When I tried to approximate $$\int_{0}^{1} (1-x^7)^{1/5}-(1-x^5)^{1/7}\ dx$$ I kept getting answers that were really close to $0$, so I think it might be true. But why? When I ask Mathematica, I get a bunch of symbols I don't understand!
Why is $\int\limits_0^1 (1-x^7)^{1/5} - (1-x^5)^{1/7} dx=0$?
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$\begingroup$
calculus
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1The Wolfram Integrator computes only *indefinite* integrals, as far as I can see, so you were getting the primitive for your integrand, which is understandably obscure (Those ${}_2F_1$ functions that show up there are hypergeometric functions (see http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/) which are a simply amazing family of functions) To compute *definite* integrals like the one you want, you can use Alpha: for example, `http://www.wolframalpha.com/input/?i=integrate+%281-x^7%29^{1%2F5}+-+%281-x^5%29^{1%2F7}+from+0+to+1` – 2010-07-29
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1(The last URL got butchered; you'll have to copy and paste, I guess) – 2010-07-29
2 Answers
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Note that if
$$ y = \left(1 - x^7\right)^{1/5} $$
then
$$ \left(1 - y^5\right)^{1/7} = x $$
This means $(1-x^7)^{1/5}$ is the inverse function of $(1-x^5)^{1/7}$. In the graph, one will be the same as the other when reflected along the diagonal line y = x.
Also, both functions
- share the same range [0, 1] and domain [0, 1] and
- monotonically decreasing,
Therefore, the area under the graph in [0, 1] will be the same for both functions:
$$ \int_0^1 \left(1-x^7\right)^{1/5} dx = \int_0^1 \left(1-y^5\right)^{1/7} dy $$
Grouping the two integrals yield the equation in the title.
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2The point that the ranges are (0,1] and the domains are [0,1) tripped me up at first - I thought x^2 and x^(1/2) would form a counterexample. I would have stated it as: "Both these functions have value 1 at x=0 and value 0 at x=1", that way the geometry becomes clearer, IMHO. – 2010-07-29
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0https://en.m.wikipedia.org/wiki/Integral_of_inverse_functions – 2017-05-09
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$\int_0^1(1-x^m)^{(1/n)}dx=(m+n)\Gamma(1/m)\Gamma(1/n)/\Gamma(1/m+1/n)$ is symmetric in $m, n$.