Need help to solve an integral problem

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

#Calculus #MathProblem #Math

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Comments

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 - 7 years, 10 months ago

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

Sebastian Garrido - 7 years, 10 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.

Jimmy Kariznov - 7 years, 10 months ago

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

Krishna Jha - 7 years, 10 months ago

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

Pranav Arora - 7 years, 10 months ago

what a great solution. Thanks!

idham muqoddas - 7 years, 10 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

sherif hesham - 7 years, 10 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??*

sherif hesham - 7 years, 10 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.

Vikas Chaudhary - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$