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Math Help - Definite Integral. Equivalency.

  1. #1
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    Definite Integral. Equivalency.

    Definite Integral. Equivalency.-screenshot-1-.pngDefinite Integral. Equivalency.-screenshot-2-.pngDefinite Integral. Equivalency.-3e3q.pngDefinite Integral. Equivalency.-3e3a.png

    My question is about a). I see that:

    \int^e_1\frac{\sqrt(ln(x))}{x}\;dx=\int^1_0\sqrt(u  )\;du

    those are equivalent because I graphed both of them. In the screenshots the area is listed in the lower left hand corner. Algebraically I am having a tougher time figuring out how those are equal. What keeps getting to me is how was it determined that the interval changes from 1 to e (to) 0 to 1.

    Thanks in advance for any responses...
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  2. #2
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    Re: Definite Integral. Equivalency.

    Quote Originally Posted by sepoto View Post
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    My question is about a). I see that:

    \int^e_1\frac{\sqrt(ln(x))}{x}\;dx=\int^1_0\sqrt(u  )\;du

    those are equivalent because I graphed both of them. In the screenshots the area is listed in the lower left hand corner. Algebraically I am having a tougher time figuring out how those are equal. What keeps getting to me is how was it determined that the interval changes from 1 to e (to) 0 to 1.

    Thanks in advance for any responses...
    \int_1^e\dfrac{\sqrt{\ln(x)}}{x}dx

    u=\ln(x)\;\;du=\dfrac{dx}{x} Do you understand this?

    now find new integration limits given the substitution

    \ln(1)=0\;\;\ln(e)=1

    so the new integral becomes

    \int_0^1\sqrt{u}\;du
    Thanks from sepoto
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  3. #3
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    Re: Definite Integral. Equivalency.

    So it appears then that the new integral holds the corresponding u values of x. I think I can follow that. Thank you for posting.
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