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Math Help - Change in integral begin/end values

  1. #1
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    Change in integral begin/end values

    I am just starting to learn about integrals. We are using dummy variables to solve integrals from say 1 to 3. When we finish the range of the integral changes and I am at a loss as to how to determine the new values.

    For example if I have Integral from 0 to 1 of x (2 + 3x^2) dx

    it becomes integral from 2 to 5 of u^1/2 du times 1/6

    PLease explain how the 0 to 1 becomes 2 to 5 and since u = 2 + 3x^2 what happened to the x in the integral that precedes the (2 + 3x^2) ^1/2?????

    Either my book is missing some basic explanations or I am just slowwwwww!

    Also how do I write an integral sign on this site????
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  2. #2
    Moo
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    Hello,

    [math ]\int_0^1[/tex]

    When doing a substitution you also have to change the values of the integral (0 and 1).

    I'll make the steps...

    \int_0^1 x(2+3x^2) dx

    u=2+3x^2

    So if x=0, u=2+3*0^2=2
    If x=1, u=2+3*1^2=5

    This explains the changes 0 and 1 to 2 and 5.

    Plus, you have to "change" dx, and express it in respect to u.

    \frac{du}{dx}=6x

    Hence dx=\frac{du}{6x}

    The integral becomes :

    \int_2^5 {\color{red} x} u \frac{du}{6 {\color{red} x}} = \int_2^5 \frac{u}{6} du
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  3. #3
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    Quote Originally Posted by Frostking View Post
    I am just starting to learn about integrals. We are using dummy variables to solve integrals from say 1 to 3. When we finish the range of the integral changes and I am at a loss as to how to determine the new values.

    For example if I have Integral from 0 to 1 of x (2 + 3x^2) dx

    it becomes integral from 2 to 5 of u^1/2 du times 1/6

    PLease explain how the 0 to 1 becomes 2 to 5 and since u = 2 + 3x^2 what happened to the x in the integral that precedes the (2 + 3x^2) ^1/2?????

    Either my book is missing some basic explanations or I am just slowwwwww!
    If u = 2 + 3x^2, then what was the beginning limit of x=0 becomes u=2+3(0)^2 \rightarrow 2, and the upper limit of x=1 becomes u=2+3(1)^2 \rightarrow 5.

    Personally, I never change the limits as I always re-sub for x back into the integrated equation. This takes away from the annoying process of re-evaluating the limits of integration.

    Quote Originally Posted by Frostking View Post
    Also how do I write an integral sign on this site????
    Type the [math ] [/tex] tags around \int like so: \int_0^1 \sqrt{2+3x^2}dx
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  4. #4
    Moo
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    Oh yeah, I think you forgot ^1/2 in

    For example if I have Integral from 0 to 1 of x (2 + 3x^2) dx
    But it doesn't change the explication ^^
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