# Three integrals I can't solve

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.

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

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

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

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

+, 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.

jbunniii
Homework Helper
Gold Member
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.

Dick
Homework Helper
+, 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.

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

I did.

Dick
Homework Helper
I did.

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

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.

http://www.quickmath.com/