I am trying to evaluate the exact value of the following definite integral:

\begin{align} \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x \end{align}

Since $ \int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x$ has symmetry, I did the following: \begin{align} \frac{91\pi}{6} \cdot \frac{2}{\pi} = \frac{91}{3} \end{align} \begin{align} \int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x = [\sin(x)]^{\frac{\pi}{2}}_0 = 1 \end{align} \begin{align} \therefore \text{Bounded Area } = 1 \cdot \frac{91}{3} = \frac{91}{3} \end{align}

Apparently, the correct answer is $ \frac{61}{2} $ which is very close to my answer. I cannot understand why my answer is wrong. Could someone please advise me?

  • $\begingroup$ I can't understand your argument of how "cosine having symmetry" is connected with what you did... $\endgroup$ – DonAntonio Oct 16 '16 at 17:39
  • $\begingroup$ @DonAntonio Cosine having symmetry means that I can just simply multiply the area by a certain number of times because the areas of the other regions are all similar to the first one. In this question, I can multiply the bounded area from $ 0 $ to $ \pi $ by 30. $\endgroup$ – LanceHAOH Oct 17 '16 at 5:15
  • $\begingroup$ @La Ok....that's exactly what I did in my answer, I just didn't understand what you mean...and the symmetry is for $\;\int|\cos x|dx\;$ . $\endgroup$ – DonAntonio Oct 17 '16 at 9:12

Your technique only works for integer multiples. After that, you're implicitly assuming that the integral will scale linearly, and that's not the case. So instead of $\frac{91}{3} \cdot 1$, it should be $$30\cdot 1 + \int_0^\frac{\pi}{6}|\cos x| dx = 30 + \sin\left(\frac{\pi}{6}\right) = \frac{61}{2}$$

A caveat - the period of $|\cos x|$ is in fact $\pi$, not $\frac{\pi}{2}$, so if the integer quotient had been odd (for instance, if the original integral had been from 0 to $\frac{47\pi}{3}$), you would need to start the integral for the remainder from $\frac{\pi}{2}$ (a region where $\cos x$ is negative), e.g.,

$$\int_0^{\frac{47\pi}{3}}|\cos x|dx = 31\cdot1+\int_\frac{\pi}{2}^\frac{2\pi}{3}|\cos x|dx = 31 - \int_\frac{\pi}{2}^\frac{2\pi}{3}\cos x dx = 31 - \frac{\sqrt{3}}{2} + 1 = 32 - \frac{\sqrt{3}}{2}$$

  • $\begingroup$ Integer multiple meaning that I can multiply my calculated area by any integer and it will give the correct result? E.g. $ 3\int^{\frac{\pi}{2}}_0dx = \int^{\frac{3\pi}{2}}_0dx $ $\endgroup$ – LanceHAOH Oct 16 '16 at 17:49
  • 1
    $\begingroup$ In the particular case, yes, $\int_0^{k\pi/2} |\cos x|dx = k\int_0^{\pi/2} |\cos x|dx$ for any integer k, for the reason you outlined in your OP. It's also the case that $\int_0^{k\pi/2} dx = k\int_0^{\pi/2} dx$, but I don't think that's what you meant to ask. $\endgroup$ – user361424 Oct 16 '16 at 17:53
  • $\begingroup$ Thanks! Indeed. I forgot to include the $ |cos(x)| $ in my integral $\endgroup$ – LanceHAOH Oct 16 '16 at 17:55
  • $\begingroup$ One caveat you should keep in mind if you encounter similar problems in the future. Since the period of $|\cos x|$ is in fact $\pi$, you only get to start the remainder from 0 because 30 is even. If the integer quotient were odd, the integral for the remainder would start from $\frac{\pi}{2}$. I'll add this to the answer with a worked example shortly. $\endgroup$ – user361424 Oct 17 '16 at 1:28
  • $\begingroup$ In short I need to check where the graph of the remainder lies right? $\endgroup$ – LanceHAOH Oct 17 '16 at 2:09

Observe that

$$\int_0^\pi|\cos x|dx=\left.2\int_0^{\pi/2}\cos x dx=2\sin x\right|_0^{\pi/2}=2\implies$$

$$\int_0^{91\pi/6=15\pi+\frac\pi6}|\cos x|dx=\left.2\cdot15+\int_\pi^{\frac{7\pi}6}(-\cos x)dx=30-\sin x\right|_\pi^{\frac{7\pi}6}=30+\frac12=\frac{61}2$$


You can't do so, but you can $\frac{90\pi}{6}⋅\frac{2}{π}+\int_0^\frac{\pi}{6}|cosx|dx=30+1/2$


for $k\in \mathbb Z$

$\int_{\frac{\pi}{2}+2k\pi}^{\frac{3\pi}{2}+2k\pi}(-cos(x))dx+$ $\int_{\frac{3\pi}{2}+2k\pi}^{\frac{5\pi}{2}+2k\pi}cos(x)=4$





Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.