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I tried to solve this problem as $$0\lt x \lt 1 \implies 0\lt x^{2018} \lt 1 \implies 0\lt (1-x^{2018}) \lt 1$$

this means

$$\int_0^1 \left[\left(1-x^{2018}\right)^{1\over 2020}- \left(1-x^{2020}\right)^{1\over 2018} \right] dx \lt \int_0^1 \left[\left(1-x^{2018}\right)^{1\over 2020}\right] dx \lt \int_0^1 1^{{1\over 2020}} dx$$

as inequality can be integrated.

So I have come as far as proving the given expression is less than 1 but I cannot proceed further.

Can someone show me how to proceed?

(Please try give answers which can be understood by those in elementary calculus courses)

Hrishabh Nayal
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    This question is very similar. Found using Approach0. – Toby Mak Apr 05 '20 at 03:49
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    For $x,y\ge 0$ we have that $y=\sqrt[2020]{1-x^{2018}}$ is equivalent to $y^{2020}+x^{2018}=1$ or $x=\sqrt[2018]{1-y^{2020}}$. Therefore, the two functions are inverse to each other and since they both map $[0,1]$ to $[0,1]$, the two areas must be the same. Hence, $$\int_0^1 \left(1-x^{2018}\right)^{1\over 2020}=\int_0^1 \left(1-x^{2020}\right)^{1\over 2018}$$. – Axion004 Apr 05 '20 at 04:06

1 Answers1

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You are thinking too hard.

Consider the family of functions $$f_{a,b} : [0,1] \to [0,1]$$ of the form $$f_{a,b}(x) = (1 - x^a)^{1/b}$$ where $a, b$ are positive even integers. Then note that the inverse function of $f_{a,b}$ is $f_{b,a}$. Since both functions are nonnegative and monotone decreasing, it immediately follows that $$\int_{x=0}^1 f_{a,b}(x) - f_{b,a}(x) \, dx = 0$$ for all such positive even integers $a, b$. In fact the identity holds for any positive reals $a, b$ but the positive even case is easy to see and is applicable to your case.

heropup
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  • Thanks! Just one thing how can we say that inverse functions have same area in same limit? – Hrishabh Nayal Apr 05 '20 at 04:17
  • @HrishabhNayal Think about it. If the domain and codomain are the same, the inverse function is simply the reflection about the line $y = x$; therefore, the area under the curve is unchanged. – heropup Apr 05 '20 at 04:19