Help to resolve I[x(t)]=∫ 4 8 [t−4x(t)] 2 dt,x(4)=1,x(8)=2

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I[x(t)]=∫₄⁸[t−4x(t)]²dt,x(4)=1,x(8)=2

Guest Jan 29, 2017

edited by Guest Jan 29, 2017
edited by Guest Jan 29, 2017

Alan · Answer 1 · 2017-01-29T06:51:24

You need to know the form of the function x(t).

For example, you could have: x(t) = t/4 or x(t) = (t^2)/48 + 2/3 or ...

These will give different results for the integral in general.

Help to resolve I[x(t)]=∫ 4 8 [t−4x(t)] ​2 dt,x(4)=1,x(8)=2