Reverse Definite Integral of Algebraic Functions

In this work, an algebraic function is considered and integrated the function with some particular boundaries to obtain the area. Through the help of the area obtained and given boundaries, we determined different functions of different degree. Also, found a relationship between them.


Introduction
We consider a function and boundaries with x-axis and calculate the area occupied by that function.What would be the case if we go reverse of it?We mean, we are given area and boundaries, can we predict the function?In my opinion yes we can predict the function.But, the degree of the function will be in our own decision.Assuming fixed area and boundaries, we are finding different functions.What can be the relationship between those functions?What can be the good interpretation of this?What may be the application of this approach?

Determination of Functions
In general any algebraic functions can be in the form of: where, m = Slope or coefficient z = Degree of the function c = Constant In the above function y = mx z + c, when z = 1 Now, Consider Area = 24.5 × m + 7c Boundaries with x-axis: (0, 0) − (7, 0) Function=??
Let one degree function be y = mx + c then, ] 7 0 + 7c, which is true. Again, Similarly, let two degree function be y = mx 2 + c then = 7 3 3 m + 7c = 114.33× m + 7c (which is not equal to 24.5 × m + 7c) Now, here m and 7c are common in both the result.Therefore, we will match the value by trying to reduce 114.33 to 24.5 and this can be done by either multiplying or dividing.We cannot match the value by addition or subtraction.It is because on derivative that will be zero.Here, to get the actual result (24.5 × m + 7c) we will divide ( x 3 3 ) by 114.3 24.5 = 4.66 i.e = x 3 3 4.66 e the required second degree function is y = 3m 14 x 2 + c Likewise, the third degree function is: y = 4m 98 x 3 + c Similarly, when z = 2 i.e., for function y = mx 2 + c On applying above method for the function y = mx 2 + c, within the boundary (0 − 7) the first three functions giving same result on integration are:

Checking if Determined Functions are True or False
(1) y = mx + c Solution: ∴ Area = 24.5 × m + 7c (which is true)

Determination of Relation Between Functions
First, three functions giving same result on integration of function y = mx 1 + c are: Now, taking only denominator part of above functions, we get: 2 m , 14 m , 98 m , ... , in G.P x n + c Since, difference of boundaries is 7 i.e 7 − 0 = 7, let us denote 7 in above relation by △N for general.then above relation can be written as, x n + c.Thus, relation is found as: Similarly, first three functions giving same result on integration of function y = mx 2 + c are: Now, taking only denominator part of above functions, we get: 3 m , 21 m , 147 m , ... , in G.P ∴ T n = a.r n−2 Here, instead of using (n − 1) we have used (n − 2).It is because in the above mentioned list of functions the power of variable x starts to increase from 2 and also in this G.P: n = 2, 3, 4, ...
x n + c.Thus, relation is found as: From above relations (1) and ( 2), we can derive more general relation as: where, ξ = degree of variable in original function.
Therefore, we can conclude that the alternative relation is true.

Conclusion
Thus, we successfully determined different functions of different degree when area and boundaries are provided.we also determined the relation between number of functions and it is found to be: y = (△N) (ξ−n) .(1+ n).m.(ξ + 1) −1 .xn + c above relation let's determine the function of power 5 within the boundary (4 − 15