I want to solve this double integral.

$$\int_0^4 \int_\sqrt x^2 \sin(x^3) \,dy\,dx$$ I tried to swap the limits of integration so I got $$\int_0^2 \int_0^{y^2} \sin(x^3) \,dx\,dy$$

But when I looked on te answers (see the link at the bottom) I saw that they also change x to y and got $$\int_0^2 \int_0^{y^2} \sin(y^3) \,dx\,dy$$

My question is what is the justification for this?

The answer

  • 3
    $\begingroup$ It's just a typo. Your version is correct. The integrand should stay the same. $\endgroup$ – quasi Jul 10 '17 at 11:18
  • $\begingroup$ It will remain $\sin x^3$, in the last expression. $\endgroup$ – MAN-MADE Jul 10 '17 at 11:21
  • $\begingroup$ Ok. But I think that the indefinite Integral of $\sin(x^3)$ is not elementary$ $\endgroup$ – roro Jul 10 '17 at 11:24
  • $\begingroup$ So how can I solve it? I tried to swap the limits (that's the ex. topic) and got stuck on this $\endgroup$ – roro Jul 10 '17 at 11:25

It's a typo. You don't switch the bound variable when switching the order of integration.

The version of the integral in the answer sheet is much easier than the original version. I'd guess (hope) that the second version is what was intended. As you said, $\displaystyle \int \sin(x^3) \, \mathrm dx$ is not elementary, unless I'm missing some trick.

Can you solve the version in the answer key?

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  • $\begingroup$ Yes, the second version is clear to me, probably a type as you said $\endgroup$ – roro Jul 10 '17 at 11:46

$$\int_{0}^{4}\int_{\sqrt{x}}^{2}\sin x^3dy\space dx=\int_{0}^{2}\int_{0}^{y^2}\sin x^3dx\space dy=\int_{0}^{2}\int_{0}^{y^2} \Big( x^3-\frac{x^{9}}{3!}+\frac{x^{15}}{5!}\dots\Big)dx=\int_{0}^{2} \Big( \frac{x^4}{4}-\frac{x^{10}}{10.3!}+\frac{x^{16}}{16.5!}\dots\Big)_{0}^{y^2}dy=\int_{0}^{2}\Big( \frac{y^8}{4}-\frac{y^{20}}{10.3!}+\frac{y^{32}}{16.5!}\dots\Big)dy=\Big( \frac{y^9}{9.4}-\frac{y^{21}}{21.10.3!}+\frac{y^{33}}{33.16.5!}\dots\Big)_{0}^{2}=\Big( \frac{2^9}{9.4}-\frac{2^{21}}{21.10.3!}+\frac{2^{33}}{33.16.5!}\dots\Big)$$.

I don't know the simplified answer.

You can see here, but that will not help you anyway.

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  • $\begingroup$ I don't think it was their intent cause the answer is simple and we didn't learn to solve the last expression $\endgroup$ – roro Jul 10 '17 at 12:03
  • $\begingroup$ @roro can you tell me the name of the book, you get this problem.(If not a book then give the link where you get it and if it is from your your assignment, then do nothing.) $\endgroup$ – MAN-MADE Jul 10 '17 at 12:06

$\sqrt x \leq y \leq 2$

$0 \leq x \leq 4$

You need to draw the graph out. Try to plot $y=2$ $y=\sqrt x$

and two more lines $x=o$ & $x=4$

We notice that $y=2$ is higher than $y=\sqrt x$

The question is with respect to x

enter image description here

Now let take Change the variable with respect to y.

We have

$$0 \leq x \leq y^2$$

$$0 \leq y \leq 2 $$

I think you can do it now

$$\int_{0}^{2} \int_{0}^{y^2}\sin(y^3) dxdy=\int_{0}^{2}[x\sin(y^3)]_0^{y^2} dy$$

$$\int_{0}^{2}[x\sin(y^3)]_0^{y^2} dy=\int_{0}^{2}y^2\sin y^3 dy $$

$$\int_{0}^{2}y^2\sin y^3 dy=-[\frac{1}{3}\cos(y^3)]_0^{2} $$

$$ \frac{1-\cos 8}{3}$$

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  • $\begingroup$ Done bro. I you can't change $y^3$ to $x^3$ in the integrand ! $\endgroup$ – Crazy Jul 10 '17 at 11:49
  • $\begingroup$ I understood everything and that's what I tried but how you can change $\sin(x^3) $ to $\sin(y^3) $? what is the justification for this? The draw helped you tp o change the limits not the integrand $\endgroup$ – roro Jul 10 '17 at 11:51
  • $\begingroup$ You are right. I think there is a typographical error in the question! Since the answer that you provided is $sin(y^3)$ so the correct integrand must be so! It is good that you notice something wrong with the question itself! I mean the integrand $\endgroup$ – Crazy Jul 10 '17 at 11:53

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