Straight Question, Spivak, Integrals

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c.teixeira
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Homework Statement



Exercice 14-14, 3[itex]^{rd}[/itex] edition, Spivak:

Find a function f such that f[itex]^{```}[/itex](x) = [itex]\frac{1}{\sqrt{1 + sin^{2}x}}[/itex].

The answer acording to the book is :

f(x) = [itex]\int\left(\int\left( \int \frac{1}{\sqrt{1+sin^{2}x}} dt \right) dz \right)[/itex]dy

With the 1[itex]^{st}[/itex], 2[itex]^{nd}[/itex] and 3[itex]^{rd}[/itex] integrals evaluated from 0 to x, y, z respectively.

My question:

aren't they forgetting the constants?

For instance:

(f``)` = [itex]\frac{1}{\sqrt{1 + sin^{2}x}}[/itex] = g(x).

Then if F(x) = [itex]\int\frac{1}{\sqrt{1 + sin^{2}x}}[/itex], with the integral evaluated from a to x.
As g is continuos, F' = g = (f``)`, thus F = f`` + C, by the Mean Value Theorem.

I gave only calculated f``, but I am guessing this is the kind of procedure they use to reach f(x).
So, to my question, what are they doing with the constants of integration?
Is this reasoning correct?

Regards,

c.teixeira
 
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The problem, as you stated it, was "find a function". The most general form of such a function would be
[tex]\int_0^x\left(\int_0^y\left(\int_0^z\frac{1}{1+ t^2}dt\right)dz\right)dy+ Cx^2+ Dx+ E[/tex]
for arbitrary constants C, D, and E. But taking C= D= E= 0 gives a[\b] function.

However, what you wrote, with "x" in the integrand rather than "t" is incorrect. I assume that was a typo.
 
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HallsofIvy said:
The problem, as you stated it, was "find a function". The most general form of such a function would be
[tex]\int_0^x\left(\int_0^y\left(\int_0^z\frac{1}{1+ t^2}dt\right)dz\right)dy+ Cx^2+ Dx+ E[/tex]
for arbitrary constants C, D, and E. But taking C= D= E= 0 gives a[\b] function.

However, what you wrote, with "x" in the integrand rather than "t" is incorrect. I assume that was a typo.


Yes, that was a typo. Regardless, my doubt is answered.

Thank you,