# Given the polynomial of second degree f prove that for m,n,p real numbers the identity is true.m*f(-1)+n*f(0)+p*f(1)=definite integral of f, if x is in the interval [-1,1]

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### 1 Answer

If the polynomial is expressed by a quadratic function, we'll have:

f(x) = ax^2 + bx + c

Since in the given constraint, there are the values f(-1), f(0) and f(1), we'll calculate them:

f(-1) = a - b + c

f(0) = c

f(1) = a + b + c

We'll calculate the definite integral of f(x), if the limits of integration are the ends of closed interval [-1,1].

Int f(x)dx = Int ax^2dx + Int bxdx + Int cx

Int f(x)dx = ax^3/3 + bx^2/2 + cx

We'll apply Leibniz Newton:

F(1) = a/3 + b/2 + c

F(-1) = -a/3 + b/2 - c

Int f(x)dx = F(1) - F(-1)

Int f(x)dx = a/3 + b/2 + c + a/3 - b/2 + c

We'll eliminate like terms:

Int f(x)dx = 2a/3 + 2c

From enunciation, we'll get:

2a/3 + 2c = m(a-b+c) + n*c + p(a + b + c)

We'll remove the brackets:

2a/3 + 2c = a(m + p) + b(-m + p) + c(m + n + p)

Comparing, we'll get:

m + p = 2/3 (1)

-m + p = 0 (2)

We'll add (1) and (2) and we'll get:

2p = 2/3

p = 1/3 => m = 1/3

m + n + p = 2

2/3 + n = 2 => n = 2 - 2/3 => n = 4/3

**The values of m,n,p, for the identity ****m*f(-1) + n*f(0) + p*f(1) = Int f(x)dx is true, are: m = 1/3 , n = 4/3 , p = 1/3.**