Question:
Calculus: Comparison Test?
manwitnowords
2010-11-11 11:39:05 UTC
For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. For example, one possible answer is BF, and another one is L.

1. integral e^-x/x^2 from 1 to infinity
2. integral 1/(x^2+3) from 1 to infinity
3. integralx/(sqrt(x^6+3)) from 1 to infinity
4. integral 5+sin(x)/(sqrt(x-0.2) from 1 to infinity
5. integral cos^2(x)/(x^2+3) from 1 to infinity



A. The integral is convergent
B. The integral is divergent
C. by comparison to . integral of 1/(x^2-3) from 1 to infinity
D. by comparison to . integral of 1/(x^2+3) from 1 to infinity
E. by comparison to . integral of cos^2x/(x^2) from 1 to infinity
F. by comparison to . integral of e^x/(x^2) from 1 to infinity
G. by comparison to . integral of -e^-x/(2x) from 1 to infinity
H. by comparison to . integral of 1/(sqrt(x)) from 1 to infinity
I. by comparison to . integral of 1/(sqrt(x^5)) from 1 to infinity
J. by comparison to . integral of 1/(x^2) from 1 to infinity
K. by comparison to . integral of 1/(x^3) from 1 to infinity
L. The comparison test does not apply
Three answers:
Mathmom
2010-11-11 12:30:33 UTC
1. ∫{1 to ∞} e⁻ˣ/x² dx



For x ≥ 1, e⁻ˣ < 1

Therefore e⁻ˣ/x² < 1/x²

∫{1 to ∞} e⁻ˣ/x² dx < ∫{1 to ∞} 1/x² dx which converges



A J



---------------------



2. ∫{1 to ∞} 1/(x²+3) dx



x²+3 > x²

1/(x²+3) < 1/x²

∫{1 to ∞} 1/(x²+3) dx < ∫{1 to ∞} 1/x² dx which converges



A J



---------------------



3. ∫{1 to ∞} x/√(x⁶+3) dx



3/x² > 0

x⁴ + 3/x² > x⁴

√(x⁴+3/x²) > x²

x√(x⁴+3/x²)/x > x²

√x²√(x⁴+3/x²)/x > x²

√(x⁶+3)/x > x²

x/√(x⁶+3) < 1/x²



∫{1 to ∞} x/√(x⁶+3) dx < ∫{1 to ∞} 1/x² dx which converges



A J



---------------------



4. ∫{1 to ∞} (5+sin(x))/√(x-0.2) dx



sin(x) ≥ -1

5 + sin(x) ≥ 4 > 1



x-0.2 < x

√(x-0.2) < √x

1/√(x-0.2) > 1/√x



(5+sin(x))/√(x-0.2) > 1/√x



∫{1 to ∞} (5+sin(x))/√(x-0.2) dx > ∫{1 to ∞} 1/√x dx which diverges



B H



---------------------



5. ∫{1 to ∞} cos²x/(x²+3) dx



x² + 3 > x²

1/(x²+3) < 1/x²

cos²x/(x²+3) < cos²x/x²



∫{1 to ∞} cos²x/(x²+3) dx < ∫{1 to ∞} cos²x/x² dx which converges



A E
2016-12-05 00:18:19 UTC
right this is one answer; i'm specific there's a extra elementary answer than this: by making use of Stirling's formula, lim(n-->infinity) n! / [(n/e)^n * (2 pi n)^(a million/2)] = a million. So, comparing sum n!/n^n with sum{[(n/e)^n * (2 pi n)^(a million/2)] / n^n} by making use of shrink assessment attempt yields: lim(n-->infinity) {n!/n^n} / {[(n/e)^n * (2 pi n)^(a million/2)] / n^n} = lim(n-->infinity) n! / [(n/e)^n * (2 pi n)^(a million/2)] = a million. for this reason, we'd desire to envision no remember if sum{[(n/e)^n * (2 pi n)^(a million/2)] / n^n} converges. notice that [(n/e)^n * (2 pi n)^(a million/2)] / n^n = (2 pi n)^(a million/2) / e^n < sqrt(2 pi) * n/e^n, because sqrt(n) <= n for n >= a million besides the incontrovertible fact that, sum n / e^n is conventional to converge (an elementary shrink assessment with sum a million/e^n, it particularly is a convergent geometric sequence, will without postpone coach this). for this reason by making use of the assessment attempt, we see that sum{[(n/e)^n * (2 pi n)^(a million/2)] / n^n} converges. for this reason, by making use of the shrink assessment attempt, the unique sequence converges. i wish that became obtainable! ------------------ replace: notice to Tabula Rasa: that's what i became finding for!
Katrina
2016-08-07 13:59:18 UTC
That's an interesting question


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