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Integral Calculus
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sreeraman nagasubramaniyan
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20 Mar 2008 19:30:28 IST
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20 Mar 2008 20:26:47 IST
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The Obvious solution is f(x) = -g(x).
But consider,
let g be any real valued,continuous function that satisfies f + g =0 for at least one value of x
(1.4)
Let H(x) =
( f + g)dx
By LMVT for H over (1,4),
H'(c) = [H(4) - H(1)]/3
But H'(c) = f + g
if f + g =0 for at least one c (1.4),
Then H'(c) = 0 for at least one c
(1.4),
implies , a solution for g can be got if ,<Though this is not the only solution>
H(4) - H(1) =0
implies [1]
[4]( f + g)dx = 0
implies
(1.4)Let H(x) =
( f + g)dxBy LMVT for H over (1,4),
H'(c) = [H(4) - H(1)]/3
But H'(c) = f + g
if f + g =0 for at least one c (1.4),
Then H'(c) = 0 for at least one c
(1.4),implies , a solution for g can be got if ,<Though this is not the only solution>
H(4) - H(1) =0
implies [1]
[4]( f + g)dx = 0implies
[1]
[4](g)dx = -3
Hence any function that satisfies [1]
[4](g)dx = -3
is one class of solution.
example g(x) = (-2x/5)
That solves the problem!
[4](g)dx = -3Hence any function that satisfies [1]
[4](g)dx = -3is one class of solution.
example g(x) = (-2x/5)
That solves the problem!
20 Mar 2008 21:13:58 IST
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good. that was what i was looking for.
If a
b f(x) = 0, then by mean value theorem, there exists c
(a,b) so that f'(c) = 0
b f(x) = 0, then by mean value theorem, there exists c
(a,b) so that f'(c) = 0Another way to understand is that if a
b f(x) = 0, then f(x) neither be +ve for all x in the interval nor -ve for all x. So there's got to be x in the interval for which it is zero.
b f(x) = 0, then f(x) neither be +ve for all x in the interval nor -ve for all x. So there's got to be x in the interval for which it is zero.










