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Math Help - Confusing integrals

  1. #1
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    Confusing integrals

    These integrals don't make any sense. I know the basics but these are out of my legaue. I don't even know how to write the integral for that it has a top number and a bottom number to it. So ill give it a shot :

    1) ∫^1 and the bottom is 0 then (4x^3 - 2x)dx

    2) ∫^pie(3.14) and the bottom is pie/2 then (sin2x dx)

    3) ∫^2 and the bottom is 0 then [3dx/ (4x +1)^1/2]

    I know its confusing to read but i really need help on this to be ready for my quiz. So thanks for the help.
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  2. #2
    Super Member angel.white's Avatar
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    I will do the first one as an example:
    Quote Originally Posted by uniquereason81 View Post
    1) ∫^1 and the bottom is 0 then (4x^3 - 2x)dx
    Initial equation:
    \int_{0}^{1} (4x^3 - 2x) ~dx

    Integrate:
    =x^4 - x^2 |_0^1

    The line on the right tells you to evaluate it from 0 to 1.
    This means that if f(x) is our function: f(x) = x^4 -x^2
    Then we are looking for f(1)-f(0)

    So:

    =[(1)^4 -(1)^2] - [(0)^4 - (0)^2]

    =[1 -1] - [0 - 0]

    =0
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  3. #3
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    Thanks but my teacher was saying we could sub x1 and x2 as u1 and u2 by finding it in the equation. Or something like that, is that correct
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    For the third one

    Quote Originally Posted by uniquereason81 View Post
    These integrals don't make any sense. I know the basics but these are out of my legaue. I don't even know how to write the integral for that it has a top number and a bottom number to it. So ill give it a shot :

    1) ∫^1 and the bottom is 0 then (4x^3 - 2x)dx

    2) ∫^pie(3.14) and the bottom is pie/2 then (sin2x dx)

    3) ∫^2 and the bottom is 0 then [3dx/ (4x +1)^1/2]

    I know its confusing to read but i really need help on this to be ready for my quiz. So thanks for the help.
    you have \int_0^{2}\frac{3}{\sqrt{4x+1}}dx\Rightarrow{\frac  {3}{4}\int_0^{2}\frac{4}{\sqrt{4x+1}}}dx...now you should see that there is a quantity and a derivative of that quantity
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  5. #5
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    Though how would you start number 3. Thats the most confusing one for me to do and also start
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Uhm

    Quote Originally Posted by uniquereason81 View Post
    Though how would you start number 3. Thats the most confusing one for me to do and also start
    Didnt I just give you a hint? I am sorry did I do the wrong one?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by uniquereason81 View Post
    Though how would you start number 3. Thats the most confusing one for me to do and also start
    Well you are not responding so I will just show you we got this far \frac{3}{4}\int_0^{2}\frac{4}{\sqrt{4x+1}}dx...now that we have an integral of the form \int{f(g(x))\cdot{g'(x)}dx} (since the derivative of [tex]4x+1{/math] is 4) then we just integrate it as though it was just f(x)...so we would get \int_0^{2}\frac{3}{\sqrt{4x+1}}dx=\bigg[\frac{3}{4}\cdot{2}\sqrt{4x+1}\bigg]\bigg|_0^{2}=3
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  8. #8
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    ooo my bad i didn't see your post.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Haha

    Quote Originally Posted by uniquereason81 View Post
    ooo my bad i didn't see your post.
    Don't worry about it
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  10. #10
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    How would you start number 2
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by uniquereason81 View Post
    How would you start number 2
    Same concept you have \int_{\frac{\pi}{2}}^{\pi}sin(2x)dx...to integrate that we need the derivative of the inside of the sine function to be on the outside...the derivative is 2...but to put a two we need to counter with a half...so we would have \frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}2sin(2x)dx\Ri  ghtarrow{\frac{-cos(2x)}{2}\bigg|_\frac{\pi}{2}^{\pi}}=-1
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  12. #12
    Super Member angel.white's Avatar
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    Quote Originally Posted by uniquereason81 View Post
    How would you start number 2
    Follow my example.

    Integrate the function exactly as you normally would for an indefinite integral (one with no values at the bottom and top). After it is integrated, evaluate it like this:

    integral from a to b of f'(x) = f(x) from a to b = f(b) - f(a)

    You can copy my example almost exactly.
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