prove $ e^{bt} \ \int_0^t \ f(s) ds=\int_0^t \ ( e^{-bs} \ f(s)-be^{-bs}\int_0^s\ f(u)\ du) \ ds $

Question

please How can I prove that
$$ e^{-bt} \ \int_0^t \ f(s) ds=\int_0^t \ ( e^{-bs} \ f(s)-be^{-bs}\int_0^s\ f(u)\ du) \ ds $$ f non-negative measurable function

I would appreciate it enormously if anyone could help best,educ

Answer 1

first Thank you dear professor for ur time

so we know that : $$(fg)'= f'(s).g(s)ds +f(s).g'(s)ds $$ $$\int_0^t(fg)'=(fg)= \int_0^t \ f'(s).g(s)ds +\int_0^tf(s).g'(s)ds$$ so we use that for the left-side term : $$ e^{-bt} \ \int_0^t \ f(s)ds$$ with $u(s)=e^{-bs}, v(s)=\int_0^s f(u)du ,u'(s)=-be^{-bs},v'(s)=f(s)ds$ now it's easy to see : $e^{-bt} \ \int_0^t \ f(s)ds=u(t)v(t)=\int_0^t \ u'(s).v(s)ds +\int_0^tu(s).v'(s)ds=\int_0^t \ ( e^{-bs} \ f(s)-be^{-bs}\int_0^s\ f(u)\ du) \ ds$ you see the details make anyone able to understand, that's why i love the details best regrads , Educ

Answer 2

Try differentiating both sides.