Every Step Calculus

Show Work Step by Step on your TI-89 Calculator Screen

You are here: Home / Archives for U Substitution

U Substitution, Completing the square, 1/√(2x-x^2)

by Leave a Comment

Transcript

Hello, Tom from everystepcalculus.com and everystepphysics.com, I’m going to show you how to do this integration problem that a student asked for help with Yahoo and so I’ll show you how my programs work on this. Index 8 to get to my menu I’m already at U substitution, I’ll show you why it’s used substitution in a little bit. We’re gonna enter the function you have to press alpha first before you enter things in these entry lines here in my program. So, I’m going to press alpha the problem is one / what put the left parenthesis in for the divisor and second and then we get the square root sign of 2 times X – X2 and then cap it off with the 2 parentheses. Press Enter and looks pretty good here we’re gonna rewrite it, bringing the X2 – 2X in the proper form because we’re going to complete the square. Something you’re probably an expert on but not me so I wanted to show you how to do that and we’re going to complete the square by taking the half of the middle term here squaring it. This is – 1, – 12 and anytime in mathematics when you add something to a formula or function you have to take away the same thing to make it easy. You can’t change things just for no reason so we subtract the – 12 here also. Okay, this sets up the identity the integral of one divided by the square root of a squared minus U2, DU is equal to the arc sine of U over A plus C. So, U is X – 2 and A = 1 here’s the U2 or use substitution area. So, pathetic calculus you know and so we add the U and the A and we come up with the answer here. Arc sine of X – 2 plus C isn’t this wonderful.

Now, notice that the any time you do the arc sine you’re finding the angle in trigonometry okay. So, now you’ve found the angle and you still have to add some arbitrary C there’s ten million answers to this problem unless somebody comes up with a letter C okay. So, again when you find these answers notice how nonsensical they are and how useless they are the answers which calculus does totally it suggested hundreds of little puzzles that professors have dreamed up to solve here and they solve nothing. They just come up with answers and we get so involved when we’re trying to pass tests that we get involved with this stuff and say oh that’s good. We just got the answer here and the answer is generally is complicated or more complicated than the original function but that’s calculus okay. Now, you can go to my site buy my programs for $40, best $40 you’ll ever spend and you already can study this stuff for hours like I have to find the steps to do it. Find the best way of doing these problems and you know you can buy my programs and pass calculus and cut down on your study time and everything else okay. So, think about that but have a good one okay.

U substitution, cos(ln(x))/x and cos(ln(x)), integration by parts

by Leave a Comment


Transcript
Hello! Tom from everystepcalculus.com and everystephysics.com. We are going to do a problem in calculus U substitution, and we are going to do a function with cosine in it. Index 8 to get to my menu. Press enter. I already had U substitution but you’ll scroll with this cursor here down or up or whatever to get to the [unclear 00:33] for the problem that you have at hand. So, U substitution is this problem and we are going to enter the function, we have to press alpha first before you enter anything in these entry lines here. So, we are going to press alpha, and then we want cosine, secant z here which is cosine, and then secant x which is log of x, close up the parentheses, and divided by x, is the function. So, it will show you what you have enter. You can change it if you want, else say it’s okay. Notice that the trick to use substitution is that whatever inside parenthesis here, you take the derivative of that and it’s got to be able to be matched to the outside somehow, okay? So, we are going to rewrite this because here we have 1/x here dx, so we are going to kind of isolate that so it’s already cleared for you. Cos(ln(x))*(1/x)dx, and then u = ln(x) du = 1/x
You have to memorize that. The integral of 1/x is ln(x) etc. so the opposite derivative of ln(x) is 1/x. So, now we are going to do the integral of cos u, du equal sin(u) + c. When you do the integral of cos u it’s sin u. so, now we have the answer of sin[ln(x)] + c, as the answer. Now, we are going to do another problem here, and we are going to press alpha, we are going to have secant. Cosine of second log of x, and without the x, divided by that. Okay? It shows you what you have entered, say it’s okay. And notice that this is not a U substitution problem but an integration by parts problem because notice that the derivatives of the parenthesis here, ln(x) is equal to 1/x but there’s nothing on the outside that equals 1/x. So, if that’s the case then you cannot integrate it by U substitution, you have to go to integration by parts. And that’s just pathetic here with the long, long here of getting the answer, more of Sudoku of math. So we have,
u = cos[ln(x)],
du = -sin[ln(x)],

U Substitution, x*e^(3x^2)

by Leave a Comment

Transcript
Hello. Tom from everystepcalculus.com and everystepphysics.com.  A step-by-step process for solving an e to the x problem.  Transcendental problem. Normally in my programs, if you see a cosine or sine sign and you’re integrating those and log, and e to the x, you are thinking about integration by parts, ok.  But I checked for you so you go to integration by parts you and you put the those things in there and and I find out if it’s u substitution or integration by parts and then I’ll direct you in those direction or solve it that way.  So, let’s do it. Index 8 to get to my menu.  Scroll down to integration by parts because they’re asking this on the test program or homework problem. And integrate by parts here and of course, these are the choices here.  You’re going to choose number two, e to the x and we’re gonna add our
function. You have to press alpha before you enter anything in this entry lines here. Alpha x times e to the 3 times x squared. Keep in mind the integral of e to the 3 x squared or e to the 3  x cubed, etc. This can be solved because of one of the tricks in calculus.   And it can be solved because because the inside the parentheses, the derivative of this matches the outside and I’ll show you how to do that. You need to know their derivatives right away in calculus you or you’ll never get through calculus. For instance, you look at this and say the parentheses, you’re going to come up with 6x right away, not even a question about it.  And 6x is close and x on the outside here so I’ll show you how to do that.  Any time you enter a problem, I show you what you’ve entered, you can change it if you want. I say it’s ok.  I ask you this because I’m checking whether the inside of the derivative matches the outside. In this case it does so we’re going to answer yes.  This is a u substitution problem and it goes to another program u substitution so I asked you the question again no big deal we can do these things to get through our tests and homework, get through calculus. Say it’s yes and then we solved the problem.  We rewrite u substitution to make it easier for instance, I took the x and put it over xdx. See it here. That’s what we’re going to match.  And this is one of the big clean tricks that I’ve learned in my programs how I tell people like yourself to do the problem. U equals 3 x squared, DU equals 6x dx , you gotta take the 6 and move it over here to make the ex dx to match the problem that we just showed you in the previous screen. And then we go into you can change the u parameters. We did. The integral of e to the u is actually e to the u. That’s the reason they use it so much in calculus tests and such is because it looks complicated and it is complicated unless you know that trick.  So here is the answer, 1 sixth, notice we we had e to the u with du divided by 6, you always the constants out of the integral. I just do one because it can be more complicated when we do it then 1, 10, 15 whatever. I do that for you here, too. And here’s the 1 sixth times e to the u and then we substitute back to  get this.  Pretty need, huh? everystepcalculus.com. Go to my site, buy my programs and pass calculus. Best deal you’ll ever make.  Believe me you’ll never throw it out. You’ll throw your calculus book but you’ll never throw these programs within your titanium out.  Have a good one. I’m all for you passing calculus.

Integral Calculator with Steps

by Leave a Comment

∫ 6x^4*(3x^5+2)^6 dx

Raw Transcript

Hello, everyone. This is Tom from everystepcalculus.com. There’s been many, there’s always requests for Integral Calculator with Steps and that’s exactly what my programs do. And Integration is one of the toughest things in Calculus. It was for me when I was in class. I hated
Calculus. You probably feel the same way and I’ve found nobody that likes Calculus. Except
maybe Professors. But anyways, I’m going to show you how my programs work on U Substitution. Index 8 to get to my menu. I’m already at U Substitution. I have scrolled there. We’re going to enter our function. You have to press Alpha before you enter anything into these entry lines, here. Alpha 6 times x to the fourth power times the quantity parentheses
3 times x to the fifth power plus 2, close off the parentheses to the sixth power. I always show you what you’ve entered. 6x 4 times 3 x quantive to the 5 plus 2 to 6. Looks good to me. I say it’s okay. And we’re going to work the problem. Busy means the program is loading. We’re going to evaluate this. First we rewrite it where all constants come out of the integral. And then we put the x to the 4. The way that you know that any problem is U Substitution is that you look immediately at what’s inside the parentheses. You take the derivative, right now you should be able to do, this is 15x to the 4. Right now in one second, you should be able to know the derivative of that. And you notice that x4 is on the outside, too. If it isn’t, it’s not a U Substitution problem. It has to be converted, okay. So I do all that for you, really. But that you have some understanding of how you do U Substitution. U is equal to this, du is equal to this and then we make this the other trick, this whole system here took me about a year to figure this out in a system that works, you know. And so, this always has to be x to the 4 dx over here so you have to take the 15 and divide the du by that on the other side using Algebra, of
course. And so it works the problem. 6, you take of course the du with the 6 here, you have a du with a 15 notice the 15 come out of the, here’s the constant again so that comes out of the integral and goes in front right here, see it right here. 6 times that, of course, with the 2 fifths. etc, etc. Here’s the answer to your problem right here. Have a good one. everystepcalculus.com

Surface of Area of Revolution-Video

by Leave a Comment

Screen shot 2015-01-28 at 9.25.10 PM

 

 

 

Transcript

A problem in calculus called Surface Area of a Revolution. So let’s do it. Index 8 to get to my menu. We’re going to go up to get to the bottom of the alphabet. And scroll up to Surface Area of Revolution.
We’re going to enter our, I’ll show you the actual formula here. Let’s write down your paper first. Press Alpha to enter anything into these entry lines here. Alpha x cubed over the range of alpa zero for a for b, Alpha 2. I show you what you’ve entered so you can change it if you want. I say it’s okay. The first derivative is 3x squared. That goes into the formula here. That’s part of the formula. Over 0 and 2 for the range. And we start doing the computations. Squaring things. This is U Substitution. U equals this. DU equals 36. You always take 36 and put it on the other side as a denominator. DU/36 equals that. And this x cubed equals the problem. X cubed dx. And then we substitute for U. In the problem here, DU 36 has to come out of the integral. So that 2 pi, there’s the 1 over 36 here and that computes to pi over 18, etc. We’re doing the integral of one half, of course you add two halves to that and Upper Range less than the Lower Range. And you get 203 squared units. Pretty neat, huh? everystepcalculus.com. Go to my site, subscribe to see more videos or go to the menu and look up what you need to learn about and pass your calculus. Have a good one.

U Substitution Solver-Video

by Leave a Comment

Raw Transcript
Hello, everyone. Tom from everystepcalculus.com and everystepphysics.com. I’m going to do a U Substitution problem. Basically a simple one but nothing is simple in Calculus unless you know what you’re doing. It certainly helps to have a photographic memory. This problem was sent to me by a student. I’m going to do it on the video and show him how to do it and show you how to do it. Index 8 to get to my menu. U Substitution. You’d scroll down to it there. Choose that. We’re going to enter the function. Alpha, left parentheses 6 times x minus 5, close off the parentheses to the 4th power. So here’s what we have. I say it’s okay. So let’s talk a little bit about U Substitution. You should do derivatives in your head within one second. What’s the derivative of this function in here? It’s 6. There’s no x’s nothing, it’s just 6. And you have dx on the outside. Okay. So there’s no x or nothing else on the
outside except dx. So let’s, that’s what you choose for U. So U is 6x minus 5, derivative is 6dx, anytime you see a constant on here. This is the trick that always screwed me up in school. You need to take this out of this side and isolate the dx. Okay, because remember the dx is by itself on the function and now this dx needs to be by itself like here. So you divide both sides by 6 through Algebra. DU divided by 6 equals dx. It variably happens like this. Don’t think it doesn’t. Almost every problem you do on U Substitution, your’e going to take the constant from this side and put it over on this side. Okay. So no U equals 6x minus 5. Therefore, the integral of u to the 4th is the function, u to the 4th times du divided by 6. Now here’s the du we just did, du divided by 6. In integration, the constants always have to come out of the integration. So you have to take one sixth out which I do here. Here’s one sixth out of the integral. And here we have the u to 4 by itself, and the du is by itself. Okay. So now we got one sixth times, and now we do the integral,u to the 5 divided by 5. Okay. So now this constant has to come out of here, too. So we have one sixth times one fifth times u to the 5 plus C. Well one sixth times one fifth is one 30th times 6 and then we substitute back in for u, 6 times x minus 5 to the 5 power plus C. Every U substitution is not in the system and if you learn this system, you’re going to be alive but the main thing is to be able to do derivatives in your head in one second. In your sleep. The derivative of this right here is 6 and you need to look and see if you can match something on the outside, etc. For U substitution. It’s all tricky though until you do it ten thousand times. Calculus is a language and most the cases when I was going to school, it was learning Latin or learning Spanish. I didn’t know what they were talking about. So, have a good one.

U-Substitution using Log Rule on TI-89

by Leave a Comment

U-Substitution using Log Rule on TI-89

U-Substitution with log rule on TI-89

Raw Transcript

Hello everyone I’m Tom from every step calculus dot com.
Im gonna do a U-substitution problem
that requires the log rule. And uh,
let’s get into it. index8() is my,
you have to put that in the end line here to get to my menu.
I’m already at U-substitution.
You scroll down with the cursor to get there.
And wait for it to load here for a second.
And we’re gonna enter our function. You have to press alpha before you enter anything in my
entry lines in my programs.
Alpha 8 times
X divided by
parentheses
X squared plus
:00
I will show you what you’ve entered. The reason you need to do a log rule is
notice in the denominator your have x squared plus one.
This is an exponent of 1. If you we’re to
transfer this up to the numerator the exponent of
one would become a -1. And in integration you always add,
the first thing you do is always add to the numerator.
And one plus minus 1 is zero so the answer will always come out to be
-1. That’s the reason anytime you see this without, without an exponent other than
one. The one is of course hidden. Um,
you know it’s a log problem.
So we’re going to press there.
Say its okay. Here’s the original
function. And all these you have to rewrite them. Notice there is a constant error
of 8. You have to bring that outside the integral, that’s very important. And we’re going to also
take the x
here and put it over by the dx.
And we’re going to put one in the numerator,
for X squared plus 1 in the denominator.
So U is x squared plus 1 the derivative of that is 2x.
Another trick to all these, took me a long time to figure out.
Was d you need to have the divisor of 2
You have to move that over as a divisor here, and leaving
this here. You’ll notice that X DX is the same as x dx.
That means it’s a U-substitution problem if it wasn’t it’s not a
U-substitution problem. I ask you that though just
to make sure you know what you’re doing. I say yes.
So we have i8 outside the integral and one over you
du number 2. Now we have a constant here at one half.
So we have to bring that outside the integral, which we do over here.
8 and 1/2 that equals 4.
And then we have four times log of U. Anything over one of U is
log of u, plus c.
And the answer is, after we substitute the u back in.
4 times log of x squared
plus 1 plus c. Pretty neat huh?
Every step calculus dot com. Go to my site buy my programs, pass your calculus
class and
have these programs for the rest of your life your kids or grandkids or
your sister, whoever might take calculus in the future,
you have these forever, that’s what’s good about them. And also subscribe to me on my
you know, so that you can see the future
movies or the other blog.

U Substitution e^(x) Solved on TI-89

by Leave a Comment

U Substitution e^(x) | Every Step Calculus  

u-substitution calculus solved on TI-89

Raw Transcript

Hello Tom from every step
Calculus dot com. Here’s a problem on u-substitution.
L,et’s get started index8() to go to my menu.
u-substitution is the subject.
Press alpha before you enterr anything in these entry lines here.
Diamond key to get to the
e(x) problems this is 7 times
X. Divided by.
Anytime you using the denominator you always have to have parentheses
around the whole function.
And then we’re going to go
to e(x) again. 7
times X plus
8.
I always show you what you’ve entered, you can change it if you want, I say it’s okay.
You have to rewrite this. Notice the idea is that the seven ex/dx is the same as
here. Otherwise it’s not a u-substitution problem.
And then I check whether that’s that’s true.
So they’re equal so we say yes.
And here’s the answer 7 times log e to the x
plus c. Pretty neat huh? Every Step Calculus dot com, go to my site buy my programs
pass calculus, also subscribe to
my channel so you can see more movies
or blogs.

U Substitution Video: Sine on the TI-89

by Leave a Comment

U Substitution Sine on the TI-89: Raw Transcript

Hello everyone, Tom from everystepcalculus.com and everystepphysics.com, Iím going to do a U substitution problem that a customer sent me regarding sine and letís get started.
Index 7 to get to both menus, both programs are installed and weíre going to go to U substitution in the menu, we are going to enter our function, we have to pres alpha before we enter anything into this entry lines, alpha square root of X times sine of X to the three halves minus one. I always show you what you have entered, here we have square root of X times sine of X to the three halves minus one, notice the parenthesis, you have to use good parenthesis in math, learn that, itís important and I say itís OK, Iíll give you a chance to change it and when weíre evaluating this we rewrite it because this over here, the square root of X over here matches what weíre going to do in the next screen. U we chose is X to the three halves minus 1, the derivative of that is three halves square root of X and we always take the *** from this side and put it over to di divide by three halves, notice when youíre dividing by three halves youíre going to invert that to two thirds, always remember that important step in algebra and math and here is the same, we wrote it over here so we know itís a U substitution problem, if that wasnít the same and you canít make it to be the same then thatís not a U substitution problem and so here is sine of U is equal to sin of this right here which we know and ** with the integral of the sine of U we have du divided by three halves, here we convert it, two third times the integral of U, two third times the derivative of the integral of the sign of U is minus cosine of U plus C, answer is minus two third cosine of this as we substitute back in for U, pretty neat, everystepcalculus.com or everystepphysics.com, go to my site, buy my programs and pass your calculus class.

U-Substitution Video using Log Rule on TI-89

by Leave a Comment

U-Substitution using Log Rule Video Example

u-substitution-log-rule-every-step-calculus-ti-89

Raw Transcript

Hello everyone I’m Tom from every step calculus dot com.
Im gonna do a U-substitution problem
that requires the log rule. And uh,
let’s get into it. index8() is my,
you have to put that in the end line here to get to my menu.
I’m already at U-substitution.
You scroll down with the cursor to get there.
And wait for it to load here for a second.
And we’re gonna enter our function. You have to press alpha before you enter anything in my
entry lines in my programs.
Alpha 8 times
X divided by
parentheses
X squared plus 1
I will show you what you’ve entered. The reason you need to do a log rule is
notice in the denominator your have x squared plus one.
This is an exponent of 1. If you we’re to
transfer this up to the numerator the exponent of
one would become a -1. And in integration you always add,
the first thing you do is always add to the numerator.
And one plus minus 1 is zero so the answer will always come out to be
-1. That’s the reason anytime you see this without, without an exponent other than
one. The one is of course hidden. Um,
you know it’s a log problem.
So we’re going to press there.
Say its okay. Here’s the original
function. And all these you have to rewrite them. Notice there is a constant error
of 8. You have to bring that outside the integral, that’s very important. And we’re going to also
take the x
here and put it over by the dx.
And we’re going to put one in the numerator,
for X squared plus 1 in the denominator.
So U is x squared plus 1 the derivative of that is 2x.
Another trick to all these, took me a long time to figure out.
Was d you need to have the divisor of 2
You have to move that over as a divisor here, and leaving
this here. You’ll notice that X DX is the same as x dx.
That means it’s a U-substitution problem if it wasn’t it’s not a
U-substitution problem. I ask you that though just
to make sure you know what you’re doing. I say yes.
So we have i8 outside the integral and one over you
du number 2. Now we have a constant here at one half.
So we have to bring that outside the integral, which we do over here.
8 and 1/2 that equals 4.
And then we have four times log of U. Anything over one of U is
log of u, plus c.
And the answer is, after we substitute the u back in.
4 times log of x squared
plus 1 plus c. Pretty neat huh?
Every step calculus dot com. Go to my site buy my programs, pass your calculus
class and
have these programs for the rest of your life your kids or grandkids or
your sister, whoever might take calculus in the future,
you have these forever, that’s what’s good about them. And also subscribe to me on my
you know, so that you can see the future
movies or the other blog.

U Substitution 6*x^2/(x^3+1)^5 on TI-89 | Every Step Calculus Video

by Leave a Comment

Raw Transcript

Ok a video on u substitution with regard to a division function. Let’s get started you press second alpha put the i n d e x letters in you need to press alpha again to put the eight and the open and closed parenthesis press enter and you’re into my menu. I want u substitution which is at the bottom it’s all alphebetical here. So I’m going to go up with the cursor which goes to the bottom and choose u substitution I’m going to enter my function which is press alpha first on anything you enter into my entry lines in my programs press alpha and we got six times x squared divided by the quantity x cubed plus one to the fifth power always use plenty of parenthesis especially in division so they know exactly what your doing what your dividing by very important I show you what you’ve entered in case you’ve made a mistake but I say it’s ok. I press ok I always rewrite things notice you have a constant here you have to bring it outside the integral here it is here times the integral and then you take the x squared what’s left and bring it to the end here where they have dx x squared dx and u equals what’s inside the parenthesis x squared plus one du equals three x squared or that’s x cubed, sorry etcetera you always bring the anytime there is a constant in front of the x or whatever you get here you always divide du divides by three by that and I always ask you if x squared equals x squared. I can’t do it automatically in the calculator and so I need to ask you that if it wasn’t a u substitution problem this would be different and would be integration by parts so u to the minus five is equal to x to the minus five and you do these calculations on your paper put it all down exactly like this and here’s your answer minus one x cubed plus one to the fourth power plus c. Pretty neat huh? everystepcalculus.com.  Buy my programs
and pass your calculus class.

U Substitution Problems | TI-89 App | Every Step Calculus Video

by Leave a Comment

 

U-Substitution on the TI-89: Raw Transcript

This is a video solving a u substitution problem step by step and
also demonstrating how my downloadable programs work in your TI 89 Titanium calculator
and other TI calculators for calculus and physics problems
and let’s get started uhm second alpha i n d e x the letters
and you press alpha to put eight and open and closed parenthesis
press enter and you’re into my menu you can scroll down or up to that u substitution
it’s all alphebetical and we’re going to get to u substitution here
enter our function you have to press alpha before you enter
anything in the entry lines here and we’re going to do
two times x times the
square root orange key
and then the over the times key x squared plus nine
press enter I always show you what you’ve entered
in case you’ve made a mistake I say it’s ok
and here’s your function we’re going to have to rewrite it
here’s the constant you have to take it outside the integral sign
take the x and put it over here by the dx and u equals x squared plus nine
and here’s the derivative anytime you have a constant here
you have to put it over and divide du by that constant
and leave the rest over here should be no numbers over here
notice this is the same as the previous x by dx
that’s why we rewrite it here’s the calculations
put em all down on your paper exactly like this
pretty complicated but when you take you notice when you have a square root
it’s to the one half and so therefore you’re using u to the one
half and you have to add one to that
which will be two halves which equals three halves and then
you have to divide by three halves that’s the way you integrate
and then of course don’t forget to uhm when you’re gonna when your dividing by
a fraction you always have to invert and multiply
here’s two thirds inverted from three halves or three over two
then you substitute back in for u and here’s your answer here
pretty neat huh every step calculus dot com
go to my site buy my programs
and pass your calculus

U Substitution Integration 5x*e^(3x^2) on the TI-89 Video

by Leave a Comment

 

U-Substitution Integration on the TI-89: Raw Transcript

This is a video on u substitution with regard to e to the x
a transcendental function and a little more difficult one
which I wanted to show you of how my programs react to that
to get started were going to press second alpha
put the i n d e x letters in this entry line of the calculator
and then press alpha and put in the eight and the
open and closed parenthesis press enter and you’re into my menu
you can scroll up or down to many things here it’s really integration by parts that your
always thinking about when you do a transcendental cosign sign natural log or
e to the x if you’re going to integrate them you’d go to think integration by parts first
but we’re going to scroll up to e to the x here
choose e to the x from that menu and we’re going to enter the function
you have to press alpha before you enter anything in the entry lines of my programs
so alpha and we’re going to do five times x times e to the three times x
squared I always show you what you’ve entered
so you can change it if you’ve made a mistake or something
and I always ask you whether this if this didn’t equal each other then it would
not be u substitution so you need to tell me that
by stating yes or no it does equal, x equals x so were going to
press yes and then I explain to you that anything inside
the parenthesis of e to the x
the derivative must match the outside and so here’s our original function
we re write it anytime you have a number or a
constant inside the integral you always bring it out and put it on the outside of
the integral before you start to integrate
which I’ve done here you take what’s left the x
and put it over here in the x dx so thats re writing it
and I choose the u for you. here is the derivative of three x squared
it’s six x and then you always take here’s another thing where
there is a constant or number is outside or next to the x dx
you always put that on the other side make du one sixth du equals x dx
transpose it by division and then you do everything
all this on your test or homework whatever exactly correct exactly as you see it
here we’re multiplying the outside of the integral
times the du of the one sixth comes of du comes out side too
gives us five sixrths and then you substitute three x squared for
u answer
five sixths times e to the 3 x^2 plus c pretty neat huh
every stem calculus dot com go to my site, buy my programs
and pass calculus

U Substitution Integral x^2/(x^3+1)^2 | TI-89 Video

by Leave a Comment

 

U-Substitution Integral on the TI-89: Raw Transcript

This video is on u substitution uhm with regard to calculus
for your tests and homework and let’s get started
you have to press second alpha first to put the letters i n d e x in here
then press alpha and put the eight and the open and closed parenthesis
press enter and you’re into my menu I’m already at u substitution
but you can see there’s many many things in here
for you to pass your calculus tests and do your homework
I was never really interested in learning this stuff
I just wanted to know how to do it for my tests and homework
ah, but were going to do u substitution that’s the way my programs are designed
all of my programs including my physics programs
uhm u substituion lets put the
you have to press alpha before you put anything in the entry lines
here in my programs
so I press alpha and let’s put in x squared divided by the
quantity one plus x cubed to the second power
I always show you what you’ve entered so you can change it in case you’ve made a
mistake I say it’s ok
we’re going to evaluate x squared divided by x cubed plus one over two
and we rewrite the program to put the x squared over by the dx here
and of course there’s one in the numerator there
and we choose, i tell you what to choose to isolate the x you take the three and move it to the other
side for the du du divided by three equals x squared dx
and then we of course change the function to u
1 over u squared same thing as one over what you entered there
and you do the calculations notice that we have du divided by three
but you always take a constant or fraction and move it to the front of the integral sign
here which I’ve done for you
and then you have you have to change u two to the numerator
you can’t integrate with a division or times sign
you have to change it to a single numerator and so we have u to the minus two
and you add one to that which becomes a minus one
and you get one third over one to the minus u
and etcetera and here’s your answer uhm pretty neat huh
every step calculus dot come go to my site
buy my programs and pass your calculus

U Substitution Calculus | TI 89 App | Every Step Calculus Video

by Leave a Comment

Raw Transcript

This is a video solving a u substitution problem step by step and also demonstrating how my downloadable programs work in your TI 89 Titanium calculator and other TI calculators for calculus and physics problems. So let’s get started you have to press second alpha before you enter the letters i n d e x in the entry line of the calculator here and then press alpha again to put the eight and the open and closed parenthesis. Press enter and you’re into my menu. Many things to choose from in here trig stuff quotient rule quadratic formula whatever to help you pass calculus and do your homework we want to do u substitution right now we have to press alpha before you enter anything in these entry lines where we enter functions press alpha x times e to the three times x squared. I always show you what you’ve entered so you can change it if you want give you that option I say it’s ok we’re evaluating this and we re-write it so we put this x here over by the dx to make us understand we need to match that by doing the derivative of the inside here, inside the parenthesis and we choose u which is three x squared du is equal to six x we take the constant always out of this and leave the x dx there’s our shows us that it is a u substitution problem and we have du divided by six transpose it to the other side by division and you take e to the u and make it equal to you know that it equals e to the three x squared and the integral of e u with du divided by six but here we have to take the six and bring it outside the integral always a constant or fraction anything like that you bring outside the integral which we’ve done here and then we do the derivative a integral of e to the u which is e to the u and here’s the answer we substitute back in the three x squared pretty neat, huh? everystepcalculus.com. Go to my site and buy my programs and pass calculus.

U substitution x*e^(x^2) on the TI 89 | Every Step Calculus Video

by 1 Comment

Raw Transcript

This is a video on integration of e to the x transcendental function uhm and let’s get started on it uhm for my programs you have to press second alpha to put the letters i n d e x in here and then press alpha again to put in the eight and the open and closed parenthesis press enter and your into my menu integration by parts is the program that does cosign sign e to the x and natural log and so in my programs which says integrate transcendentals we’re going to choose e to the x and to enter anything in my lines here the entry lines, you have to press alpha first so were gonna press alpha and put uhm x times let’s say e to the x squared I always show you what you’ve entered so you can change it in case you’ve made a mistake you have that option and I also need you to tell me whether this is equal or not if it is, I haven’t been able to program the calculator to do that in a certain way it’s too complicated for you to understand but and so if it is equal you press yes if it isn’t then I tell you what else to do with it and I explain to you that the derivative of the inside function in parenthesis matches in some way the outside and if that happens then that’s a u substitution problem and so then we’re evaluating this you re write it by putting this x here over by the x dx and isolating that choose x squared for u du the derivative two x and then when ever you have anything in front of the x here you have to transfer it to the other side by division so here’s on half du equals x dx and that was the same as the original function when we re wrote it and so you do the calculation write everything you see on your paperfor your test and homework to get an a on this problem and the answer is one half e to the u plus c then you substitute back in x squared for u so you get one half e to the x squared plus c that’s the answer pretty neat huh everystepcalculus.com.  Go to my site, buy my programs enjoy them and pass calculus. 

U Substitution Definite Integral on TI-89 | Every Step Calculus Video

by Leave a Comment

 

Definite Integral Example on the TI-89: Raw Transcript

This is a video on u substitution and this one is about a definite integral
where when you actually compute the integral you are going to compute it over a range
with a lower limit and a upper limit so anytime you have a problem like I’m going
to show you here that’s the way you use my programs
you go to u substitution let’s get started here
we’re going to press 2nd alpha and put i n d e x
into the entry line of the calculator then press alpha and put the eight and the
open and closed parenthesis press enter and you’re into my menu
you have many, many things on here notice graphing by hand
and all kinds of things that are perfectly done
and wonderfully done on these programs to help you with your tests and homework
and that’s what we’re interested in is passing calculus
and never to do it again so anyways you can scroll up or down
to u substituion this is all alphebetical and this case you
want u substitution so we might press the upper cursor to go up
and to the bottom menu, which is u substituion
notice you have trig d dx integrals for trig
half angle formulas all kinds of things you need for calculus
one for sure and then I have other programs for calculus
two and three so if I press enter we’re into u substitution
and generally you press alpha and put your function in here
you have to press alpha first and then put the function in
but I’ve already done that to speed up the video so i’m going to
I always show you, this is the function we’re doing
I haven’t put the limits in yet so you have to recognize that
you are going to do the integral first and then at the end I ask you if you want
to do the range for the area or limits
now, I always show you what you’ve entered so you can change it if you want
I say it’s ok and we’re going to evaluate this integral
here now you’ll notice that the derivative of three
x to the five is really fifteen x to the four
and here’s an x to the four so if you didn’t have my calculator
that is what you are really looking at everytime you look at an integral
you say well is the inside, can you do the derivative
to make it equal to the outside somehow and
but I do that for you in my programs we always rewrite it
anytime you have a number or constant inside the integral
you take it outside the integral like I’ve done here in my programs
and then you evaluate the integral inside but also when you rewrite it you need to take
this x to the four and put it over here by the
dx and then you choose the u which is the three
x to the five plus two and the derivative of that is fifteen x to
the four anytime you have a number here before the
x you need to transpose it to the other side
by division so you are going, which I’ve done here now
du divided by 15 equals x to the four dx you’ll notice that this equals what we re-wrote
in the beginning here and so we know that that is a u substitution
problem and I ask you that too
because I need to check that before and so I ask you if x to the four equals x
to the four if it says yes, if they’r equal you say yes
of course if it’s not equal you say no if you say no then you are into log rhythms
because in integrals you can’t do integrals with a times sign or
division sign you need to seperate it into plus and minus
parts so that you can integrate it
and that’s what log rhythms do for instance in times you are taking a log
of one factor plus the log of the other factor in division you are taking the log of one
factor minus the other.
so anyway they are equal so we’re going to get on with this
you write all of this on your paper and write it kind of sloppy
students and people that really know this stuff write sloppy
that’s been my experience, including professors they scribble because they want to make everyone
know that they are genius’s at this stuff.
so you do the same don’t write it clean like this write it sloppy
and uhm, here’s the answer after you do all the u
you’ll notice the du divided by fifteen you need to bring that to the outside of integral
which I do one fifteenth times six here
multiply that together to get two fifths and then you do the integral of u to the 6
which you add one to the six and get seven and divide by seven
u seven divided by seven well then you have to do the computation there
seven times five is thirty five so you have two over thirty five
now you substitute back in the u which is thirty five plus two
and you have your answer plus c now the really tricky part here
I ask you if you want to evaluate the range you can press one here and do it
let’s do the limit which I’ve put in there as number one
for the lower range or limit and number two for the upper limit
and so here’s the integral that I showed you at the
beginning of the video. which is called a definite integral
where you find the area under the curve and again if there is a mistake in adding
the limits I ask you that so you can change them if you
want now you’re going to substitute
if x equals one then u equals
remember we’ve decided that u equals this and then you add the one in there and mulitiply
it out to get five ok?
if x equals two you put this exactly on you paper, here’s
what your doing u equls three x two
and that equals ninety eight so you have u substituted here and you have
the upper limit ninety eight and five notice you can’t use one and two for the limit
because you’ve changed it to the u system, so
and then with u equals ninety eight the upper limit
you have to go back to the original function and put it in here the original integral that
you’ve found and that equals four point nine six times
e to the twelve and then when, the lower limit is five so
you are adding that into the original u funtion to get four point four three time e to the
3 you substract the upper limit
the lower limit from the upper limit here’s the area under the curve
four point nine six times e to the twelve square units
uhm, notice how fabulous these programs are and how they will help you so much
even in learning this stuff let alone passing tests
or for homework so you can buy my programs at
every step calculus dot com and enjoy my programs
and pass calculus

Integration U Substitution on the TI 89 | Every Step Calculus Video

by Leave a Comment

Raw Transcript

Okay this is a video on u substitution. A little more difficult problem you find in calculus 2 and let’s get started. Here after you buy my program and installed in your calculator, you have to press second alpha to put the I n d e x in here the letters and for push alpha and put the 9 in the open close parentheses. for calculus 2, it’s a 9 there. For calculus 1 its an 8. To get my menu press Enter and all these different options to go you can go up ok we need closer to u so there all alphabetical so I’m going to go up here Notice we have work lifting object, work lifting springs, work putting a shuttle in space, work pumping oil, velocity, vectors, It’s all calculus of course but I do sell physics programs also. everystepphysics.com. We’re gonna press u substitution here. Get into our u substitution program. And we’re going to press alpha before we enter anything in the entry lines in my programs. So we’re gonna push alpha X plus 3 divided by quantity X squared plus 6 times x minus 5 to to the second power. We check our formula to see if we put it in righ. and you have a chance to change if you want. I say it’s okay so press okay and we’re going to evaluate this function integrated. We always rewrite it You put X plus 3, you notice the x plus 3 in the numerator here over the, by the DX because this derivative of what’s inside the parentheses must equal x plus 3 what we just rewrote it over there. And so I checked in my programs if it doesn’t and all hell can break loose. But I say it I see it that equals. We’re into the u of course u to the minus 2 and that’s what in the parentheses because we brought it form the denominator to the numerator. I don’t need to teach, this stuff is for passing tests and doing your homework quickly and not learning it but anyways you you are u integrals. Notice that DU divided by two has to come out. Any time you have a constant fraction or pie or anything, a decimal has to come outside the integrals. So that’s the first thing you do. Put in on here nice and clear with the parentheses. And then you do the integration of -2 which you’re gonna a 1 there, right. So you divide by that same exponent that you found. Here’s the answer. And you substitute back in the what’s in the parentheses at the beginning plus c. Pretty neat, huh? everystepcalculus.com. Go to my site and buy my programs and pass calculus.

Integration by U Substitution on TI-89 | Every Step Calculus Video

by Leave a Comment

Raw Transcript

This is a video solving a u substitution problem step by step and also demonstrating how my downloadable programs work in your TI-89 Titanium calculator and other TI calculators, for calculus and physics problems. Let’s get started you have to press second alpha to put the i n d e x letters in there and then you have to press alpha to put the eight and the open and closed parenthesis press enter you’re into my menu here’s the menu I happen to be at the bottom right now but it’s all alphabetical we’re doing u substitution click on that you have to press alpha before you put anything into my entry lines here so were going to add the function. I go two times the quantity two times x plus one to the fourth power I always show you what you’ve entered. So you can change it if you want I say it’s ok and because we have a constant here we need to put it outside the integral so we put it outside and then we find u u is two x plus one which is inside the parenthesis derivative is two then you always take the two if there is a constant in this part take it and move it over here du divided by two dx equals dx and then you have u four equals two x minus one to the four or two x plus one to the power four and then we do our integral here here’s the two on the outside and the integral u four du over two divided by two and we have to multiply that times this, etc. Here’s your answer. Pretty neat huh? everystepcalculus.com. Go to my site buy my programs and pass calculus.

U Substitution Practice | TI 89 App | Every Step Calculus Video

by Leave a Comment

Raw Transcript

This is a video solving a u substitution problem step by step and also demonstrating how my downloadable programs work in your TI 89 Titanium calculator and other TI calculators for calculus and physics problems and let’s get started here to access my programs you have to press second alpha and put the i n d e x letters in here and then press alpha again to put the eight and the open and closed parenthesis press enter you’re into my menu, many things to choose from here as you can see definite integral critical points graphing by hand chain rule we’re going to go up here with the curser to get down to the bottom of the menu which is u substitution and we’re going to put in our function you have to press alpha again to put anything in these entry lines remember to do that alpha let’s go five times x times sign of three times x squared. I always show you what you’ve entered you can change it if you want. I say it’s ok and we’re going to evaluate this of course you have to bring the constants outside the integral which I do for you any time there’s that plus we want to bring this x now over by the x dx to make sure it’s a u substitution problem somehow this inside of this parenthesis has to be made to match this I choose that for you here’s the du you always take a constant out of here and du divided by six and then isolate the x dx which you can see is the same so this is a valid u substitution if it wasn’t it’d be an integration by parts problem. I make sure you know that so I have to ask you this does x equal x it checks whether that is a u substitution problem and here’s your answer you write all of this down on your paper exactly like I’ve got it here you’re bringing constants outside the integral you’re multiplying times what you’ve put outside the integral before, and here’s the answer right here minus five sixths times cosign of three x squared plus c. Pretty neat, huh? everystepcalculus.com. Go to my site buy my programs and pass calculus.