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# Need help to solve an integral problem

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

Note by Idham Muqoddas
4 years, 2 months 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, $$\displaystyle \int_{0}^{1}\left[\sqrt[4]{1-x^7} - \sqrt[7]{1-x^4}\right]\,dx = \displaystyle\int_{0}^{1}\sqrt[4]{1-x^7}\,dx - \displaystyle\int_{0}^{1}\sqrt[7]{1-x^4}\,dx = 0$$.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

what a great solution. Thanks!

- 4 years, 2 months ago

That's a nice one Jimmy, thanks for sharing!

- 4 years, 2 months ago

Nyc... this is an elegant solution..short and elegant.

- 4 years, 2 months ago

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

- 4 years, 2 months ago

* 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??*

- 4 years, 2 months ago

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.

- 4 years, 2 months ago