Solving a Weird Integral Homework Problem: Tips and Attempted Solutions

[Selfluminous]

Homework Statement

evaluate

Homework Equations

The Attempt at a Solution

i don't think there is an elementary function as anti derivative for this integral

i tried taylor expansion, doesn't seem to work.

Can anyone give me a hint ?

 

[SammyS]

Homework Statement

evaluate

Homework Equations

The Attempt at a Solution

i don't think there is an elementary function as anti derivative for this integral

i tried taylor expansion, doesn't seem to work.

Can anyone give me a hint ?

WolframAlpha shows the indefinite integral to include the Hypergeometric function ₂F₁. http://www.wolframalpha.com/input/?i=Integrate+%281-x^5%29^%281%2F7%29-%281-x^7%29^%281%2F5%29dx

It shows the definite integral to be zero.

 

[DivisionByZro]

Although I really don't think there's an easy way to solve this one, it's neat to notice the weird symmetry of the powers on x and of the roots (1/5 and 1/7 compared with 7 and 5 for powers). Notice that you can factor both (1-x^5) and (1-x^7). It's kind of taunting when you do factor them and realize they're very similar. Try playing around with this.

 

[Curious3141]

There's a neat way to do this. Start by observing that:

f(x) = (1-x^5)^{\frac{1}{7}} and g(x) = (1-x^7)^{\frac{1}{5}}

are in fact inverse functions. This means that they are symmetrical when reflected in the line y = x.

Over the domain [0,1], they both have the range [0,1].

Can you now draw a conclusion about the areas under each curve from 0 to 1?

This general method applies to a lot of functions of that form. Nothing special about the 5 or 7 here.

 

[DivisionByZro]

There's a neat way to do this. Start by observing that:

f(x) = (1-x^5)^{\frac{1}{7}} and g(x) = (1-x^7)^{\frac{1}{5}}

are in fact inverse functions. This means that they are symmetrical when reflected in the line y = x.

Over the domain [0,1], they both have the range [0,1].

Can you now draw a conclusion about the areas under each curve from 0 to 1?

Bah, I was getting there! Nice find Curious.

 

[Selfluminous]

There's a neat way to do this. Start by observing that:

f(x) = (1-x^5)^{\frac{1}{7}} and g(x) = (1-x^7)^{\frac{1}{5}}

are in fact inverse functions. This means that they are symmetrical when reflected in the line y = x.

Over the domain [0,1], they both have the range [0,1].

Can you now draw a conclusion about the areas under each curve from 0 to 1?

This general method applies to a lot of functions of that form. Nothing special about the 5 or 7 here.

Wow thank you and everyone.
mathematics is awesome
\m/

 

[SammyS]

There's a neat way to do this. Start by observing that:

f(x) = (1-x^5)^{\frac{1}{7}} and g(x) = (1-x^7)^{\frac{1}{5}}

are in fact inverse functions. This means that they are symmetrical when reflected in the line y = x.

Over the domain [0,1], they both have the range [0,1].

Can you now draw a conclusion about the areas under each curve from 0 to 1?

This general method applies to a lot of functions of that form. Nothing special about the 5 or 7 here.

Yes, this is very interesting!

So, under what conditions is \displaystyle \int_0^1 \left(f(x)-f^{-1}(x)\right)\,dx=0\,?

Of course the domains & ranges have to all be [0,1].

As it so happens, map 1 → 0 , and 0 → 1 . Is that important ? Yes it is.

In the case where f(0) = 0 = f ^-1(0) and f(1) = 1 = f ^-1(1), we find that \displaystyle \int_0^1 \left(f(x)+f^{-1}(x)\right)\,dx=1 instead. Notice that the functions are added here.

 

[Curious3141]

Yes, this is very interesting!

So, under what conditions is \displaystyle \int_0^1 \left(f(x)-f^{-1}(x)\right)\,dx=0\,?

Of course the domains & ranges have to all be [0,1].

As it so happens, map 1 → 0 , and 0 → 1 . Is that important ? Yes it is.

In the case where f(0) = 0 = f ^-1(0) and f(1) = 1 = f ^-1(1), we find that \displaystyle \int_0^1 \left(f(x)+f^{-1}(x)\right)\,dx=1 instead. Notice that the functions are added here.

Yes, it depends on whether the functions are increasing or decreasing over the interval.

 

[clamtrox]

You can also follow the hints that you were given: first substitute, say in the second integral, t=(1-x^7)^{1/5}. You'll get something like \int dt t f(t). Then do partial integration with this. Since the integration limits are 0 and 1 in both cases, the remaining integrals should cancel and you're left with a boundary term which is easy to evaluate.