A tricky integral

[Tolaso J Kos]

Evaluate \( \displaystyle \int_0^1[(1-x^7)^{1/5} - (1-x^5)^{1/7}]dx \).

[Papapetros Vaggelis]

We define \(\displaystyle{f:\left[0,1\right]\longrightarrow \mathbb{R}}\) by \(\displaystyle{f(x)=\sqrt[5]{1-x^7}}\) .

The function \(\displaystyle{f}\) is continuous and \(\displaystyle{1-1}\) at \(\displaystyle{\left[0,1\right]}\)

and differentiable at \(\displaystyle{\left[0,1\right)}\) .

Futhermore, \(\displaystyle{0<1\implies f(0)=1>f(1)=0}\) , so \(\displaystyle{f}\) is strictly decreasing at \(\displaystyle{\left[0,1\right]}\)

with \(\displaystyle{f\,\left(\left[0,1\right]\right)=\left[f(1),f(0)\right]=\left[0,1\right]}\) . Let \(\displaystyle{y\in f\,\left(\left[0,1\right]\right)}\) .

Then, there is \(\displaystyle{x\in\left[0,1\right]}\) such that \(\displaystyle{f(x)=y\iff 1-x^7=y^5\iff x^7=1-y^5\iff x=\sqrt[7]{1-y^5}}\) ,

so \(\displaystyle{f^{-1}:\left[0,1\right]\longrightarrow \left[0,1\right]\,,f^{-1}(x)=\sqrt[7]{1-x^5}}\) .

We have to evaluate the integral \(\displaystyle{A=\int_{0}^{1}\left(f(x)-f^{-1}(x)\right)\,\rm{dx}}\) .

Let \(\displaystyle{B=\int_{0}^{1}f^{-1}(x)\,\rm{dx}}\) . The function \(\displaystyle{f^{-1}}\) is continuous,

so, by substituting \(\displaystyle{x=f(t)}\) we get :

\(\displaystyle{B=\int_{1}^{0}t\,f^\prime(t)\,\rm{dt}=-\int_{0}^{1}x\,f^\prime(x)\,\rm{dx}}\) and then

\(\displaystyle{A=\int_{0}^{1}\left(f(x)-f^{-1}(x)\right)\,\rm{dx}=\int_{0}^{1}\left(f(x)+x\,f^\prime(x)\right)\,\rm{dx}=\left[x\,f(x)\right]_{0}^{1}=0}\) .