Solving 2 Tricky Math Problems: Bounded Area & Convergence of Sequence

[yeuVi]

There are two problems I got stuck...

1. Is the area in the first quadrant bounded between the x-axis and the curve

$$\ y= \frac{x} {2*(x^2+2)^{7/8}}$$

finite? This one, I used the Area formula... but then I cannot integrate it... and then how to determine if it's finite or not?

2. Does the sequence

$$\ an= \frac{cos(n)+1}{n^3}$$

converge?

How do you do this one? I need some hints to find the limit...

Thank a lot guys ^_^

 

[shmoe]

1) Try using the comparison test with a simpler function that you can integrate (or know about it's convergence).

2)Hint: -1<=cos(n)<=1

 

[dextercioby]

You can integrate this

$$\frac{1}{2}\int \frac{x}{\left(x^{2}+2\right)^{\frac{7}{8}}} \ dx$$

Make the substitution

$$x^{2}+2=u$$

Daniel.

 

[yeuVi]

But the first one, after I get the integrate, what should I do next? I mean, it's is finite if the integrate limit is converge?

The secon one, I do not really understand the hint... can u explain a little more detail? I guess you want me to apply the comparion limit test for a sequence? I though that test is only for series...

Thanx every1 for helping me... ^_^

 

[shmoe]

But the first one, after I get the integrate, what should I do next? I mean, it's is finite if the integrate limit is converge?

Yes, if the integral converges you'd say the area was finite. If the integral diverges then you'd say the area was "infinite". (note since the integeand is positive there can only be one kind of divergence, growing without bound).

The secon one, I do not really understand the hint... can u explain a little more detail? I guess you want me to apply the comparion limit test for a sequence? I though that test is only for series...

$$0\leq a_n \leq \frac{2}{n^3}$$. Think squeeze theorem.

 

[yeuVi]

Thanx a lot. I tried ur hints and got some results. Can u guys check if they correct?

The 1st one, I got the area is finite...

The 2nd one, the sequence is convergence, since

$$\ \lim{0} = \lim{ \frac{2}{n^3}} =0$$

so $$\ \lim{a_n}=0$$

therefore the sequence is converge to 0...

 

[shmoe]

The 1st one, I got the area is finite...

How about posting the details of your work?

The 2nd one, ... therefore the sequence is converge to 0...

Looks good.

 

[yeuVi]

The 1st one, I integrated it by substitution u= x^2+2

so the integration is

1/4 * 8* (u^(1/8))

It's converge to 0, so the area 's finite...

 

[Data]

The function is always positive when x>0, so how could the area be 0?

HINT: What bounds did you integrate over? Remember, this is an area, so you should be taking a definite integral!

 

[yeuVi]

Oh, it's not the area is 0, but the limit of the integral is 0...

 

[Data]

what limit of the integral?

 

[yeuVi]

Yes, if the integral converges you'd say the area was finite. If the integral diverges then you'd say the area was "infinite". (note since the integeand is positive there can only be one kind of divergence, growing without bound).

Oh, I was following his hint...

 

[Data]

The hint is fine, but I don't see how you found the integral to be convergent! Was the integral that you did

$$\int_0^\infty \frac{x}{2(x^2+2)^{\frac{7}{8}}} \ dx$$

? Do you see why this is the correct integral, if not?