2u^1/2

    2(9)^1/2 - 2(1)^1/2 = 4

    This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem . The answer should be 10/3
    Let \zeta=1+2x\implies x=\frac{1}{2}(\zeta-1)

    Thus, \,d\zeta=2\,dx\implies\,dx=\frac{1}{2}\,d\zeta

    Limits change as well.. \zeta(4)=9, \ \zeta{0}=1

    Thus, our integral is

    \frac{1}{4}\int_1^9 \frac{\zeta-1}{\sqrt{\zeta}}\,d\zeta=\frac{1}{4}\int_1^9 \bigg[\sqrt{\zeta}-(\zeta)^{-\frac{1}{2}}\bigg]

    =\frac{1}{4}\left.\bigg[\frac{2}{3}\zeta^{\frac{3}{2}}-2\sqrt{\zeta}\bigg]\right|_1^9=\frac{1}{4}\bigg(12+\frac{4}{3}\bigg)=  \color{red}\boxed{\frac{10}{3}}
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  11. #11
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    Yippee~ So the new boundaries will still remain the same as what I originally had above right?

    New lower = 1 and New upper = 9?
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Let \zeta=1+2x\implies x=\frac{1}{2}(\zeta-1)

    Thus, \,d\zeta=2\,dx\implies\,dx=\frac{1}{2}\,d\zeta

    Limits change as well.. \zeta(4)=9, \ \zeta{0}=1

    Thus, our integral is

    \frac{1}{4}\int_1^9 \frac{\zeta-1}{\sqrt{\zeta}}\,d\zeta=\frac{1}{4}\int_1^9 \bigg[\sqrt{\zeta}-(\zeta)^{-\frac{1}{2}}\bigg]

    =\frac{1}{4}\left.\bigg[\frac{2}{3}\zeta^{\frac{3}{2}}-2\sqrt{\zeta}\bigg]\right|_1^9=\frac{1}{4}\bigg(12+\frac{4}{3}\bigg)=  \color{red}\boxed{\frac{10}{3}}
    I say we reform! No more using x or y as a variable. Viva la \varphi!
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  13. #13
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    Quote Originally Posted by JonathanEyoon View Post
    Yippee~ So the new boundaries will still remain the same as what I originally had above right?

    New lower = 1 and New upper = 9?
    Yep...
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  14. #14
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I say we reform! No more using x or y as a variable. Viva la \varphi!
    Haha. I'm gonna use this one from now on. Viva la \varrho!
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