# Find $\int_1^a \sqrt{x^5-1}\ dx + \int_0^b \sqrt{x^5+1}\ dx$, where $a^5-b^5 = 1$

So I am preparing to go to this olympiad, the Latvian Sophomore's dream calculus olympiad. I received previous years problems and the toughest problem in the definite integral section was this

$$\text{Find } \int_1^a \sqrt{x^5-1}\ dx + \int_0^b \sqrt{x^5+1}\ dx$$ $$\text{where }\ a^5-b^5 = 1$$

I tried substituting the whole root sign in the respective integrals but that led to nowhere. I don't see how trigonometric substitution could be used, dummy variables or the DI method. I am really at a loss here.

Any ideas?

I added a picture of all the problems. • Hint: the functions are one the inverse of the other; using this, try drawing a graph and remember the interpretation of an integral as the area under a curve. Dec 29, 2019 at 15:42
• Do you have a pdf or something with this Latvian olympiad by any chance? Dec 31, 2019 at 12:08
• I don't have a pdf. Best I can do is provide a picture of the problems Dec 31, 2019 at 12:10
• I added a photo of the problems. Jan 1, 2020 at 17:55

Lemma. Let $$[c,d]\subset\mathbb R$$ be an interval and $$f:[c,d]\to f([c,d])$$ be a bijective function with integrable derivative $$f'$$. Let $$f^{-1}$$ be the inverse of $$f$$. Then $$\int_c^d f(x)\,\mathrm dx+\int_{f(c)}^{f(d)} f^{-1}(x)\,\mathrm dx=df(d)-cf(c).$$
Proof. Since $$f$$ is continuous, $$f([c,d])$$ also is an interval. Hence by substitution and integration by parts, $$\int_{f(c)}^{f(d)} f^{-1}(x)\,\mathrm dx=\int_c^d x f'(x)\,\mathrm dx=\big[xf(x)\big]_{c}^d-\int_c^d f(x)\,\mathrm dx=df(d)-cf(c)-\int_c^d f(x)\,\mathrm dx.$$
This completes the proof. $$\square$$
In our particular case, we have $$f:[1,a]\to[0,b]$$ with $$f(x)=\sqrt{x^5-1}$$ which indeed suffices all conditions of the lemma.
So $$\int_1^a \sqrt{x^5-1}\,\mathrm dx \ +\ \int_0^b \sqrt{x^5+1}\,\mathrm dx=af(a)-f(1)=a\sqrt{a^5-1}-0=ab.$$
$$a^5-b^5=1\Rightarrow a=\sqrt{b^5+1}$$ $$\text{let } \sqrt{x^5-1}=t\Rightarrow x=\sqrt{t^5+1}\Rightarrow dx=(\sqrt{t^5+1})'dt$$ $$\Rightarrow \color{blue}{\int_1^{\sqrt{b^5+1}}\sqrt{x^5-1}\,dx}=\int_0^b t (\sqrt{t^5+1})'dt\overset{IBP}=t \sqrt{t^5+1}\bigg|_0^b-\color{red}{{\int_0^b \sqrt{t^5+1}\,dt}}$$ $$\overset{\color{red}{t=x}}\Rightarrow \color{blue}{\int_1^a\sqrt{x^5-1}\,dx}+\color{red}{\int_0^b \sqrt{x^5+1}\,dx}=b\sqrt{b^5+1}=ab$$