CK asked in Science & MathematicsMathematics · 3 weeks ago

Integral help.....?

Please help me !

Thanks in advance !

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3 Answers

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  • 3 weeks ago
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    (Definitely graph all the functions to get a better feel for what's happening)

    We have a variable of integration itself as a function of x, and it's also discontinuous. Might want to take it step by step. (no?)

    Let's first compute the integral only on [0, 1].

    Here, the variable of integration is simple: x - [_x_] = x - 0 = x and so we have just a "normal integral"

    int[0 to 1] e^(2x - [_2x_]) dx

    The integrand is discontinuous at x=0.5 so we split into the sum of integrals from 0 to 0.5, and from 0.5 to 1.

    -------On [0, 0.5]:

    2x - [_2x_] = 2x - 0 = 2x

    int[0 to 0.5] e^(2x)dx

    = (1/2)e^(2x) [0 to 0.5]

    = (1/2)(e - 1)

    -------On [0.5, 1]

    2x - [_2x_] = 2x - 1

    int[0.5 to 1] e^(2x - 1) dx

    = 1/(2e) e^(2x) [0.5 to 1]

    = 1/(2e)(e^2 - e)

    = (1/2)(e - 1)

    The sum is e - 1

    But we already sniffed out x - [_x_] has a period of 1, and 2x - [_2x_] of 0.5, and 0.5 fits a whole number of times into 1 meaning the integral is the same on any interval of length 1.

    Since we found it on [0, 1], just multiply with 10.

    = 10(e - 1)

    • CK2 weeks agoReport

      I solved like this sir https://imgur.com/a/LtsdrKG . 

  • 3 weeks ago

    Bernie Sanders is my favourite hobbit

  • 3 weeks ago

    Did you mean [0 → 10] ∫ e^(2x – ⌊2x⌋) (x – ⌊x⌋) dx which is

    10 × ([0 → 1] ∫ e^(2x – ⌊2x⌋) (x – ⌊x⌋) dx), where ⌊a⌋ = floor(a)?

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    • CK3 weeks agoReport

      Thank you so much sir for the answer ! :)) 

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