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hey show the existence of m,n,p f(x)=ax^2+bx+c if integral (sup lim 1)(inf lim...
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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` .
Posted by sciencesolve on August 2, 2012 at 2:50 PM (Answer #1)
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