Hi Jagdish,

If you note, the maximum value of the integral will depend on the nature of the function f(x).

Without knowing the nature of the function, it would not be possible to find the maximum numerical value of the integral.

At the most the maximum value of the integral could only be found only interms of f(a), where a is some known number in the range [0,1].

The other maxima and minima question you had posted cannot be viewed on the forum as you have added a period at the end of the Subject Heading. Kindly repost without any periods in the header.

Regards,

Ashwin (IIT Madras).

Thanks ashwin for giving me a responce

actually here we have to calculate minimum of given expression

thanks

Hi Jagdish,

If we have to find the minimum, then it is fine.

Consider this problem geometrically (I am not attaching a figure, hope you can visualise it).

Consider a square of unit area, enclosed by the 4 lines (x-axis, y-axis, x=1,y=1).

All my working will be in the domain [0,1], as that is the range where our problem is defined.

Consider at some point in [0,1] f²(x) is less that 1, then there has to be a point in [0,1] where f²(x) is more than 1.

That is to say f²(x) cannot be all less than 1 in the interval [0,1]. (the reverse is also true, that is not all of f²(x) in [0,1] can be more than 1).

Now let us say there is some point a in [0,1] where f²(a) < 1, and so there must be a point b in [0,1] where f²(b) is more than 1.

Now f⁴(x) for a point where f²(x)>1, will be more than 1.

And for the point where f²(x)<1, √[f²(x)] (=f(x)) will be more than f²(x).

So, in case f(x) is positive, then the minimum value of the integral will occur for f(x) = 1, for all x in [0,1].

Now in case, we need to minimise this further (then f(x) will have to be -1, entirely in the given interval).

So f(x) = -1 in the entire domain [0,1], satisfying the 1st given integral.

Also the minimum value of the 2nd integral will be 0.

Hope it helps.

Regards,

Ashwin (IIT Madras).