11thank you very much sir...
Given a function f : [0,4] ----> R is differentiable , then for some \alpha , \epsilon (0 , 2) , \int_{0}^{4}{f(t) dt} equals to :
a) f(\alpha2) + f(2)
b)2\alpha f(2) + 2
f(\alpha2)
c)\alpha f(2) +
f(\alpha2)
d)f(\alpha)f() (f(\alpha)+ f(
))
( ans given is b)
-
UP 0 DOWN 0 1 5
5 Answers
341Nice prob, at least its different from the run of the mill stuff.
First consider \int_0^{z^2} f(x) \ dx
With the substitution x=y^2, we have that this integral is equivalent to \int_0^z 2y f(y^2) \ dy
So consider F(t) = \int_0^t 2y f(y^2) \ dy
Remember that F(2) = \int_0^4 f(x) \ dx which is what we are seeking.
From Rolle's Theorem, \exists \ \alpha \in (0,1) such that F'(\alpha) = \frac{F(1) - F(0)}{1-0} = F(1) \ (\because \ F(0) = 0)
Similarly \beta \in (1,2) satisfying
F'(\beta) = \frac{F(2) - F(1)}{2-1} = F(2) - F(1)
Adding we obtain F'(\alpha) + F'(\beta) = F(2)
But from Leibnitz Theorem F'(\alpha) = 2 \alpha f(\alpha^2) and F'(\beta) = 2 \beta f(\beta^2)
Hence
\int_0^4 f(x) dx =F(2) = F'(\alpha) + F'(\beta) = 2 \alpha f(\alpha^2) + 2 \beta f(\beta^2)
341Midway through the solution I remembered a problem posed in goiit some months back: http://www.goiit.com/posts/list/differenciation-pls-solve-97873.htm