Proving Integral A & Finding Antiderivative f(x)

[omni]

how i Prove A?

and how i find the antiderivative of f(x) ?

thanks.

 

[betel]

Are you sure you wrote the complete equation? Because in A for sure 0.2*2^8 is not smaller than 2.

Well you could explicitly find the anti derivative if you expanded the bracket. This will give you a polynomial with powers of x and Sqrt[x] which can be integrated.

But in this question you are supposed to find lower and upper estimates for the integrand to approximate it by something you can easily integrate.

I guess the correct equation should be <2^8 on the right hand side.

 

[omni]

well look again is not 0.2*2^8 smaller than 2

is0.2*2^8 smaller than 2^8

and this is correct .

thanks.

 

[Char. Limit]

well look again is not 0.2*2^8 smaller than 2

is0.2*2^8 smaller than 2^8

and this is correct .

thanks.

No, it definitely says 2. Perhaps you should look again?

 

[omni]

:) YES truth my wrong but the correct A is this:


i sorry about my hasty . ;)

 

[Char. Limit]

Well, the tedious way you could do it is by expanding and then evaluating the integral. That's not fun, though.

What's the value of the integral of 2^8 from 0 to 1? Can you prove that this MUST be greater than the integral of [itex]\left(1+\sqrt{x}\right)^8[/tex] from 0 to 1?[/itex]

 

[omni]

i will try to do something.

i got Hint that say's who more big 1 or sqrt x
and also i not must Calculate the integral to prove A truth?

 

[Char. Limit]

Well, since (1+sqrt(x))^8 < 2^8 at all points between 0 and 1, it stands to reason that $$\int_0^1 \left(1+\sqrt{x}\right)^8 dx < \int_0^1 2^8 dx = 2^8$$ as well. Can you see why?

 

[betel]

If you would calculate the integral there would be no need for the inequalities.
For the upper bound follow Char.
For the lower bound you could think what is bigger,x or sqrt x in the area under consideration.

 

[omni]

ok i think i Understand it thanks to all of you.