Alright, just had my Calc II final today, and there were two questions I wasn't too sure of.
The first: ∫ ln(x)/(x^(1/2)) on the interval of 0 to 1 - does it converge? I'm fairly certain that l'hopitals' doesn't work here, as plugging in 0 would give negative infinity over 0. If it does converge, the definite integral should be around -4 if I did my math correctly.
The second problem: does the sequence (1 + 3/n)^2n converge? If so, what value does it converge to? I wasn't sure how to approach this one, though I figured that the number in the parenthesis is always going to be > 1, therefore raising it to an increasing power will give a number that doesn't converge...although as the sequence continues on it will head towards 1... heh I think I might have answered my own question there, I must've been thinking of series while doing the problem at first.
2007-12-13 03:22:19 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I was thinking about the questions, and the comparison rule works for the first question - ln(x)/x^(1/2) > 1/x^(1/2) - which diverges, therefore the original diverges. For the second question, I forgot that it's a sequence, not a series, and the sequence will definitely go to 1 after n > 3.
2007-12-13 04:08:12 · update #1
Dont forget all of your tests that you were shown. Sounds like the Comparison test would work on the second problem. Either the Ratio or Comparison test will work on the first. I am not absolutely certain.
2007-12-13 03:48:03 · answer #1 · answered by james w 5 · 1⤊ 0⤋
Looks more like Calc III, if there is such a thing. I couldn't begin to solve the integral.
I only know l'hopitals rule as applied to differentiation.
If you try to draw a graph of f(x) = ln(x)/x^(1/2) you see there is an undefined value for x = 0. Examining the graph tell us the integral is minus, if it exists. counting squares could allow us to get a guesstimate.
The y axis is an asymptote.
I cheated by integrating between 1 X 10^-8 and 1 using a calculator, and got -3.99 something.
2007-12-13 11:41:13 · answer #2 · answered by Sciman 6 · 0⤊ 1⤋
Substitute x=y^2, this converts the integral to
4*int(ln(y),y=0..1)
which then evaluates to -4.
Hint: use the fact that the indefinite integral of ln(y) is y*ln(y)-y+C,
then use L'Hopitals Rule to complete the solution.
2007-12-13 12:16:26 · answer #3 · answered by Anonymous · 0⤊ 0⤋