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Hello, Tom of everystepcalculus.com and everystepphysics.com. A definite integral over X and Y. Let’s do it. Index 8 index to get to my menu. Scroll down to D section. Definite Integral X over XY. Double Integral. Enter the function. Alpha before you enter anything into the entry lines. This one on the internet. X times Y squared divided by parentheses, you always put parentheses in denominators. X squared plus one. I don’t know who dreams up these functions but somebody does. Remember, we’re not learning calculus for life long desires unless you’re a math major. We’re learning Calculus to pass a test. To get through Calculus. Never to touch it again. Keep that in mind, okay. Press enter here. And see if that’s what you entered. I say it is. And we’re do it over to DY, DX. And remember that D you integrate with respect to Y first when it’s DY here and DX second. And so DY is these limits of integration. And DX is these limits. So the limits given in this problem were X for 0. Alpha 0, Alpha 1. And for AY was Alpha minus 3. And Alpha 3. And here’s the Integral right here. And we’ve done the integral with respect to Y. Here’s the answer: X times Y cubed divided by 3 times x squared plus 1. And we’re going to do that over -3 and 3. So when Y equals 3, you enter these quotation marks are where you enter parentheses on your paper. Okay. Because that’s substituting Y for anything in these functions, these answers. Another one, -3 is when Y equals minus 3, these is the, you put parentheses around quotations marks. This is the answer. This is the answer for when you put three up there and you subtract the upper limit from the lower. and you get 18 X for x squared. Now we’re doing the integral of that answer over 0 in 1 limits. So we do on the integral of that which equals 9 times log of X squared plus 1. At x equals 1, this is the answer. 9 times log over 2. log 2 and x equals 0. 9 times log over zero x squared plus one which equals zero. Upper limit from the lower is this minus the 0. And so this is the answer also but when you work it out, it comes up to 6.24 square units. Pretty neat, huh? Go to my site and subscribe it you want to see more movies.

Hello everyone. Tom from come everystep calculus.com and

everystepphysics.com

I’m gonna do it Definite Integral for you.

this came up on my Facebook and I want to answer it here in a video.

Let’s get started. Index 8 to go to my menu.

We’re going to scroll down to Definite Integral. It’s all alphabetical so it

should be in D’s here

and here’s Definite Integral.

And we’re going to enter the function. You have to press Alpha before you enter anything

into these entry lines here.

parentheses one

plus cosine

of two times x,

and close of the main parentheses,

divided by two. I always show you what you’ve entered. You can change it if you want.

I say it’s okay. Now we’re going to enter

the range. At a range of, the lower limit was

was alpha 0

and the upper limit was

alpa pi

divided by two. Again I show you, this is the problem that you’re asking

and was asked.

Now we do the derivative of that which is sin of 2x plus 2x

divided by four and

and pi over two is the upper limit and the lower limit is zero.

We substitute the upper limit for all of x’s.

Here were substituting pi over two for x, pi over two for x.

We come up with pi over four

or approximately .785. Substitute the lower limit

for x. Notice we substitute for x 0

and 0. Here’s 0 and the answer is 0.

So the Definite Integral is this

and we take the upper limit and minus the lower limit.

Pi over four minus 0 answers Pi over four square units or

.785 square units. Remember and integral is

an area under a curve.

So go to my site. Buy my programs if you want to

get help with calculus and subscribe to me also

if you wanna see future movies

or whatever, Thank You.

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

Raw Transcript

This video’s going to be on definite integral…where you take the integral of the function and then compute it at a lower and upper parameter and ah, really your finding the volume of something within a certain range and so let’s get started here you have to press second alpha this shows you’re gonna enter letters into the calculator i_n_d_e_x press Alpha again to get back to the number system on the calculator and press here comes my menu all the things you want to do chain rule area a parallelogram okay