# Help with Integration algebra!

1. Aug 8, 2005

### 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 plz show me how to get this to equal to $$A=\frac{n-1}{n+1}$$

Cheers

2. Aug 8, 2005

### hypermorphism

Are there limits on those integrals ?

3. Aug 9, 2005

### bayan

The limits are from 0 to 1

4. Aug 9, 2005

### 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?

Last edited: Aug 9, 2005
5. Aug 9, 2005

### bayan

I am preety sure mine was wrong as I ended up with a diffrent answer. Can some one plz 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 appriciate if anyone could help me.

6. Aug 9, 2005

### bayan

I am preety sure mine was wrong as I ended up with a diffrent answer. Can some one plz 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 appriciate if anyone could help me.

7. Aug 9, 2005

### bayan

Also the limits of all these integrations are 0 to 1

8. Aug 9, 2005

### 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,

9. Aug 9, 2005

### 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 $$=\frac {n}{n+1}$$

10. Aug 9, 2005

### 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,