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    u=2x+1 works fine.

    What do you want v for?

    Post some working.
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    (Original post by sohail.s)
    How would i integrate the following function :

    \int^1_0 \sqrt {2x+1}\ dx

    I'v never seen a problem like this before and my only guess was to use the substitution  u=2x+1 and v=u^{1/2} but that isn't getting me anywhere
    u=2x+1 is enough. Are you familiar with integration by substitution? You need to now differentiate the substitution and find the equivalent of dx in terms of du. Don't forget to change your limits.
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    (Original post by PerigeeApogee)
    That substitution should work.


     u = 2x+1 \to du = 2\,dx

    Limits:

     x = 1 \to u = 3
     x = 0 \to u = 1

    Hence:

     \displaystyle I = \int_1^3 u^{\frac{1}{2}} .\, 2\,du = 2\int_1^3 u^{\frac{1}{2}} \, du
    You mean

     \displaystyle I = \frac{1}{2}\int_1^3 u^{\frac{1}{2}} du
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    (Original post by PerigeeApogee)
    That substitution should work.


     u = 2x+1 \to du = 2\,dx

    Limits:

     x = 1 \to u = 3
     x = 0 \to u = 1

    Hence:

     \displaystyle I = \int_1^3 u^{\frac{1}{2}} .\, 2\,du = 2\int_1^3 u^{\frac{1}{2}} \, du
    Woah slow down, that's not correct.

    du/dx = 2 therefore du = 2dx which is correct. However we are replacing dx with du so we need dx in terms of du, dx = du/2

    So we have the integral of \displaystyle\frac{u^{1/2}}{2} du, which is very different entirely.
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    You could aslo use the chain rule
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    (Original post by sohail.s)
    is .2 before du supposed to mean multiply by 2 or a half?
    It's multiplying by a half. I recommend that you go and re-read the chapter on integration by substitution. It makes a lot more sense when you actually know what the method is.
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    (Original post by soup)
    You could aslo use the chain rule
    This is integration and NOT differentiation.
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    For some reason, questions like this came up in my C3 before I'd done integration by substitution (which is in C4 on OCR). They can be solved by inspection without the need for a substitution (as such): if you know that (ax+b)^n differentiates to an(ax+b)^{n-1} (by the chain rule), then going backwards reveals that \displaystyle \int (ax+b)^n\, dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + C. Here, you have (2x+1)^{\frac{1}{2}}, so you can integrate it using that result. Note that you can't apply this result to any other type of function; for example you wouldn't be able to integrate \sqrt{x^2+1} using this result.
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    (Original post by Farhan.Hanif93)
    This is integration and NOT differentiation.
    Yeah I meant the "chain rule" for integration. See nuodai's post
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    (Original post by soup)
    Yeah I meant the "chain rule" for integration. See nuodai's post
    nuodai's post is the reverse chain rule (I know everyone hates that phrase), which is not the same thing as the chain rule itself. I think the preferred name for it is integration by inspection/recognition. Saying it's the chain rule is a bit misleading.
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    (Original post by Farhan.Hanif93)
    nuodai's post is the reverse chain rule (I know everyone hates that phrase), which is not the same thing as the chain rule itself. I think the preferred name for it is integration by inspection/recognition. Saying it's the chain rule is a bit misleading.
    I didn't know what to call it. I just refer to it as the chain rule for integration
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    (Original post by Farhan.Hanif93)
    nuodai's post is the reverse chain rule (I know everyone hates that phrase), which is not the same thing as the chain rule itself. I think the preferred name for it is integration by inspection/recognition. Saying it's the chain rule is a bit misleading.
    The reverse chain rule is just integration by substitution:
    \displaystyle \frac{d f(g(x))}{dx} = f'(g(x))g'(x) \implies \int f'(g(x))g'(x)\, dx = f(g(x))+C
    (...at least this result holds for A-level stuff)

    Technically, I did use the chain rule, because my result was derived by differentiating something using the chain rule, but I did it in a way that kind of required me to already know the result before I used it. As such, saying "use the chain rule" isn't particularly helpful unless you already know this method... so you're kind of both right.
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    (Original post by nuodai)
    For some reason, questions like this came up in my C3 before I'd done integration by substitution (which is in C4 on OCR). They can be solved by inspection without the need for a substitution (as such): if you know that (ax+b)^n differentiates to an(ax+b)^{n-1} (by the chain rule), then going backwards reveals that \displaystyle \int (ax+b)^n\, dx = \dfrac{(ax+b)^{n+1}}{a(n+1)} + C. Here, you have (2x+1)^{\frac{1}{2}}, so you can integrate it using that result. Note that you can't apply this result to any other type of function; for example you wouldn't be able to integrate \sqrt{x^2+1} using this result.

    how would you calculate the last equation you wrote then?
 
 
 
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