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The question is: Integrate the function $\left(\dfrac{9}{(6x-5)}\right)^{1/2}$, with no limits given. I can get the integral, which is $(6x+5)^{1/2}+c$, but the second part is , hence, find the range of values for the integral to be valid, showing your working clearly. I would like to know under what conditions an integral is valid, as I get the answer $x$ is more than or equal to $\frac{5}{6}$ but my teacher says it is just $x$ is more than $\frac{5}{6}$.

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  • $\begingroup$ What are you integrating when $x = \frac{5}{6}$? Is it valid to integrate that? $\endgroup$
    – Alex
    Mar 11 '14 at 3:53
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$$\sqrt{\frac9{6x-5}}$$ will remain real if $\displaystyle6x-5\ge0$

but will remain finite if $\displaystyle6x-5\ne0$

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  • $\begingroup$ umm.... I was asking about the validity of the integral, in this case (6x+5)^(1/2). I am a newbie at integration, so can you please explain how knowing the thing to be integrated is real or finite will help? also can you explain what finite means in this context? Thanks. $\endgroup$
    – dadadok
    Mar 11 '14 at 3:29
  • $\begingroup$ $$I=\int\sqrt{\frac9{6x-5}}dx=3\int\frac{dx}{\sqrt{6x-5}}$$ Setting $\sqrt{6x-5}=u\implies 6x-5=u^2\implies6dx=2udu\cdots$ $$I=3\frac{2\sqrt{6x-5}}6+K$$ for $6x-5>0$ $\endgroup$ Mar 11 '14 at 3:33
  • $\begingroup$ The questions is why $x \gt \frac{5}{6}$ and why not $x \geq \frac{5}{6}$. I do not know how to succinctly explain this. $\endgroup$
    – Brad
    Mar 11 '14 at 3:35
  • $\begingroup$ So, is 6x-5=0 in this case valid? $\endgroup$
    – dadadok
    Mar 11 '14 at 3:38
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    $\begingroup$ @Brad, the integral must be finite $\endgroup$ Mar 11 '14 at 3:39
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Because the original function doesn't exist at $ x = 5/6 $. Sure the integral can still be evaluated, but it won't be a standard integral (I'm assuming you haven't got to improper integrals yet).

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