Ineq. involving integral: Use sqrt[3]+(x/4)<=sqrt[x+3]<=2 for 0 <=x<=1 to prove...

[e_glaud]

Ineq. involving integral: Use sqrt[3]+(x/4)<=sqrt[x+3]<=2 for 0 <=x<=1 to prove...

Use the fact that:

. . . . .\(\displaystyle \sqrt{\strut 3\,}\, +\, \dfrac{x}{4}\, \leq\, \sqrt{\strut x\, +\, 3\,}\, \leq\, 2\, \mbox{ for }\, 0\, \leq\, x\, \leq\, 1\)

to show that:

. . . . .\(\displaystyle \displaystyle \sqrt{\strut 3\,}\, +\,\dfrac{1}{20}\, \leq\, \int_0^1\, \sqrt{\strut x^4\, +\, 3\,}\, dx\, \leq\, 2\)

How do I solve this problem? Specifically, how do I prove the integral is greater than or equal to what is on it's left? I was told to give the top part the same integral, but I don't see how that will help. Thanks!

 

[stapel]

Use the fact that:

. . . . .\(\displaystyle \sqrt{\strut 3\,}\, +\, \dfrac{x}{4}\, \leq\, \sqrt{\strut x\, +\, 3\,}\, \leq\, 2\, \mbox{ for }\, 0\, \leq\, x\, \leq\, 1\)

to show that:

. . . . .\(\displaystyle \displaystyle \sqrt{\strut 3\,}\, +\,\dfrac{1}{20}\, \leq\, \int_0^1\, \sqrt{\strut x^4\, +\, 3\,}\, dx\, \leq\, 2\)

How do I solve this problem? Specifically, how do I prove the integral is greater than or equal to what is on it's left? I was told to give the top part the same integral, but I don't see how that will help. Thanks!

What did you get, when you applied integrals to the first inequality? I get:

. . . . .\(\displaystyle f(x)\, =\, \sqrt{\strut 3\,}\, +\, \dfrac{x}{4}\)

. . . . .\(\displaystyle g(x)\, =\, \sqrt{\strut x\, +\, 3\,}\)

. . . . .\(\displaystyle h(x)\, =\, 2\)

Since f(x) < g(x) < h(x) on the interval [0, 1], then so are their integrals. Thus:

. . . . .\(\displaystyle \displaystyle \int_0^1\, f(x)\, dx\, \leq\, \int_0^1\, g(x)\, dx\, \leq\, \int_0^1\, h(x)\, dx\)

. . . . .\(\displaystyle \left(\left(\sqrt{\strut 3\,}\right)\, x\, +\, \dfrac{x^2}{8}\right)\bigg|_0^1\, \leq\, \left(\dfrac{2}{3}\, \sqrt{\strut (x\,+\, 3)^3\,}\right)\bigg|_0^1\, \leq\, 2x \bigg|_0^1\)

. . . . .\(\displaystyle \sqrt{\strut 3\,}\, +\, \dfrac{1}{8}\, \leq\, \dfrac{2}{3}\, \left(\sqrt{\strut 4^3\,}\, -\, \sqrt{\strut 3^3\,}\right)\, \leq\, 2\)

. . . . .\(\displaystyle \sqrt{\strut 3\,}\, +\, \dfrac{1}{8}\, \leq\, \dfrac{2}{3}\,\left(8 -\, 3\, \sqrt{\strut 3\,}\right)\, \leq\, 2\)

Can you see any way to relate the proposed inequality to the above?

Please be complete. Thank you!

 

[Ishuda]

Use the fact that:

. . . . .\(\displaystyle \sqrt{\strut 3\,}\, +\, \dfrac{x}{4}\, \leq\, \sqrt{\strut x\, +\, 3\,}\, \leq\, 2\, \mbox{ for }\, 0\, \leq\, x\, \leq\, 1\)

to show that:

. . . . .\(\displaystyle \displaystyle \sqrt{\strut 3\,}\, +\,\dfrac{1}{20}\, \leq\, \int_0^1\, \sqrt{\strut x^4\, +\, 3\,}\, dx\, \leq\, 2\)

How do I solve this problem? Specifically, how do I prove the integral is greater than or equal to what is on it's left? I was told to give the top part the same integral, but I don't see how that will help. Thanks!

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

Hint: If \(\displaystyle 0\, \le\, x\, \le\, 1\) then \(\displaystyle 0\, \le\, x^4\, \le\, 1\)