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I am trying to calculate $$\int_{\pi^2}^{4\pi^2} \frac{\cos(\sqrt{x})}{\sqrt{x}} \,dx.$$ I calculated the integral and got $2\sin(\sqrt{x})$ as a result, but for $x=\pi^2$ and $x=4\pi^2$ we get that $2\sin(\sqrt{\pi^2})=0$ and $2\sin(\sqrt{4\pi^2})=0$ So the Riemann integral will be $0-0=0$ which is not true, as you can see from ploting $2\sin(\sqrt{x})$.

Any help will be much appreciated!

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  • $\begingroup$ I don't think there's anything wrong with what you did...I even plotted $2\sin\sqrt{x}$ as well as the integrand and it looks fine. And Maple agrees with 0. $\endgroup$ Jan 8 '13 at 20:28
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    $\begingroup$ Even the graph of the function seems to agree with the result: wolframalpha.com/share/… $\endgroup$
    – Andrew D
    Jan 8 '13 at 20:32
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$$\int_{\pi^2}^{4\pi^2} \frac{\cos(\sqrt{x})}{\sqrt{x}} \,dx.$$

Letting $u=\sqrt{x}$, we have $du=\large\frac{1}{2\sqrt{x}}\,dx\;$ or $\;2\,du = \large\frac{dx}{\sqrt{x}},\;$ so the integral becomes $$ \int_\pi^{2\pi} 2\cos u \,du\;=\;2\sin u\Big|_\pi^{2\pi} = 2\sin (2\pi)-2\sin(\pi)\,=\,0 - 0 = 0 $$

I simply changed the bounds of integration, so there's no need to "back-substitute". So it seems, as you proceeded in your evaluation of the definite integral, that your answer is indeed correct.

See, e.g. Wolfram|Alpha's computation:

$\quad\quad\quad\quad$enter image description here


Visual representation of the integral:

enter image description here

*If we are looking to calculate the area between the x-axis and the curve, then we need to split the integral to compute the area below the x-axis, and the area above the x-axis: for $u$, the dividing point will be $\large\frac{3\pi}{2}$ (for $x$: $\large\frac{9\pi}{4}).$

$$\Big|\int_\pi^{\large\frac{3\pi}{2}} 2\cos u \,du\;\Big|\;\;+ \;\;\Big|\int_{\large\frac{3\pi}{2}}^{2\pi} 2\cos u \,du\;\Big|\;\; = \;\;\Big|2\sin u|_\pi^{3/2\pi}\Big|\;+\; \Big|2\sin u|_{3/2\pi}^{2\pi}\Big|\; =\; 2 + 2 = 4$$

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    $\begingroup$ You answered the OP in detailed. + $\endgroup$
    – Mikasa
    Jan 9 '13 at 3:09
  • $\begingroup$ I see.. Thank you so much! $\endgroup$
    – Belial
    Jan 9 '13 at 17:09
  • $\begingroup$ Your welcome, Belial! If you find an answer to be especially helpful, you can "accept" it by clicking on the "greyed-out" checkmark to the left of the answer. $\endgroup$
    – amWhy
    Jan 9 '13 at 17:15
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You are right. Here is a slightly different approach, if you want. Using the substitution $u=\sqrt{x}$, we have $du=\frac{1}{2\sqrt{x}}dx$ so the integral becomes $$ \int_\pi^{2\pi} 2\cos u du=2\sin (2\pi)-2\sin(\pi)=0 . $$

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Although @julien got the right fact about the definite integral, we should find a positive value for that. I mean we should consider the following integrals as well to find a real area: $$ \Big|\int_\pi^{3\pi/2}\Big|\;\;+\;\;\Big|\int_{3\pi/2}^{2\pi}\Big| $$

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  • $\begingroup$ @BrianM.Scott: I edited it. Thanks. $\endgroup$
    – Mikasa
    Jan 8 '13 at 20:40
  • $\begingroup$ Nice! I'll edit to make bigger absolute value signs +1 $\endgroup$
    – amWhy
    Feb 21 '13 at 0:10
  • $\begingroup$ @amWhy: Thanks for the time. $\endgroup$
    – Mikasa
    Feb 21 '13 at 5:47

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