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IrishBoy123

  • one year ago

order of integration

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  1. IrishBoy123
    • one year ago
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    show that \( \huge \int_{ 0}^{x} [ \int_{0}^{t} f(p) \ dp] \ dt = \int_{0}^{x} (x − p) \ f(p) \ dp\) by changing order of integration have a real blind spot on this, simply cannot visualise or draw it.

  2. ganeshie8
    • one year ago
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    Hmm so \(x\) is a constant here.

  3. ganeshie8
    • one year ago
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    lets sketch the curves representing the bounds |dw:1438709430741:dw|

  4. IrishBoy123
    • one year ago
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    question

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  5. IrishBoy123
    • one year ago
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    solution

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  6. ganeshie8
    • one year ago
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    that format files wont open on my system

  7. Astrophysics
    • one year ago
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    |dw:1438709761438:dw| |dw:1438709783170:dw|

  8. IrishBoy123
    • one year ago
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    sorry i will try and convert it the RHS looks like a convolution from laplace, not that that helps me. the LHS has me seriously befuddled. i was thinking Fubini and double integrals, which i can usually handle, but no idea really

  9. IrishBoy123
    • one year ago
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  10. ganeshie8
    • one year ago
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    |dw:1438709868443:dw|

  11. IrishBoy123
    • one year ago
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  12. IrishBoy123
    • one year ago
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    thank you @Astrophysics , very kind.

  13. Astrophysics
    • one year ago
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    No problem :)

  14. ganeshie8
    • one year ago
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    changing order of integration, do we get : \[\int\limits_{0}^x\int\limits_0^t f(p) dp\,dt~~=~~\int\limits_{0}^x\int\limits_p^x f(p) dt\,dp\]

  15. Astrophysics
    • one year ago
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    Yes, that's what the solution shows

  16. ganeshie8
    • one year ago
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    you may shoot an arrow and see the starting and leaving curves to setup the bounds : |dw:1438710530659:dw|

  17. Astrophysics
    • one year ago
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    |dw:1438710651083:dw| When I used to do double integrals I did the same thing, it's easier to see