Convergence of Integral Using Limit and Direct Comparison Tests | Homework Help

[whatlifeforme]

Homework Statement

Use direct comparison test or limit comparison test to determine if the integral converges.

Homework Equations

[itex]\displaystyle\int_0^6 {\frac{dx}{9-x^2}}[/itex]

The Attempt at a Solution

If i were to use the limit comparison test, would these integrals fit the criteria.
** if the positive functions f and g are continues on [a,∞)

Note: what does it mean by positive functions?

limit[itex]_{x->infinity} \frac{f(x)}{g(x)} = L [/itex] 0 < L < ∞

then [itex]\displaystyle\int_a^∞ {f(x) dx}[/itex] and

[itex]\displaystyle\int_a^∞ {g(x) dx}[/itex]

both converge or diverge.

f(x) = [itex]\displaystyle\int_0^6 {\frac{dx}{9-x^2}}[/itex]

g(x) = [itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex]

since these functions are not continuous at a=0, then is the limit comparison test not an option here?
If not, how would i go about choosing a function for the direct comparison test?

 

[Dick]

No need to hyperbold on the font. Your integral already has a problem way before infinity. Where is it?

 

[whatlifeforme]

No need to hyperbold on the font. Your integral already has a problem way before infinity. Where is it?

i was stating the theorem from the book. it states must be continuous from [a,∞), where the a=0 in the case, and (sorry forgot to include the comparing function).

[itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex] is not continuous at a=0.

f(x) = [itex]\displaystyle\int_0^6 {\frac{dx}{9-x^2}}[/itex]

g(x) = [itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex]further, i don't think it is stating the integral must be to infinity, just that it must be continuous on, or am i incorrect?

how should i solve with direct comparison test, if easier. please help. test in 2 days.

 

[Dick]

i was stating the theorem from the book. it states must be continuous from [a,∞), where the a=0 in the case, and (sorry forgot to include the comparing function).

[itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex] is not continuous at a=0.

f(x) = [itex]\displaystyle\int_0^6 {\frac{dx}{9-x^2}}[/itex]

g(x) = [itex]\displaystyle\int_0^6 {\frac{1}{x^2} dx}[/itex]further, i don't think it is stating the integral must be to infinity, just that it must be continuous on, or am i incorrect?

how should i solve with direct comparison test, if easier. please help. test in 2 days.

Do direct comparison. The problem with your integral happens at x=3, doesn't it? It's not continuous there. Is 1/(9-x^2) integrable on the interval [0,3]? Try to think of a comparison function on just that interval. Factor 9-x^2.

 

[whatlifeforme]

is (3+x) or (3-x) < (3+x)(3-x) meaning [itex]\frac{1}{3+x} or \frac{1}{3-x} > \frac{1}{9-x^2}[/itex] ??

thus, since

[itex]\displaystyle\int_0^6 {\frac{1}{3+x} dx} [/itex] converges to a finite number.

and [itex]\frac{1}{3+x}> \frac{1}{9-x^2} [/itex]

[itex]\displaystyle\int_0^6 {\frac{1}{9-x^2} dx} [/itex] it would conclude that this also converges?

By Direct Comparison Test.

 

[Dick]

is (3+x) or (3-x) < (3+x)(3-x) meaning [itex]\frac{1}{3+x} or \frac{1}{3-x} > \frac{1}{9-x^2}[/itex] ??

thus, since

[itex]\displaystyle\int_0^6 {\frac{1}{3+x} dx} [/itex] converges to a finite number.

and [itex]\frac{1}{3+x}> \frac{1}{9-x^2} [/itex]

[itex]\displaystyle\int_0^6 {\frac{1}{9-x^2} dx} [/itex] it would conclude that this also converges?

By Direct Comparison Test.

The comparison you want is something like ##\frac{1}{9-x^2} > C \frac{1}{3-x}## on the interval 0<x<3 for some constant C. Because ##\frac{1}{3-x}## diverges on that interval allowing you to conclude the original integral diverges. Can you find such a constant C?

 

[whatlifeforme]

The comparison you want is something like ##\frac{1}{9-x^2} > C \frac{1}{3-x}## on the interval 0<x<3 for some constant C. Because ##\frac{1}{3-x}## diverges on that interval allowing you to conclude the original integral diverges. Can you find such a constant C?

i'm sorry. i am still confused. is what i stated above incorrect? the integral is from 0 to 6 not 0 to 3.

what if i choose 9 for c? i just chose that at random.

 

[Dick]

i'm sorry. i am still confused. is what i stated above incorrect? the integral is from 0 to 6 not 0 to 3.

what if i choose 9 for c? i just chose that at random.

Yes, what you stated before is incorrect. I know your integral is 0 to 6. If you can show it diverges on 0 to 3 then it will diverge on 0 to 6. Why don't you try making a nonrandom choice of C? Use the factorization and figure out what range of values 1/(3+x) takes on [0,3].

 

[whatlifeforme]

Yes, what you stated before is incorrect. I know your integral is 0 to 6. If you can show it diverges on 0 to 3 then it will diverge on 0 to 6. Why don't you try making a nonrandom choice of C? Use the factorization and figure out what range of values 1/(3+x) takes on [0,3].

are we working with 1/(3-x) or 1/(3+x) or does it matter because we have switched up a few times?


if we are using 1/(3+x), do i need to integrate on [0,3] to find the range of values?

 

[iRaid]

Why don't you just split this integral into two parts and solve for both using limits from the right and left.
[tex]\int_0^6 \frac{dx}{9-x^{2}}=\lim_{t \to 3^{-}}\int_0^t \frac{dx}{9-x^{2}}+\lim_{t \to 3^{+}}\int_t^6 \frac{dx}{9-x^{2}}[/tex]
See what I'm saying?

 

[whatlifeforme]

if it were, 1/(9+x^2) i would choose 1/x^2 and if it converged i would know the other converged.the -x^2 is throwing me off in 1/(9-x^2).

(-1/16) < (1/16) when we get into negatives, correct, i know this sounds elementary, but in this application when we are determining smaller and larger functions are negatives always smaller than positives?

 

[whatlifeforme]

Why don't you just split this integral into two parts and solve for both using limits from the right and left.
[tex]\int_0^6 \frac{dx}{9-x^{2}}=\lim_{t \to 3^{-}}\int_0^t \frac{dx}{9-x^{2}}+\lim_{t \to 3^{+}}\int_t^6 \frac{dx}{9-x^{2}}[/tex]
See what I'm saying?

i want to know how to solve using a limit comparison test or direct comparison test, not by evaluating.

 

[Dick]

if it were, 1/(9+x^2) i would choose 1/x^2 and if it converged i would know the other converged.


the -x^2 is throwing me off in 1/(9-x^2).

(-1/16) < (1/16) when we get into negatives, correct, i know this sounds elementary, but in this application when we are determining smaller and larger functions are negatives always smaller than positives?

Negatives are always smaller than positives, but when you doing a comparison test you usually want to compare things of the same sign. Just think about it this way. 1/(9-x^2)=(1/(3+x))(1/(3-x)). If you replace the 1/(3+x) with its minimum on [0,3], then you can also replace the '=' sign with '>='.

 

[iRaid]

i want to know how to solve using a limit comparison test or direct comparison test, not by evaluating.

Oh my bad sorry.

 

[whatlifeforme]

Negatives are always smaller than positives, but when you doing a comparison test you usually want to compare things of the same sign. Just think about it this way. 1/(9-x^2)=(1/(3+x))(1/(3-x)). If you replace the 1/(3+x) with its minimum on [0,3], then you can also replace the '=' sign with '>='.

what do you mean by the minimum on [0,3] ? 1/6 would be the minimum, or do you mean integrate on [0,3] ?

 

[Dick]

what do you mean by the minimum on [0,3] ? 1/6 would be the minimum, or do you mean integrate on [0,3] ?

Yes, I mean 1/6. So 1/(9-x^2)>=(1/6)(1/(3-x)). Does the integral of 1/(3-x) converge on [0,3)?

 

[whatlifeforme]

[itex]\displaystyle\int_0^3 {\frac{1}{3-x} dx}[/itex]

[itex]\lim_{b \to 3^-} ln|3-x|]^{0}_{b}[/itex]

[itex]\lim_{b \to 3^-} ln|3-b| - ln|3-0| --> ln|3-3| = ln|0| = ∞ [/itex]

Thus, the smaller diverges, so the original function diverges.

 

[Dick]

[itex]\displaystyle\int_0^3 {\frac{1}{3-x} dx}[/itex]

[itex]\lim_{b \to 3^-} ln|3-x|]^{0}_{b}[/itex]

[itex]\lim_{b \to 3^-} ln|3-b| - ln|3-0| --> ln|3-3| = ln|0| = ∞ [/itex]

Thus, the smaller diverges, so the original function diverges.

Right!

 

[whatlifeforme]

so explain one last time how you got: 1/(9-x^2)>=(1/6)(1/(3-x))

 

[Dick]

so explain one last time how you got: 1/(9-x^2)>=(1/6)(1/(3-x))

It's all in post 13. I factored 1/(9-x^2) and then replaced one of the factors by something smaller.

 

[whatlifeforme]

so when the limit comparison test says positive functions, what does it mean?

 

[Dick]

so when the limit comparison test says positive functions, what does it mean?

I think it means what it says. You can work around that condition like when they are both negative, or when they don't violate that condition 'too much'. So if they are both positive or both negative, then you don't have to think about it much. Just apply the test. If they aren't, then you have to think about it harder.