# Three integrals I can't solve

1. Mar 13, 2010

### Dustinsfl

j=$$\int\sqrt{1-x^{4}}$$

k=$$\int\sqrt{1+x^{4}}$$

l=$$\int\sqrt{1-x^{8}}$$

I am trying to figure out the order for example j<k<l. I don't know how to integrate any of these.

2. Mar 13, 2010

### Dustinsfl

I forgot to mention 0 to 1 are the bounds of all 3.

3. Mar 13, 2010

### Dick

Don't even try to integrate them. Can't you order the functions you are integrating on [0,1]?

4. Mar 13, 2010

### Dustinsfl

I am trying to determine the order but I don't know how to do that without solving them.

5. Mar 13, 2010

### Dick

Which is largest, sqrt(1+x^4), sqrt(1-x^4) or sqrt(1-x^8)?

6. Mar 13, 2010

### Dustinsfl

+, but for the x to the 8th and 4th it depends on if 0<x<1 or if x is outside that range. If x is between 0-1, the order would be +, 8th power, 4th. If not in that range, +, 4th, and 8th.

7. Mar 13, 2010

### jbunniii

I don't think these are integrable in terms of elementary functions.

But if you just want to sort them from lowest to highest, that shouldn't be too hard.

For example, compare the integrands of j and k:

$$\sqrt{1-x^4}$$

and

$$\sqrt{1+x^4}$$

Clearly the first one is $\leq$ the second one for all $x \in [0,1]$, and the inequality is strict for $x \in (0, 1]$, so that implies $j < k$.

Comparing the integrand for L shouldn't be too much harder - give it a try and let us know if you get stuck.

8. Mar 13, 2010

### Dick

Didn't you say the range of integration is 0<=x<=1?

9. Mar 13, 2010

### Dustinsfl

I did.

10. Mar 13, 2010

### Dick

Hence, why are you worried about values outside that range?

11. Mar 14, 2010

### yusukered07

http://www.quickmath.com/