\(\int\limits_{0}^{1}{\sqrt[4]{1-{{x}^{7}}}-\sqrt[7]{1-{{x}^{4}}}}\text{ }dx=...\)

\(\int\limits_{0}^{1}{\sqrt[4]{1-{{x}^{7}}}-\sqrt[7]{1-{{x}^{4}}}}\text{ }dx=...\)

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TopNewestConsider the area in the first quadrant bounded by \(x^7+y^4 = 1\) with \(x,y \ge 0\) and the coordinate axes.

Since the curve is given by \(y = \sqrt[4]{1-x^7}\), \(x \in [0,1]\), the area is \(\displaystyle\int_{0}^{1}\sqrt[4]{1-x^7}\,dx\).

Since the curve is given by \(x = \sqrt[7]{1-y^4}\), \(y \in [0,1]\), the area is \(\displaystyle\int_{0}^{1}\sqrt[7]{1-y^4}\,dy\).

But both methods must yield the same area. Therefore: \(\displaystyle\int_{0}^{1}\sqrt[4]{1-x^7}\,dx = \displaystyle\int_{0}^{1}\sqrt[7]{1-y^4}\,dy\).

Thus, \(\displaystyle \int_{0}^{1}\left[\sqrt[4]{1-x^7} - \sqrt[7]{1-x^4}\right]\,dx = \displaystyle\int_{0}^{1}\sqrt[4]{1-x^7}\,dx - \displaystyle\int_{0}^{1}\sqrt[7]{1-x^4}\,dx = 0\). – Jimmy Kariznov · 4 years ago

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– Sebastian Garrido · 4 years ago

I love this solution! Still, it left me wondering if there is more standard method (perhaps in multivariable calculus) that formalizes this?Log in to reply

Let \(\Omega = \{(x,y) | 0 \le x \le 1 , 0 \le y \le \sqrt[4]{1-x^7}\} = \{(x,y) | 0 \le y \le 1 , 0 \le x \le \sqrt[7]{1-y^4}\}\).

Then, by Theorem III in this Wikipedia article, we have:

\(\displaystyle\iint\limits_{\Omega} \,dA = \int_{0}^{1} \int_{0}^{\sqrt[4]{1-x^7}} \,dy\,dx = \int_{0}^{1}\sqrt[4]{1-x^7} \,dx\)

\(\displaystyle\iint\limits_{\Omega} \,dA = \int_{0}^{1} \int_{0}^{\sqrt[7]{1-y^4}} \,dx\,dy = \int_{0}^{1}\sqrt[7]{1-y^4} \,dy\)

It's not hard to see that the conditions of that theorem are satisfied. The conclusion follows. – Jimmy Kariznov · 4 years ago

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– Idham Muqoddas · 4 years ago

what a great solution. Thanks!Log in to reply

– Pranav Arora · 4 years ago

That's a nice one Jimmy, thanks for sharing!Log in to reply

– Krishna Jha · 4 years ago

Nyc... this is an elegant solution..short and elegant.Log in to reply

– Sherif Hesham · 4 years agoplease do you mean it is a function and it's inverse the function and it's mirror by the line of angle 45 so the bounded areas for the same integral for the same period is equal as they will be in mirror mode alsoLog in to reply

* please do you mean it is a function and it's inverse the function and it's mirror by the line of angle 45 so the bounded areas for the same integral is equal as they will be in mirror mode also??*– Sherif Hesham · 4 years agoLog in to reply

Since the curve is given by y=1−x7−−−−−√4, x∈[0,1], the area is ∫101−x7−−−−−√4dx.

Since the curve is given by x=1−y4−−−−−√7, y∈[0,1], the area is ∫101−y4−−−−−√7dy.

But both methods must yield the same area. Therefore: ∫101−x7−−−−−√4dx=∫101−y4−−−−−√7dy.

Thus, ∫10[1−x7−−−−−√4−1−x4−−−−−√7]dx=∫101−x7−−−−−√4dx−∫101−x4−−−−−√7dx=0. – Vikas Chaudhary · 4 years ago

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