two intregal problems

**soyeahiknow** · July 14th 2010, 04:10 PM

Hi all,

can someone solve these two problems for me?

The first one is to proof is that is true or not

and the second one, I just totally forgot how to do the anti derivative of that....

$two intregal problems-math.jpg$

Thanks!!!

[IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]
[IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]

[IMG]file:///C:/DOCUME%7E1/UR_PUB%7E1.UR/LOCALS%7E1/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]

**pickslides** · July 14th 2010, 04:38 PM

Here's a hint

$\int \frac{1}{t^2+1}~dt = \arctan (t) +C$

**Jskid** · July 14th 2010, 05:42 PM

With respect to the first question: I think the first step is to integrate both functions
for $\int_0^1 \sqrt{1+x^2}$ sinces it's of the form $\sqrt{a^2+x^2}$ let $x=\arctan\theta$
and for $int_0^1 \sqrt{1+x}$ use u-substitution with $u=1+x$

I'm not sure but integration may note be nescecary...because the one with $x^2$ is obviously always going to be greater than or equal to the other.

Does this help?

**TheCoffeeMachine** · July 14th 2010, 06:49 PM

Originally Posted by Jskid

I'm not sure but integration may note be nescecary...because the one with $x^2$ is obviously always going to be greater than or equal to the other.

But after you integrate that will change.

**mohammadfawaz** · July 14th 2010, 09:06 PM

Originally Posted by J
I'm not sure but integration may note be nescecary...because the one with [tex

x^2[/tex] obviously always going to be greater than or equal to the other.

Does this help?

yeah...clearly integration is not needed. Simply notice that x is between 0 and 1 and hence $x^2\leq x$ . This directly leads to the given inequality since everything else is the same in both sides.

**Ackbeet** · July 15th 2010, 04:54 AM

jskid and TheCoffeeMachine are both incorrect on some points; mohammadfawaz is correct. On the interval from 0 to 1, $x^{2}\le x.$ That's true regardless of whether you integrate or not.

**chisigma** · July 15th 2010, 05:33 AM

The first question doesn't require any integration tool because for $0 \le x \le 1$ is $\sqrt{1+x^{2}} \le \sqrt{1+x}$ ...

Kind regards

$\chi$ $\sigma$

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