order of integration

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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.
Hmm so \(x\) is a constant here.
lets sketch the curves representing the bounds |dw:1438709430741:dw|

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that format files wont open on my system
|dw:1438709761438:dw| |dw:1438709783170:dw|
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
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|dw:1438709868443:dw|
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thank you @Astrophysics , very kind.
No problem :)
changing order of integration, do we get : \[\int\limits_{0}^x\int\limits_0^t f(p) dp\,dt~~=~~\int\limits_{0}^x\int\limits_p^x f(p) dt\,dp\]
Yes, that's what the solution shows
you may shoot an arrow and see the starting and leaving curves to setup the bounds : |dw:1438710530659:dw|
|dw:1438710651083:dw| When I used to do double integrals I did the same thing, it's easier to see