6.8 Improper Integrals Flashcards
what does convergent mean
finite answer, exists
what does divergent mean
infinite answer, kinda does not exist
imagine the graphs of
x
1/x
x^2
square root of x
a^x
ln x
sin x
cos x
tan x
hehe
how to rewrite the improper integral from a to infinity of f of x
lim as t approaches infinity of the integral from a to t of f of x
how to rewrite the integral from negative infinity to b of f of x
lim as t approaches negative infinity of the integral from t to b of f of x
how to rewrite the integral from negative infinity to positive infinity
integral from negative infinity to a plus the integral from a to infinity
then you can rewrite them with the lim but just one since if the first one is convergent or divergent, the second one is the same
Simple steps to solve improper integrals with infinite bounds
- rewrite with the limit
- find antiderivative and evaluate at the bounds
- pass the limit
- figure out its behavior
- answer or divergent
what are the two types of improper integrals and their differences
over infinite bounds and finite bounds
both are improper just for different reasons
infinite - bounds are infinite
finite - vertical asymptote
rewrite an improper integral from a to b with a vertical asymptote c
lim as t approaches vertical asymptote (c) from left of integral from a to t
plus
lim as s approaches vertical asymptote(c) from the right of integral from s to b
simple steps for solving a finite improper integral - meaning bounds are finite but there is something weird aka a vertical asymptote at a bound
- add the limit based on where the vertical asymptote is AND coming from the right direction
- find antiderivative and evaluate at the bounds
- pass the lim (CAREFUL WITH THE DIRECTION OF THE LIMIT)
- find how it behaves
- answer or divergent
when do you use p-integrals
when you cannot integrate
OR
they JUST ask you if it is convergent or divergent
what are the two p-integrals
integral from 1 to infinity of 1/ x^p
converges for p > 1
diverges for p <= 1
integral from 0 to 1 of 1/x^p
converges for p < 1
diverges for p >= 1
when to use the comparison tests
when we cannot integrate the integral we came up with
we just need to know if it and can find if it converges or diverges
what is the comparison test and simple steps
both positive function with f being less than g for all of x
if g converges, f converges
if f diverges, g diverges
- make sure bound 1 is greater than 1 or 1
- make sure f (the integrand) is greater than 0
- simplify f and g into an equality
- g is one we are making to compare to f (given)
-many ways you can do it
f >= or<= g
- start with f as the most complicated thing in integrand and then modify both side to try to fit og function
- 4 > 3 so 1/4 < 1/3 - compare like test says
- in explanation
- the g you came up with diverges/converges bc p integral
- since inequality
- then given integral diverges/converges by comparison test
what is the limit comparison test and simple steps
f and g are positive continuous function on a to infinity
if lim as x approaches infinity of f / g = a number L greater than 0
then f and g both converge or diverge
- f is the integrand
- find g
- remove things that are small or not doing anything
- -g should be convergent - divide f by g
- pass lim
it - IF you get a number use the test
- if g diverges or converges, so does f
–in explanation
- got a number L between 0 and infinity then integral from 1 to 0 of f and g both converge or diverge by limit comparison test
- we know g diverges/converges bc p integral
- conclusion f also diverges/converges
