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

Note by Idham Muqoddas
4 years ago

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Consider 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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@Jimmy Kariznov I love this solution! Still, it left me wondering if there is more standard method (perhaps in multivariable calculus) that formalizes this? Sebastian Garrido · 4 years ago

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@Sebastian Garrido I'm not sure what you mean by formalize, but if you want to use multivariable calculus:

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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@Jimmy Kariznov what a great solution. Thanks! Idham Muqoddas · 4 years ago

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@Jimmy Kariznov That's a nice one Jimmy, thanks for sharing! Pranav Arora · 4 years ago

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@Jimmy Kariznov Nyc... this is an elegant solution..short and elegant. Krishna Jha · 4 years ago

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@Jimmy Kariznov 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 for the same period is equal as they will be in mirror mode also Sherif Hesham · 4 years ago

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* 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 ago

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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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