let f(x)=t, following the given equation we have t^3+t=x, differentiate both sides, we have dx=(3t^2+1)dt, again from t^3+t=x, we plug 0 and 2 into x, we get t=0 and 1 as real solutions, so our new bounds are from 0 to 1, and integral becomes tdx=3t^3+tdt (to be rigorous, f(x) can be proven to be integrable from the cubic roots formula)
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Can you solve this question
f is real valued function.
f^3(x)+f(x)=x.
Find integration of f(x) from 0 to 2.
let f(x)=t, following the given equation we have t^3+t=x, differentiate both sides, we have dx=(3t^2+1)dt, again from t^3+t=x, we plug 0 and 2 into x, we get t=0 and 1 as real solutions, so our new bounds are from 0 to 1, and integral becomes tdx=3t^3+tdt (to be rigorous, f(x) can be proven to be integrable from the cubic roots formula)