27
$\begingroup$

I have built this integral with the purpose of presenting a question. I find interesting and pleasant to readers MSE for the New Year 2016 and expecting to see different methods of solution.

I can confirm that this has been the case by the welcome that has been given and the two motivated answers it have had.

Calculate:

$$\large \int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-x} }{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x-2016}}\mathrm dx$$

$\endgroup$
7
  • 3
    $\begingroup$ After a standard change of variables, the answer seems to be 2016. $\endgroup$
    – Hmm.
    Dec 31 '15 at 18:46
  • 2
    $\begingroup$ @SoumyaSinhaBabu: Go ahead dear friend. $\endgroup$
    – Piquito
    Dec 31 '15 at 18:48
  • 4
    $\begingroup$ Observation - $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x)dx$. $\endgroup$
    – Hmm.
    Dec 31 '15 at 18:51
  • 1
    $\begingroup$ Is this a challenge, where you know the answer? Please provide more context. $\endgroup$ Dec 31 '15 at 20:57
  • 2
    $\begingroup$ It was in fact a challenge, a pleasant and pertinent one I feel. Tomorrow I´ll try to edit with my deficient English (I built the question myself for the New Year). $\endgroup$
    – Piquito
    Jan 1 '16 at 1:56
47
$\begingroup$

Define $$I:=\int_{k}^{3k}\frac{\left(3k-x\right)^{1/5}dx}{\left(3k-x\right)^{1/5}+\left(x-k\right)^{1/5}},\,k:=2016.$$

The substitution $x\mapsto 4k-x$ gives$$I:=\int_{k}^{3k}\frac{\left(x-k\right)^{1/5}dx}{\left(3k-x\right)^{1/5}+\left(x-k\right)^{1/5}}.$$Halving the sum of these expressions gives$$I=\frac12\int_k^{3k}dx=k.$$

$\endgroup$
5
  • 3
    $\begingroup$ A more complete answer than the one posted above, as you have generailsed to a limit $k$. nice =1 $\endgroup$
    – user284001
    Oct 18 '16 at 10:16
  • $\begingroup$ I see the notation := often; what does it mean? $\endgroup$
    – user71207
    Apr 3 at 11:59
  • 1
    $\begingroup$ @user71207 "is defined as". You can also write a definition backwards, viz. $\prod_{j=1}^nx=:x^n$, but that's a lot rarer. $\endgroup$
    – J.G.
    Apr 3 at 12:23
  • $\begingroup$ Ok thanks. Do you still have to say "Define $I := blah blah$" If := already means "define this to be"? $\endgroup$
    – user71207
    Apr 4 at 4:51
  • $\begingroup$ @user71207 I suppose I could change "Define" to "For", in which case the reader would understand the "is" disappears, but "Define" is also OK because the point is the mechanistic meaning of the symbols, not an attempt to parse everything as English grammar. $\endgroup$
    – J.G.
    Apr 4 at 6:13
31
$\begingroup$

Let, $$I=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-x}}{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x- 2016}}\ dx\tag 1$$ Now, using the property of definite integral: $\int_a^bf(x)\ dx=\int_{a}^bf(a+b-x)\ dx$, one should get

\begin{align*} I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-(4\cdot2016 -x)}}{\sqrt[5]{3\cdot 2016-(4\cdot2016 -x)}+\sqrt[5]{(4\cdot2016 -x)- 2016}}\ dx\\[3ex] I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{x-2016}}{\sqrt[5]{x-2016}+\sqrt[5]{3\cdot2016 -x}}\ dx\\[3ex] I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\ dx\tag 2\\[6ex] \end{align*}

Now, adding (1) & (2), one should get

\begin{align*} I+I&=\int_{2016}^{3\cdot 2016}\left(\frac{\sqrt[5]{3\cdot 2016-x}}{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x- 2016}}+\frac{\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\right)\ dx\\[3ex] 2I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\ dx\\[3ex] I&=\frac12\int_{2016}^{3\cdot 2016}\ dx\\[3ex] &=\frac12(3\cdot 2016-2016)\\[3ex] &=\color{red}{2016} \end{align*}

$\endgroup$

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.