# hey show the existence of m,n,p f(x)=ax^2+bx+c if integral (sup lim 1)(inf lim -1)f(x)dx=af(-1)+bf(0)+cf(1)

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You need to evaluate f(-1),f(0) and f(1) such that:

`f(-1) = a-b+c ; f(0) = c ; f(1) = a+b+c`

Evaluating the definite integral yields:

`int_(-1)^1 f(x)dx = int_(-1)^1 (ax^2 + bx + c)dx `

`int_(-1)^1 f(x)dx = (ax^3/3 + bx^2/2 + cx)|_(-1)^1 `

`int_(-1)^1 f(x)dx = a/3 + b/2 + c + a/3 - b/2 + c `

`int_(-1)^1 f(x)dx = 2a/3 + 2c`

Notice that the problem provides the information that`int_(-1)^1 f(x)dx = mf(-1) + nf(0) + pf(1)`  such that:

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

You need to open the brackets such that:

`2a/3 + 2c = ma - mb + mc + nc + pa + pb + pc`

You need to move all terms to the left such that:

`2a/3 + 2c- ma+ mb- mc- nc- pa- pb- pc = 0`

You should factor out a,b,c such that:

`a(2/3 - m - p) + b(m - p) + c(2 - m - n - p) = 0`

`2/3 - m - p = 0 =gt m+p = 2/3`

`m - p = 0 =gt m = p`

`m + n + p = 2`

Since `m=p =gt 2p = 2m = 2/3 =gt m = p = 1/3`

`2/3 + n = 2 =gt n = 2 - 2/3 =gt n = 4/3`

Hence, evaluating m,n,p under the given conditions yields `m = p = 1/3`  and `n = 4/3` .

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