Question

Let f be a continuously differentiable real-valued function on [0,1] such that ∫2/3 1/3f(x)dx=0. Find the minimum value of ∫f′(x)^2dx/(∫f(x)dx)^2, 0 and 1 are the upper and lower bound of the question

Answer 1

Solution:

It is Given that f be a continuously differentiable real-valued function on [0,1] such that ,$\int_{1/3}^{2/3}f(x) dx$ = 0

→ $[f'(x)]_{1/3}^{2/3}$ =0

→f'(2/3) - f'(1/3)=0

→f'(2/3) = f'(1/3)

Now, As area under the curve can't be equal to zero. So

→f(x)=1 , why i have written 1 because a portion of line segment is considered for that part we have to find the area.

Differentiating both sides we get

f'(x) = 0

now,→ $\frac{∫{f'(x)}^{2}dx}{[∫f(x)dx]^{2}}$ =0/1=0,as there is no effect of upper and lower limit.

because , $\int$[f'(x)]²dx = 0 as f'(x)=0