How to Prove the Area Between Two Curves with Integration Algebra?

[bayan]

Howdy felles.

I have a question which I find rather confusing, and that is to prove that the area between two curves is given by A=\frac{n-1}{n+1} where n is the power of X.


The equation is A=\int X^\frac{1}{n} dx - \int X^n dx

Could someone please show me how to get this to equal to A=\frac{n-1}{n+1}

I had a very diffrent answer. Plz help

Cheers

 

[hypermorphism]

Are there limits on those integrals ?

 

[bayan]

The limits are from 0 to 1

 

[lurflurf]

First do the integrals
=\int_0^1 x^\frac{1}{n} dx
and
=\int_0^1 x^n dx
Then simplify and get a common denominator.
What did you get for the integrals?

 

[bayan]

I am preety sure mine was wrong as I ended up with a diffrent answer. Can some one please check where I made mistake>

A=\int X^\frac {1}{n} dx - \int X^n dx
A=\frac {[X^\frac {1}{n}+1]}{\frac {1}{n}+1} - \frac {[X^n+1}{n+1]} Cant seem to make any progress from here onwards. All I do is make a massive mess of it and I have to do it again without any good signs.

Would really appreciate if anyone could help me.

 

[bayan]

I am preety sure mine was wrong as I ended up with a diffrent answer. Can some one please check where I made mistake>

A=\int X^\frac {1}{n} dx - \int X^n dx
A=\frac {X^\frac {1}{n+1}}{\frac {1}{n}+1} - \frac {X^n+1}{n+1} Cant seem to make any progress from here onwards. All I do is make a massive mess of it and I have to do it again without any good signs.

Would really appreciate if anyone could help me.

 

[bayan]

Also the limits of all these integrations are 0 to 1

 

[VietDao29]

Okay, you should remember that:
\int^{\beta}_{\alpha} f(x)dx = F(\beta) - F(\alpha)
So I'll do for you the first one. And you can follow it to do the second:
\int^{1}_{0} x ^ {\frac{1}{n}}dx = \left[ \frac{x ^ {\frac{1 + n}{n}}}{\frac{1 + n}{n}} \right] ^{1}_{0} = \frac{1 ^ {\frac{n + 1}{n}} - 0 ^ {\frac{n + 1}{n}}}{\frac{1 + n}{n}} = \frac{n}{1 + n}
So can you do the second integral?
Viet Dao,

 

[bayan]

I am still having abit trouble .

I am sposed to have A=\frac {n}{1+n} - \int^{o}_{1} X^n dx

which I ended up getting \int^{1}_{0} X^n dx = \frac {1^n^+^1 - 0^n^+^1}{n+1}
does it =\frac {n}{n+1}

 

[VietDao29]

Nope. Your integral is 'nearly' correct. The last part is wrong.
Remember : 1 ^ m = 1, m \in \mathbb{R}
0 ^ m = 0, m \in \mathbb{R} - \{0\}
So 1 ^ {n + 1} = 1
Viet Dao,