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Math Help - New to Integrating, I need a lot of help

  1. #1
    Junior Member R3ap3r's Avatar
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    New to Integrating, I need a lot of help

    I need a lot of help understanding this as I am completely lost at what to do
    Evaluate to 4 significant digits: \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by R3ap3r View Post
    I need a lot of help understanding this as I am completely lost at what to do
    Evaluate to 4 significant digits: \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx
    the derivative of the inside ofthe quanityt is \frac{3}{5} of whats on the outside
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    Quote Originally Posted by R3ap3r View Post
    I need a lot of help understanding this as I am completely lost at what to do
    Evaluate to 4 significant digits: \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx
    You can actually find an exact value for this integral. Use the fact that the derivative of \sqrt{x^3 + 3} is \frac{3x^2}{2 \sqrt{x^3 + 3}}.
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    Junior Member R3ap3r's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    the derivative of the inside ofthe quanityt is \frac{3}{5} of whats on the outside
    I'm going to need a little bit more help than that. Step by step process if possible as I have a lot of these to do and one step by step process will teach me a lot.
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    Quote Originally Posted by R3ap3r View Post
    I need a lot of help understanding this as I am completely lost at what to do
    Evaluate to 4 significant digits: \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx
    Ok, first step is to do a little rearranging of constants inside and outside the integral. Since you can pull out a constant, begin with

    \frac{10}{3}\int_{-1} ^5\;\frac{3x^2}{2\sqrt{x^3+3}}\;dx.

    Then you have inside the integral the derivative of \sqrt{x^3 + 3}, so you can evaluate to \frac{10}{3}(\sqrt{5^3 + 3}-\sqrt{(-1)^3 + 3}).
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    Junior Member R3ap3r's Avatar
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    Quote Originally Posted by icemanfan View Post
    Ok, first step is to do a little rearranging of constants inside and outside the integral. Since you can pull out a constant, begin with

    \frac{10}{3}\int_{-1} ^5\;\frac{3x^2}{2\sqrt{x^3+3}}\;dx.

    Then you have inside the integral the derivative of \sqrt{x^3 + 3}, so you can evaluate to \frac{10}{3}(\sqrt{5^3 + 3}-\sqrt{(-1)^3 + 3}).
    alright good start, but again i'm new to all this so how exactly did you rearrange it like that? I know you said pull out a constant but where did you get that from?
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    Quote Originally Posted by R3ap3r View Post
    alright good start, but again i'm new to all this so how exactly did you rearrange it like that? I know you said pull out a constant but where did you get that from?
    Originally, you had the constant of 5 in the numerator. I rearranged as 5 = \frac{10}{3}*\frac{3}{2} because I wanted to get 3x^2 in the numerator, as 3x^2 is the derivative of x^3 + 3.
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  8. #8
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    Quote Originally Posted by R3ap3r View Post
    I need a lot of help understanding this as I am completely lost at what to do
    Evaluate to 4 significant digits: \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx
    It might help to view this as \int_{-1} ^5\;{5x^2}(x^3+3)^{\frac{-1}{2}}\;dx

    Solving integrals a lot of times means being aware of the form \int_{a}^{b}\; u^n du where in this case u = x^3 + 3, n = \frac{-1}{2}, du = 3x^2 dx
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    Junior Member R3ap3r's Avatar
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    Ok I understand that whats the next step after that?
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    Quote Originally Posted by R3ap3r View Post
    Ok I understand that whats the next step after that?
    If you have an integral in the form \int_{a}^{b} u^n du then its evaluation is \frac{1}{n+1}b^{n+1} - \frac{1}{n+1}a^{n+1}.
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  11. #11
    Junior Member R3ap3r's Avatar
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    So what I'm getting from that is \frac{1}{-\frac{1}{2} + 1}\; 5^{-\frac{1}{2} + 1} - \frac{1}{-\frac{1}{2} + 1}\; -1^{1\frac{1}{2} + 1}

    Is this right so far?
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  12. #12
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    Quote Originally Posted by R3ap3r View Post
    So what I'm getting from that is \frac{1}{-\frac{1}{2} + 1}\; 5^{-\frac{1}{2} + 1} - \frac{1}{-\frac{1}{2} + 1}\; -1^{1\frac{1}{2} + 1}

    Is this right so far?
    Don't forget what your u is.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by R3ap3r View Post
    I need a lot of help understanding this as I am completely lost at what to do
    Evaluate to 4 significant digits: \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx
    if you can't see what to multiply by to get everything in the proper form, it is ok. just solve for it.

    Let u = x^3 + 3

    \Rightarrow du = 3x^2~dx

    \Rightarrow \frac 13 ~du = x^2~dx ................(you could also have said \frac 53~du = 5x^2 ~dx)

    So now you just replace x^2~dx with \frac 13~du and you integral now becomes:

    \frac 53 \int \frac 1{\sqrt{u}}~du = \frac 53 \int u^{- 1/2}~du


    Alternatively, you could let u^2 = x^3 + 3 and that would result in the integral \frac {10}3 \int 1 ~du
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  14. #14
    Junior Member R3ap3r's Avatar
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    uh my answer i got is 6.472. Am I even in the ballpark?
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  15. #15
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    I got 32.9983 as an answer. Hope that helps.
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