#Proof that uses mean value theorem for integrals
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It's between min f and max f times int g
@tropic mulch i don't understand what you mean
What don’t you understand? The notation or the mathematical reasoning? I can write it in Latex if needed but for me it is 3h50 am, I am using a tablet, I won’t switch on my computer now.
what's the it's part?
f(a) < f(c) < f(b) where a is the value where f(x) is at its minimum and b being the value where f(x) is at its maximum
yes I get that part. but how does g(x) get incorporated in?
Yes now consider the function h(t)=f(t) * int(g(x)) you know that here h(a)<=int(f(x) *g(x)) <= h(b) with h(a) the minimum and h(b) the maximum now what can you conclude using the intermediate value theorem ?
Oh that (1/int g)*int(fg) = f(c)
Yes well you can conclude that there exists a c such that h(c)=int(f(x) g(x)) using the intermediate value theorem
Ah ok thanks for your help @wild pebble @harsh lantern @tropic mulch
@finite lagoon has given 1 rep to @wild pebble @tropic mulch @harsh lantern
+close