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 Getting Natural With Derivative And Integral (Posted on 2010-02-11)
The function G is such that each of G, G’ and G” exists and is continuous on the interval [0, e].

It is further known that G’(e) = G(e) = G’(1) = G(1) = 1, and:

e
∫ G’(y)* y-2 dy = 0.5
1

Evaluate:

e
∫ G”(y)*ln y dy
1

Note: ln y denotes the natural logarithm of y.

 No Solution Yet Submitted by K Sengupta No Rating

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 Just a thought | Comment 3 of 4 |
Just a thought..

Here’s a fabricated, possible form for G(y). It contains a parameter k, but satisfies all four boundary conditions and the integral constraint for all values of k. (The exact form in Maple contains lots of e’s!)

G(y) =
(1.185193911 - 0.08062379520*k)y5 + (-8.485808618 + 0.7348992616*k)y4
+ (21.06791156 - 2.559419754*k)y3 + (-21.41917276 + 4.236636925*k)y2
+ (8.651876020 - 3.331492634*k)y + k

The required integral ‘evaluates’ to   0.0007799*k + 0.586745 ; in other words, to any value we like by choice of k.

Should we be suspicious?

 Posted by Harry on 2010-02-21 13:37:57

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