Comparison theorem of $int_2^4fracxsqrtx-1sqrtx-2dx$Does $int_1^inftysin(xlog x)dx $ converge?Does $int_1^2fracdxsqrt(2-x)ln(x)$ converge?How to prove that $int_0^inftyfrace^-nxsqrtx}mathrm dx$ existsConvergence of $int_0^infty {fracsin(x)(x+4)sqrtx^3(x+1)^2$.Determine convergence of the improper integral $int_-infty^inftyfrac1sqrtx^10+2 dx$$int_0^inftyfracsin^2(x)x,dx$ how to analyze it?Comparison TheoremUse the Comparison TestLimit Comparison Test for integralcomparison theorem of integral from negative infinity to positive infinity.
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Comparison theorem of $int_2^4fracxsqrtx-1sqrtx-2dx$
Does $int_1^inftysin(xlog x)dx $ converge?Does $int_1^2fracdxsqrt(2-x)ln(x)$ converge?How to prove that $int_0^inftyfrace^-nxsqrtxmathrm dx$ existsConvergence of $int_0^infty fracsin(x)(x+4)sqrtx^3(x+1)^2$.Determine convergence of the improper integral $int_-infty^inftyfrac1sqrtx^10+2 dx$$int_0^inftyfracsin^2(x)x,dx$ how to analyze it?Comparison TheoremUse the Comparison TestLimit Comparison Test for integralcomparison theorem of integral from negative infinity to positive infinity.
$begingroup$
Determine if $$int_2^4fracxsqrtx-1sqrtx-2dx$$ converges or diverges.
When I attempted this question I thought it diverged since it has a VA at x=2.
By bounding it below and got $$frac1sqrtxsqrtx-2$$
However it turns out this converges. I am trying to prove this using comparison theorem. Could I get some help on how to bound this from above?
Thanks.
calculus improper-integrals
asked Mar 16 at 23:10
GeraltGeralt
9417
$endgroup$
add a comment |
$begingroup$
Determine if $$int_2^4fracxsqrtx-1sqrtx-2dx$$ converges or diverges.
When I attempted this question I thought it diverged since it has a VA at x=2.
By bounding it below and got $$frac1sqrtxsqrtx-2$$
However it turns out this converges. I am trying to prove this using comparison theorem. Could I get some help on how to bound this from above?
Thanks.
calculus improper-integrals
asked Mar 16 at 23:10
GeraltGeralt
9417
$endgroup$
add a comment |
$begingroup$
Determine if $$int_2^4fracxsqrtx-1sqrtx-2dx$$ converges or diverges.
When I attempted this question I thought it diverged since it has a VA at x=2.
By bounding it below and got $$frac1sqrtxsqrtx-2$$
However it turns out this converges. I am trying to prove this using comparison theorem. Could I get some help on how to bound this from above?
Thanks.
calculus improper-integrals
asked Mar 16 at 23:10
GeraltGeralt
9417
$endgroup$
Determine if $$int_2^4fracxsqrtx-1sqrtx-2dx$$ converges or diverges.
When I attempted this question I thought it diverged since it has a VA at x=2.
By bounding it below and got $$frac1sqrtxsqrtx-2$$
However it turns out this converges. I am trying to prove this using comparison theorem. Could I get some help on how to bound this from above?
Thanks.
calculus improper-integrals
calculus improper-integrals
asked Mar 16 at 23:10
GeraltGeralt
9417
asked Mar 16 at 23:10
GeraltGeralt
9417
asked Mar 16 at 23:10
GeraltGeralt
9417
asked Mar 16 at 23:10
GeraltGeralt
9417
asked Mar 16 at 23:10
GeraltGeralt
9417
9417
add a comment |
add a comment |
1 Answer
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$begingroup$
$frac x sqrt x-1sqrt x-2 leq frac 4 sqrt 1 sqrt x-2$ and $int_2^4 frac 1 sqrt x-2dx=2(x-2)^1/2|_2^4=2sqrt 2$. Hence the integral converges.
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
How would you initially tell that it converged?
$endgroup$
– Geralt
Mar 16 at 23:22
$begingroup$
@Geralt A well known simple fact is $int_0^a frac 1 sqrt xdx$ is finite. So we expect $int_2^a frac 1 sqrt x-2dx$ also to be finite (by the change of variable $y=x-2$).
$endgroup$
– Kavi Rama Murthy
Mar 16 at 23:33
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$frac x sqrt x-1sqrt x-2 leq frac 4 sqrt 1 sqrt x-2$ and $int_2^4 frac 1 sqrt x-2dx=2(x-2)^1/2|_2^4=2sqrt 2$. Hence the integral converges.
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
How would you initially tell that it converged?
$endgroup$
– Geralt
Mar 16 at 23:22
$begingroup$
@Geralt A well known simple fact is $int_0^a frac 1 sqrt xdx$ is finite. So we expect $int_2^a frac 1 sqrt x-2dx$ also to be finite (by the change of variable $y=x-2$).
$endgroup$
– Kavi Rama Murthy
Mar 16 at 23:33
add a comment |
$begingroup$
$frac x sqrt x-1sqrt x-2 leq frac 4 sqrt 1 sqrt x-2$ and $int_2^4 frac 1 sqrt x-2dx=2(x-2)^1/2|_2^4=2sqrt 2$. Hence the integral converges.
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$begingroup$
How would you initially tell that it converged?
$endgroup$
– Geralt
Mar 16 at 23:22
$begingroup$
@Geralt A well known simple fact is $int_0^a frac 1 sqrt xdx$ is finite. So we expect $int_2^a frac 1 sqrt x-2dx$ also to be finite (by the change of variable $y=x-2$).
$endgroup$
– Kavi Rama Murthy
Mar 16 at 23:33
add a comment |
$begingroup$
$frac x sqrt x-1sqrt x-2 leq frac 4 sqrt 1 sqrt x-2$ and $int_2^4 frac 1 sqrt x-2dx=2(x-2)^1/2|_2^4=2sqrt 2$. Hence the integral converges.
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
$frac x sqrt x-1sqrt x-2 leq frac 4 sqrt 1 sqrt x-2$ and $int_2^4 frac 1 sqrt x-2dx=2(x-2)^1/2|_2^4=2sqrt 2$. Hence the integral converges.
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Mar 16 at 23:16
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
69.9k53170
$begingroup$
How would you initially tell that it converged?
$endgroup$
– Geralt
Mar 16 at 23:22
$begingroup$
@Geralt A well known simple fact is $int_0^a frac 1 sqrt xdx$ is finite. So we expect $int_2^a frac 1 sqrt x-2dx$ also to be finite (by the change of variable $y=x-2$).
$endgroup$
– Kavi Rama Murthy
Mar 16 at 23:33
add a comment |
$begingroup$
How would you initially tell that it converged?
$endgroup$
– Geralt
Mar 16 at 23:22
$begingroup$
@Geralt A well known simple fact is $int_0^a frac 1 sqrt xdx$ is finite. So we expect $int_2^a frac 1 sqrt x-2dx$ also to be finite (by the change of variable $y=x-2$).
$endgroup$
– Kavi Rama Murthy
Mar 16 at 23:33
$begingroup$
How would you initially tell that it converged?
$endgroup$
– Geralt
Mar 16 at 23:22
$begingroup$
How would you initially tell that it converged?
$endgroup$
– Geralt
Mar 16 at 23:22
$begingroup$
@Geralt A well known simple fact is $int_0^a frac 1 sqrt xdx$ is finite. So we expect $int_2^a frac 1 sqrt x-2dx$ also to be finite (by the change of variable $y=x-2$).
$endgroup$
– Kavi Rama Murthy
Mar 16 at 23:33
$begingroup$
@Geralt A well known simple fact is $int_0^a frac 1 sqrt xdx$ is finite. So we expect $int_2^a frac 1 sqrt x-2dx$ also to be finite (by the change of variable $y=x-2$).
$endgroup$
– Kavi Rama Murthy
Mar 16 at 23:33
add a comment |
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