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I'm having a difficult time dealing with improper integrals. I understand the main concept to a certain extent, until I reach the part where I have to test the integrals for convergence and/or divergence. The Direct Comparison tests and the Limit comparison tests perplex me. Can some math whiz briefly explain them to me? Much appreciated.

2007-04-12 04:28:29 · 3 answers · asked by Odyssey 1 in Science & Mathematics Mathematics

3 answers

I assume you mean improper fractions. I am not sure what you want. I will give a shot.

3 2/5 is a mixed number

convert a mixed number to a improper fraction

The proper approach to solve from a mixed number to a improper fraction is as follows

Multiply 5 times 3 = 15 and add 2

5 x 3 = 15

15 + 2 = 17

17/5

17/5 is the improper fraction

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adding imopropoer fractions

3 2/5 + 2 5/7 =

17/5 + 19/7 =

The common denominator is 35

Divide 5 into 35 = 7. Then multiply 7 times 17 equals 119

35 / 5 = 7

17 x 7 = 119

use the same procedure for 19/7

The fraction becomes

119/35 + 95/35 =

add the numerator. the denominator remains the same

119/35 + 95/35 =

214/35

- - - - - - - s-

2007-04-12 04:46:28 · answer #1 · answered by SAMUEL D 7 · 0 1

Improper integrals use the concept of limits to evaluate integrals. Examples of improper integrals are:

Integral (1 to infinity, (1/x) dx )

We would change this into the form of a limit;
lim Integral (1 to t, (1/x) dx )
t -> infinity

Another less-obvious example would be this one.

Integral (-3 to 5, 1/(x^2) dx )

The function inside this integral is not defined at 0, so this too must become a limit. In fact, it would be the sum of limits.

lim (-3 to t, 1/(x^2) dx) + lim (t to 5, 1/(x^2) dx )
t -> 0- . . . . . . . . . . . . . . . . t -> 0+

If the limit exists, the improper integral converges. Otherwise, it diverges.

As for the comparison tests and limit comparison tests, that is a broad broad topic that is better read from a textbook than here.

2007-04-12 20:14:52 · answer #2 · answered by Puggy 7 · 1 0

i think of it converges to 0 the required is comparable to -a million/(x-2) and as x procedures infinity, the denominator gets extra beneficial and larger, inflicting the entire ingredient to get closer and closer to 0. a million/infinity constantly equals 0. why did you narrow up it up at x=4?

2016-12-09 01:06:15 · answer #3 · answered by Anonymous · 0 0

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