1. ## two intregal problems

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....

Thanks!!!

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2. Here's a hint

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

3. 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?

4. 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.

5. 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.

6. 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.

7. 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

$\chi$ $\sigma$