Question

Prove that the definite integral [sqrt(x^4+1)]dx , 1 13 years ago

Answer 1

\[\int\limits_{1}^{3}\sqrt{x^{4}+1} dx \ge \frac{ 26 }{ 3 }\]

Answer 2

First, try a few things.

\(\int\limits_{1}^{3}x^{4}+ 1\;dx = \dfrac{252}{5}\) Okay, that's REALLY large.

\(\int\limits_{1}^{3}\sqrt{2}\;dx = 2\sqrt{2}\) Okay, that's prety small.

\(\int\limits_{1}^{3}\dfrac{\sqrt{82} - \sqrt{2}}{2}(x-1) + \sqrt{2}\;dx = \sqrt{2}+\sqrt{82}\) That's a little large.

I think we might need to break it up into convenient chunks.

Answer 3

like a cake!

Answer 4

Well, maybe, but don't try circular wedges unless you are particularly hardy.

Answer 5

all I know is i've tried to apply numerous properties but have gotten nowhere

Answer 6

Did you try left-hand rectangles at x = 1 and x = 2?

Answer 7

unfortunately, the problem calls for usage of only properties, not useful methods like Riemann sums

Answer 8

Fair enough. Kind of hair-splitting, but let's see where it goes.

Please name the "Properties" we get to use.

Answer 9

essentially these:
http://www.emathhelp.net/notes?nid=481

Answer 10

Okay, two things come immediately to mind:

1) The "Adjacency of Integral" property essentially defines the fundamentals of finite Reimann Sums, so I still like my first answer.

2) Comparison Property 5, with Example #3 is almost exactly your original problem. Why not try that?

Answer 11

Comparison property 5 was my first thought as well, and I couldn't get an answer large enough to prove it's >= 26/3. And using the adjacency property would require me to break it up into LOTS of parts because the real value of the integral is 8.98, only 0.31 large than 26/3

Answer 12

Okay, then how about Comparison Property #3

\(\sqrt{x^{4} + 1} > x^{2}\)

\(\lim\limits_{x\rightarrow\infty}\sqrt{x^{4}+1} = x^{2}\)

The key to this solution was the '3' in the denominator. It seemed suspicious or arbitrary.