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Integral Calculus

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20 Mar 2008 18:26:59 IST

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Contrived Problem

None

Ok one of my feeble attempts at making up a problem (Atleast I have not encountered one like this). Lets see how quickly this gets destroyed.

A function f(x) is such that ₁⁴f(x) dx = 3. Find a function g(x) such that f(x)+g(x) has a zero in (1,4).

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sreeraman nagasubramaniyan

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20 Mar 2008 19:30:28 IST

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$[tex] \\ \mbox{Consider} \ f(x) = 1 \ \mbox{. So we get } \int^4_1 f(x)\,dx = 3 \\ \\ g(x) = x^2- \frac{5}{2}x \ \mbox{such that} \ f(x)+g(x) = x^2- \frac{5}{2}x + 1 \\ \\ \mbox{Now this has a root 2 which belongs to } \ (1,4)$

Hari Shankar

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20 Mar 2008 19:56:30 IST

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Don't assume f(x).

I am looking for a one-size-fits-all g(x) for all f(x) which satisfy the given condition

amaron

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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](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!

Hari Shankar

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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

Another 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.

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