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 ∫10f′(x)2dx/(∫10f(x)dx)2, 2/3 and 1/3 are the upper and lower bound of the question

Answer 1

Solution: It is given that  f is a continuously differentiable real-valued function on [0,1] such that

→∫2/3 1/3f(x)dx=0

Differentiating both sides with respect to x, we get

→f(x)= 0 × x +k= k , where k is any constant.

Differentiating both sides with respect to x,

f '(x)= 0,

Now,[f' (x)]²= (0)²=0  .................(1)

→∫10×0 .d x  =0  →Using (1)

→∫10 [f'(x)]^{2} dx =0

→$\frac{0}{(\int 10 f(x)dx)^{2}}=0$ , there is no effect of upper and lower bound as numerator is equal to zero.

Answer 2

Given:

f - Continuously differential real-valued function on [ 0, 1 ]

∫ 2/3 1/3 f ( x ) dx = 0

Upper bound = 2/3

Lower bound = 1/3

To find:

The minimum value of ∫ 10 f′ ( x ) 2 dx / ( ∫ 10 f ( x ) dx ) 2

Solution:

Differentiating with respect to x,

f ( x ) = x + k

Where,

k - constant.

Differentiating with respect to x,

f '( x ) = 0

We get,

[ f' ( x ) ]^2 = ( 0 )^2

[ f' ( x ) ]^2 = 0

Substituting the value,

We get,

∫ 10 ( 0 ) d x  = 0

Hence,

∫ 10 [ f' ( x ) ]^ { 2 } dx =0

During differentiation as the numerator becomes 0, the upper and lowe bounds have no value.

Hence,  The minimum value of ∫ 10 f′ ( x ) 2 dx / ( ∫ 10 f ( x ) dx ) 2 is 0.