Integral with square root

• Jul 19th 2010, 11:27 AM
highc1157
Integral with square root
Hi there,

I am new to the forums and would appreciate a little help on this problem :D

I have a homework problem that states the following:

You are not allowed to use a calculator. Just use logic. Show all work.

Using the properties of integrals prove that : ∫ √((x^4)+1)dx >= 26/3
Definite integral from 1 to 3 (3 being upper limit, 1 being lower)
• Jul 19th 2010, 11:49 AM
Ackbeet
Try this:

$\displaystyle \int_{1}^{3}\sqrt{x^{4}+1}\,dx\ge\int_{1}^{3}\sqrt {x^{4}}\,dx=...$
• Jul 19th 2010, 12:10 PM
highc1157
well what I did was make a U-Substitution

u = (x^4) + 1 ----> dx = du/4(x^3)

giving me (1/4x^3) ∫ u^(1/2) du

but from there i am lost. Using my ti-83 i found that the definite integral from 1 to 3 √((x^4)+1)dx = 8.98.... , which is greater than 26/3

I just dont know how I am messing up, after integrating this I get: (u^(3/2)) / (6(x^3))

but when i do F(b) - F(a) I am not getting the answer 8.98.. and do not know how to do this without my calculator :(
• Jul 19th 2010, 12:29 PM
Ackbeet
I wouldn't recommend trying to compute the exact value of the integral. It involves elliptic functions.

Here's the beauty of inequalities: you have loads of freedom. You can throw away stuff left and right, simplify, and yet keep the inequality. If you follow the train of reasoning I gave in Post #2, what can you do with the integral to the right of my inequality?
• Jul 19th 2010, 12:39 PM
Also sprach Zarathustra
f(x)=sqrt(x^4+1) is monotonic increasing on [1,3] so, sqrt(x^4+1)>sqrt(x^4)=x^2, hence:

int sqrt(x^4+1) dx> int sqrt(x^4)dx=int x^2 dx =x^3/3 +C

[x^3/3]{1-->3} = 9-1/3=26/3