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Math Help - Something simple about integral theorey but I don't get it.

  1. #1
    s3a
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    Something simple about integral theorey but I don't get it.

    Question:
    "If f is continuous and if this (integral (f(x)dx) from 0 to 4 - Wolfram|Alpha) integral = 10. Then find, this(integral (f(2x)dx) from 0 to 2 - Wolfram|Alpha) integral.

    I have no idea how to do this. Someone please help me!

    Answer:
    5

    Any input would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Hi

    We know that \int_{0}^{4} f(x) dx = 10 and we want to calculate \int_{0}^{2} f(2x) dx

    \int_{0}^{2} f(2x) dx
    Let u = 2x
    Then x going from 0 to 2, u is going from 2x0=0 to 2x2=4
    and du = 2dx or dx = du/2
    Therefore
    \int_{0}^{2} f(2x) dx = \int_{0}^{4} f(u) \frac{du}{2} = \frac12 \int_{0}^{4} f(u) du = \frac12 \times 10 = 5
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  3. #3
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    Let u = 2x and du = 2dx, then
     \int_0^2 f(2x) dx = \frac{1}{2}\int_0^4 f(u) du = 5.
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  4. #4
    s3a
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    These reponses have helped get further along the problem but I am still stuck on the last step. I do not understand how f(u)du = f(x)dx (since the integral is being evulated over the same interval, etc and that is the only thing that differs).
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  5. #5
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    Quote Originally Posted by s3a View Post
    These reponses have helped get further along the problem but I am still stuck on the last step. I do not understand how f(u)du = f(x)dx (since the integral is being evulated over the same interval, etc and that is the only thing that differs).
    If you are asking why
    \int_a^b f(x)dx= \int_a^b f(u)du, that's true because the variable inside the integral is a "dummy variable". \int_a^b f(x)dx is number, not a function of x. Similarly \int_a^b f(u)du is a number not a function of u. Since it doesn't matter what we call the variable, the two numbers must be equal.

    The same thing happens with sums:
    \sum_{i=1}^3 i^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14
    \sum_{n=1}^3 n^2= 1^2+ 2^2+ 3^2= 1+ 4+ 9= 14

    so \sum_{i=1}^3 i^2= \sum_{n=1}^3 n^2.
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