Question

definite integration

Answer 1

\[\int\limits_{0}^{1}dx/\sqrt{1+x ^{4}}=I\] then
a) I>pi/4
b)I<pi/4
c)i=pi/4
d)none of these

Answer 2

\[\int\limits_{0}^{1} \frac{dx}{\sqrt{1+x^4}}\]

Answer 3

yes

Answer 4

well, wolfram used a gamma function; but i got no idea what that is :) the answer they gave is:


http://www4c.wolframalpha.com/Calculate/MSP/MSP41119fe0hefdegbcb8d000036e7b90hdib4d907?MSPStoreType=image/gif&s=5&w=188&h=49

Answer 5

its greater than pi/4 tho :)

Answer 6

\(\sqrt{1+x^4}>\sqrt{1+x^2}>\sqrt{1-x^2}\)
Then the integral is greater than
\(\int_0^1\sqrt{1-x^2}\,dx=\text{one quarter of a circle}=\pi/4\)
So \(I>\pi/4\).

Answer 7

Oh sorry, I miss read the problem :(

Answer 8

But the idea is the same.
Notice that \(0\leq (x-x^3)^2=x^2+x^6-2x^4\)
Hence \(x^2+x^6\geq 2x^4\geq x^4.\)
Now
\(\sqrt{1+x^4}\sqrt{1-x^2}=\sqrt{1+x^4-x^2-x^6}< 1\) for \(x>0\)
Hence
\(\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\)
Hence we can use the same argument as above to show that \(I>\pi/4\).

Answer 9

how did you get this watchmath\[0\leq (x-x^3)^2=x^2+x^6-2x^4\]

Answer 10

Fir every square is non-negative and the rest you just expand \((x-x^3)(x-x^3)\)

Answer 11

no thats alright. why (x-x^3)^2??

Answer 12

Because I want to show that \(x^2+x^6-x^4> 0\) and we can use \((x-x^3)^2\geq 0\) to prove that.

Answer 13

m sorry bt m rly not gettin it..
why do you wanna show x^2+x^6-x^4> 0

Answer 14

watchmath i need ur help again look at my comment i made on my question please

Answer 15

:) To read a solution you need to read it backward sometimes.
Obviously we want to show that \(\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\).
The inequality is equivalent to
\(\sqrt{1+x^4-x^2-x^6}<1\)
and I can show that if know that
\(x^4-x^2-x^6<0\)
and luckily I can prove the last inequality from \((x-x^3)^2\geq 0\).

Answer 16

ok i understand most of it. but can you tell me once what led you to show this: \[\frac{1}{\sqrt{1+x^4}}>\sqrt{1-x^2}\]

Answer 17

watchmath need ur help again sorry about this

Answer 18

Because I know (from my past experience) that the integral
\(\int_0^1\sqrt{1-x^2}\,dx\) represent a quarter of the area of a circle with radius 1.
Sketch the graph of \(y=\sqrt{1-x^2}\) and see what is the area under the graph from \(x=0\) to \(x=1\).

Answer 19

okay its pi/4...
but the doubt persists..:-(

Answer 20

what doubt?

Answer 21

if integrating \[\sqrt{1-x^2}\] gives pi/4
then how does integrating \[1/\sqrt{1+x ^{4}}\] give a value greater than pi/4

Answer 22

Well remember that if the graph of \(f\) is above the graph of \(g\) then the area under the curve of \(f\) is greater than the area under the graph of \(g\).
Now for \(f=1/\sqrt{1+x^4}\) and \(g=\sqrt{1-x^2}\) we have \(f>g\). So the integral (the area under the \(f\)) is bigger then the area under the graph of \(g\). But the area under the graph of \(g\) is \(\pi/4\).So \(\int_0^1 f>\pi/4\).

Answer 23

ok hold on.. if we know " that if the graph of f is above the graph of g then the area under the curve of f is greater than the area under the graph of g." then we dont need to calculate anything right?? that itself answers the question.

Answer 24

how will you answer the question if you per se dont know that the graph of f is greater than that of g

Answer 25

Well first we need to make a reasonable hypothesis. For example after observing the the graph of the above \(f\) and \(g\) (by sketching the two in calculator)we believe that \(f<g\) or \(f>g\), then we try to prove that rigorously.

When you see a theorem or a solution to a problem you won't see many untold story behind that proof/solution. It is possible before arrive to that solution, the author revise the hypothesis, try a lot of things that doesn't work before he can write a neat and beautiful solution.

In this problem for example we have three choices: the integral is less, equal or greater than pi/4. the beginning that we believe that the answer is pi/4. But after several attempts we might see that maybe this is not true. Then we try to show that it is less that pi/4 but maybe realize after many attempts that it is unlikely to be true. Then we try to prove it greatern that pi/4. And the above is one possible scenario how to come up with the prove for that.

Answer 26

hmm....i see what your point is.

Answer 27

any ideas on this question
http://openstudy.com/groups/mathematics/updates/4dd42d14d95c8b0b0f5156c4#/groups/mathematics/updates/4dd58012d95c8b0bcd095bc4