Personally the best way to practice integrating in my opinion is all-variable expressions such as ....do this as you normally would it gives you a great concept of how to work these...the above statement is worked out below
To evaluate this integral the preferred method is Integration by parts(discussed earlier in the thread)
so to start we define our individual parts
therefore using the appropriate formula we get
...
now to evaluate the second integral we utilize the above technique to rewerite the second integral as ...
which with some constant manipulation
so combinging we get
This post is for those ubiquitous acceleration/velocity/etc. problems
Ok we will start with the basics
Next up is this...
Q.where does a particle change direction?
A.Max or mins
Q.when is a particle at rest?
A. when
Q.How do I find Max's and Min's for S(t)?
A. Same method as you would for f(x)..utilizing the first derivative test or second derivative test(see earlier thread for more on those)
Q.What is the total distance traveled by a particle from a to b?
A. total distance traveled from a to b by a particle is
Q.What does it mean when it says "Find the intervals when the particle is moving to the right on the x-axis"
A. This is asking you to find increasing intervals which are the intervals for which every element of that interval ...to find these intervals use the first derivative test...but instead of finding max or min stop after evaluating points in the first derivative and just look for the intervals that ...when asking "...moving to the left on the x-axis" this applies to decreasing intervals which is the same concept as increasing except
Q.What exactly is speed?
A.Speed is the absolute value of velocity? |v(t)|
Q.How do I find whether or not speed is increasing at a point c?
A. There are two methods to solving this:
Physics Method...speed is increasing IFF a(c) and v(c) share the same sign then speed is increasing
Math method:Easiest way to do this is graphically. Graph your point (c,V(c))..once this is done start plotting points based on your a(t)(slope of v(t)) if the line is approaching the x-axis it is decreasing...if it is going away from the x-axis it is increasing
That is it for now
The previous post brings to light an interesting topic...now I am sure you can do ...but can you do ?
Here is how you evaluate
first you set
next set up test intervals based on the values that make and test an element of each interval in
if on the interval then on that interval
and if on the iterval then on that interval
based on these splits break the integral apart and evaluate each change in sign of seperately
Example:
...you didnt excpect me to come up with another example did you?
so first we solve ..obviously getting ...if you can't get that you probably shouldnt be reading quick tip calculus notes
ok so we set up those test intervals based on that number that makes the inside of the absolute value 0
so we could have ...nah...why include points not between the limits of integration? Don't
so we have
now we need to test an element of each interval in the function inside of the absolute value sign...so we select -2.5 for the one interval and 0 for the other...so and
thererfore on
and on
ok so now what do we do with this info?
we break up the integral!
We use the fact that assuming and f(x) is continous on (a,c)
so we break up th integral to look like this
now applying the knowledge that
and we can rewrite the integral to look like
then by standard integration techniques we get
Remember if you have ANY Calc I or II(Ab or BC also) questions feel free to ask!! I will help with the general questions on how to do something instead of the specific problem questions you usually ask! all are welcome
This time I will talk about partial fraction decomposition if conjuction with integrating...say you had the expression ...well I am sure that not many beginners would look at that and go "Hey thats obviously !"
The way we go about this is the typical way we do partial fractions...if you don't know how to do partial fractions you might be able to use this but it will be kind of hard...I might come back later and do a segment on just how to do partial fractions...but to the task at hand!
we want to decompose
so we set up our partial fractions template for repeated quadratic factors...so we have
now multiplying through by we get
expanding and grouping we get
now by equating coefficents we get and
so
then
so now we know that
therefore
now by manipulating constants we get
now it is obvious that because we have the derivative of the quantity the answer is
Another general case integral so that you can see how they are done more in depth
First step make an "inverse substitution"
Let ...
Therefore
So our integral is rewritten as
now by obvious cancellation we have
Now we multiply by
this is by the stipulation of the integral
so we have
and by constant manipulation we have
Then by using standard integration techniques we get
Now by back substituting
So we have
Multipling by
we get
Next we will discuss trigonomtric substitution
First off the basics
When you see the follwoing substitute with the following
Let
When you have
Let
And finally when we have
Let
An optional approach is to memorize the following formulas
But this is painful...instead try to just apply the substitutions and figure it out algebraically
To cement how to do these...I will do...who would have guessed it another general case integral
First identifying it is
now we must find dx
Now by substiuting we get
Now by simplifyig and cancelling we get
Now by making the useful substitution
we get
Now all we need to do is substitute
therefore we have
and now by using reference triangles we get
and by distributing we get
Next thing I will touch on is how to do trigonometric integrals...not trigonometric substitution but integrating trig functions
Let's start with the basics
Now that that is over with lets get on to some of the most useful trig indentities
These are the most of the useful ones for INTEGRATION...all trig identites are useful but these are the best for integrating
Next I will describe how to take the integral of things such as
Since it is an integral with sine to an odd power make a substitution
Then since the derivative of the quantity is wiht the second term we can just integrate to get
This works for all odd powers of sine or cosine(if it is cosine make the sub )
Next we will do
Uh-oh the exponent is even...this means that we cant use the recently discussed method...so we must use our indentities
Uh-oh...looks like we need to make another trig substitution
....so we have
To do odd powers of tan(x) or sec(x) make a similar move as with the sine and cosines
and for even powers of tan(x) or sec(x) use the pythagorean indentitys with tan(x) and sec(x) as listed in the indentities...hope this has been helpful!
It sure took a long time to write!!
Next I will start series....for this I will assume you know sequences and all the terminology I am about to use...if not just post a question in this thread!
In this thread I will provide you with a detailed explanation of each convergence test
The first one and most overlooked is the n-th term test which states
this test only proves divergence!!! If the series limit is 0 it could still be divergent
Examples...
Next example ..therefore the n-th term test tells us nothing
The second test is Geometric series.....This is a special case where you have
This series converges only if
Examples since series converges
Another example ...which is divergent becaue
The next test is the integral test which states that if is postive,continous, and decreasing then the series converges only if diverges...where
Examples ...you can establish that it is decreasing and psotive....so the test does apply
so
..which converges since therefore the series converges
To see an example of a divergent consider
The next test is a P-series...similar in nature to the Geometric series in that it is a specific case...a P-series given by
is divergent iff ...to validate this try using the integral test and consider values for p that are less than one
...convergent
...divergent
The next test is the Direct comparison test...this test states that if
If converges then converges
and if diverges then diverges
For example consider ....since we can condlude that converges since converges by p-series
Another Example would be ...selecting as our comparison series we see that therefore since is a divergent "harmonic series" is divergent
Next thread will include...Limit comparison test, Alternating series test, Ratio test, and Root test!!
As promised I will continue with the rest of the convergence tests
The next one that was in line was the limit comparison test that states that if
then both and converge or diverge
Examples
We will use a convergent P-series as our comparison series...the best way to determine what series to use is as what does the series closely resemble?
obviously as n goes to infinity the -1 is inconsequential leaving you with
so taking
therefore since it converges to value greater than one but less than ∞ they share convergence/divergence....and since our comparison series is converget this means our series in question is also convergent
Another example would be
Now since as we see as this resembles a convergent geometric series
so
Once again you use the fact that the limit is one to deduce that both series converget
Next up is the Alternating Series Test which states that in ...if
and then the series converges
Examples:
Determine the convergence of the series
We must test two things first off is ?
The answer is yes...obviously because since the larger n is the larger is..and since the ln is on the bottom we use the fact that to conclude that .
Next we must test to see if which it obviously does
Thererfore is conditionally convergent or that is at least what the test tells us! It is actually absolutely convergent but this test does not conclude that!
Next test which in my opinion is the best is the Ratio Test which states that if the series converges...if the series is absolutely convergent
If the limit is greater than one the series is divergent
if the limit is equal to one it is inconclusive
Examples
Ok here is how you would do it since ...so since thsi is fraction instead of dividing we multiply the reciporcal
to get
Therefore the series diverges
An easier method would have been simplifying making the original series ...and using the n-th term test we see the series diverges
The last thing to be discussed in this thread would be the Root test that states if the series converges if the limit is greater than one the series diverges..if the limit equals one the test is inconclusive
Example
So we apply our test to get therefore teh series diverges
Thats it for this post....next I will do some derivative integral stuff...but I will get back and do taylor series and maclaurin series...especially their application to integration
Remember any questions just ask!
Thanks is appreciated!