Integrals
Raw Transcript
This video is going to be on integrals, rewriting an integral, and also evaluating one. Lets get started – you have to press second alpha to put the letters of this code in – you just do this once and up comes the menu for you – and you push alpha and then eight and closed parenthesis, and then press enter, and up comes my menu of all the items – I know that what I want to go to is the letter p, they go from numbers here, but then they go p, so you can scroll down like this, all the way down – or if you know what your after you can go alpha and then p, and I’ll go to – I want to rewrite the integrand first – so I’m gonna – one of the most important parts – I tell you one of the most important parts in here – and so
you kinda go to this list here – and these are the tough ones – so you go to your list and like for instance let’s do number six here – press number six – n over x squared – you need to convert the denominator always to a numerator, and that in algebra becomes x to the minus two, n times x to the minus two, and then you integrate – you’re going to go integration of n to the x to the minus two – which you can do it now, and here you have – n to – notice you add one to the exponent up here and then all this you divide by that also – minus two plus one – that equals n to the minus 1 – minus one. So that equals minus n over x – and you can go to another one here, lets try number 5, here’s the cube root – well cube root of course is exponent of, to the one third, and to integrate x to the one third, you have to add one, well one in one third is three thirds, so you add that and divide, and put that in the denominator also, and you have x to the four thirds over four thirds, and when you divide by a fraction you invert and multiply, and so here’s the answer, three x to the four thirds, over four. Ahh pretty neat huh, well I wanna do, I’m going to go here and quit, lets scroll down here – and we can quit on either end of it – press enter – now if you wanna, I’m gonna go alpha and p again to get back to the program, I’m gonna – I want to evaluate, I’m gonna press number two, and evaluate – and for demonstration let’s just do uhm – you have to press alpha before you enter anything in this line here – alpha, and let’s go uhm – t times uhm – sometimes they use the variable t – times – let’s use e to the x here, e to the x of three times, t squared. And it shows you what you’ve entered here – I show you – and you can go ok or change it – I’m going to say it’s ok, so – and I don’t do it step by step particularly – uh – because this is u substitution – uhm – but anyways – ah – this is the exact answer. And so that will help you immensely – ya know – checking your integrals with any type of integral – quickly – and the step by step I have in other sections of my programs like for instance u substitution – so – fabulous programs, check em out at everystepcalculus dot com
Limits Calculus
Limits in Calculus on the TI-89: Raw Transcript
This is a video on limits as it applies to calculus, generally calculus one, and let’s
get started. You have to press second alpha to get to my menu. You can scroll down from the menu many things you want, all in alphabetical order, we’re going to go down to limits and press enter. Here’s the choices in the limit program you have with regards to limits. You can compute the limit, see the definition of a limit formula, complete a table of limits, and prove that a limit is equal to the computed limit. You can also find delta given epsilon. We’re going to do the first choice now finding delta given epsilon. You have to press alpha before you enter anything in the boxes that come up in my programs. We’re going to try two times x minus five equals one as x approaches three and the epsilon they give you, They will give you these things on the test problem they give you, so epsilon equals point zero one. So I show you what you’ve entered. The limit of two times x minus five equals one, as x approaches 3, while epsilon equals point zero one. The next screen you have the choice of changing it or saying that it is ok. I say it’s ok so I’ll press one for the choice. Here’s the first formula, absolute f of x minus L is less than epsilon. I took the one over here and added it to the minus five to get minus six equals zero. That’s an absolute value sign not a one, and that all is less than point zero one. You’ll keep writing this stuff down on your paper exactly
as it shows. This is the way it done on youtube videos, or at least what I could find, and
delta equals point zero zero fine. Now we can go back to choose main menu or new problem, we want to do a new problem so I’ll choose one again and then go to number two and compute the limit. You can put anything in you want within reason. Alpha again, anytime you are doing division you have to put parenthesis in — in any problem — so we want to make sure we get the parenthesis in there. Let’s go parentheses x minus two closed parentheses, divided by, parenthesis x squared minus x minus two, closed parenthesis. As x goes to alpha two. Here’s the problem you’ve entered and you want to compute the limit. You say it’s ok. Now at x equals two, f of x equals zero and therefore doesn’t exist, so you use the following to find the limit. You add point zero one to every x in the formula so you are a very little away from the limit given and then compute it. The limit is point three, I give you other ways of showing that, because some tests require other answer forms.
Let’s do another problem, let’s do a table of limits — number four — let’s clear
that out and do another one. Alpha closed parenthesis x minus closed parenthesis divided
by x squared minus four, as x approaches two, and here’s the table you’d mark down on
your paper. Let’s do another problem, this is always fun. Prove that the limit is L — the limit
— this is done in always calculus one and throws every student off. Youre panicked because
they are talking about epsilon and delta and the definition of it, which sounds quite complicated, and of course it is useless for the rest of calculus in your life. Let’s put a problem in here. Alpha three times x plus five equals thirty-five as x approaches ten, You have to press alpha again to enter the ten. and here’s what your problem is. Here’s the
formula again. F of x minus L is less than epsilon. You write this stuff down exactly
as shown on your paper. You factor it and then x minus ten is less than epsilon divided
by three. Epsilon becomes delta, and here’s the proof. If zero is less that absolute x
minus c — c is a constant — and is less that delta, then zero is less that absolute
x minus 10 is less than epsilon over three and therefore the original function is less
than epsilon. Put this on your paper and get one hundred percent on that problem. EveryStepCalculus.com check it out and check out the blogs also.