One more integral

sry sir. i meant same as prophet sir meant, thought people would get it. he explained it better way.

Sometimes when you see a problem, it helps to get inside the mind of the author of the problem. Of course, you guys putting x=a etc. you get the result. Do you think the author had to do this.

So lets see how this is immediately obvious:

Lagrange Interpolation Formula:

I will give the formula for the quadratic case and its easily extended to any degree polynomial.

The theorem says that if we know if P(x₁) = a, P(x₂) =b and P(x₃) = c, then a quadratic P(x) satisfying the given conditions is uniquely determined as

P(x) = \frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)} P(x_1) + \frac{(x-x_3)(x-x_1)}{(x_2-x_3)(x_2-x_1)} P(x_2)+ \frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)} P(x_3)

So, for an nth degree polynomial we will need the values taken by the polynomial at n+1 points

Now you can see that all the author has done is to put

P(x_1) = P(x_2) = P(x_3) = 1

hehe....

btw, bhargav.....how do u remember all those identities??? mind-boggling stuff, really!!

Dirtiest method:
Why not expand the function and see where we end up????? I have a feeling it will get too dirty for most ppl's liking [12]

HMMMMM

ALL V get frm

f(a)=f(b)=f(c)................... provided f is a poly of degree 2 is dat

f(x) = COSTANT........

but to get it as 1 we need to sub a value ....... (zero wud b easiie)

if i am not wrong, there is much more deepansh.

what mathie meant was:

Call the given expression as f(x). Let g(x) = f(x) - 1.

g(x) has degree at most two and hence by fundamental theorem of algebra has at most two roots.

But g(x) has three distinct roots a,b and c (oops i missed stating that a,b and c are distinct). This can happen iff g(x) is identically zero.

that means f(x) is identically equal to 1

Discussion:

Can anyone tell me why is the polynomial written in this peculiar way, instead of taking the coefficients to the numerator?

put x=a or x=b or c u get it as 1. where as given is a quadratic.

please

answer is 1.

thats because watever is inside the integral is an identity and is equal to 1 always for all values of x. [why?!]

so it is \int_{0}^{1}{1}=1