Integration? No no its ...

Calculus Level 5

$\int_{1}^{x}A(x)B(x)dx*\int_{1}^{x}C(x)D(x)dx-\int_{1}^{x}A(x)C(x)dx*\int_{1}^{x}B(x)D(x)dx=f(x)$

If f(x) is a nth degree polynomial and satisfies above equation for all real x, then area bounded by f(x) and the line y=x-1 can be represented as

$\tfrac{a}{b*c}$

Find the value of a-b+c

Assumptions:

n is an even natural number.

a,b may not be numbers.

c is a number.

A, B, C, D are non constant continuous functions of x.

1 n 0 2

1 solution

Prakash Chandra Rai
Jan 7, 2015

It is easy to see that 1 is root of above equation.

Now,let's differentiate it. By applying product rule and other mess, when we finally analyse our differentiated equation, we observe that every term of the equation has at least one integral term in which integration is performed from 1 to x. So we can conclude that 1 is root of $\frac{\mathrm{d} }{\mathrm{d} x}f(x)$ also.

Now if we again carefully analyse, then we observe that if we differentiate it again and again n-1 times, then in each derivative equation, at least one such integral will be there in every term of equation.

From this,we can conclude that f(x), $\frac{\mathrm{d} }{\mathrm{d} x}f(x),\ \frac{\mathrm{d^{2}} }{\mathrm{d} x}f(x)\,..................\frac{\mathrm{d^{n-1}} }{\mathrm{d} x}f(x)$ each derivative has 1 as it's root.

This shows that Function is $(x-1)^{n}$ .

Now,intersection points can be easily determined and integration is too easy.

a comes out to be n-1 , c=2 and b=n+1

Integration? No no its ...

1 solution

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