# 10.3 Usubstitution Definite Integralsap Calculus

- 10.3 U-substitution Definite Integralsap Calculus Algebra
- 10.3 U-substitution Definite Integralsap Calculus Solver
- 10.3 U-substitution Definite Integralsap Calculus Integrals

Let's take a look at an integral problem. Consider the indefinite integral of x times the square root of x squared minus 9. Though you'd probably solve this using the method of substitution, you'd probably think substitute for the x squared minus 9 part. But how would you solve the, the definite integral from 3 to 5. Same integram? It turns out that you can use the method os substitution for a definite integral as well. I want to show you how to do that.

So let's copy this down, the integral from 3 to 5 x times the square root of x squared minus 9. So we'll substitute for the x squared minus 9. w=x squared minus 9. And dw will equal 2x dx. Now I don't have a 2x dx but I do have an x dx. So I can just divide both sides by 2 and get one half dw equals x dx. That'll take care of everything inside the integral but it does not take care of the limits. These limits are x values. When I change this integral over to w, I really should change the limits to w values as well. And you can do that using this substitution. Right? x=3 for example this bottom limit, if you plug 3 in here you get 9, minus 9 is 0. w=0. So the 3 is going to turn into a 0 and x=5. If you plug that in here you get 25-9. w is going to be 16. So my new integral's going to be the integral from 0 to 16. The square root of x squared minus 9, that will be root w the x dx, one half dw. So I'll pull the one half outside. Integral from 0 to 16 w to the one half dw. This is one half and now I need an anti-derivative for w to the one half. I raise the exponent by 1 to w to the 3 halves and divide by 3 halves which is the same as multiplying by 2 thirds. So this is 2 thirds w to the 3 halves. And I want to evaluate this from 0 to 16. Now one half times 2 thirds is one third. So it's one third 16 to the 3 halves minus one third 0 to 3 halves. Now, of course this is going be 0 but what's 16 to the 3 halves. Whenever I have a fractional exponent depends on what the input is. As to whether I want a cube first or take the square root first but this number is a perfect square. So I'm going to take the square root first.

The square root is 4 then raised to the third power is 64. So this is 64 thirds. And that's my answer. So the new trick here is switching over the limits of integration using your substitution. Remember this formula tells you how. These are x values. Plug them in here and you can get the corresponding w values.

The Substitution Method for Integration. Integration by substitution is one of the many methods for evaluation of integrals in calculus. Other methods of integration include the use of integration. 5 Integration 5.2 The Definite Integral 5.4 The Fundamental Theorem of Calculus 5.3 Riemann Sums In the previous section we defined the definite integral of a function on a, b to be the signed area between the curve and the x -axis.

### Definite Integral Example on the TI-89: Raw Transcript

## 10.3 U-substitution Definite Integralsap Calculus Algebra

## 10.3 U-substitution Definite Integralsap Calculus Solver

## 10.3 U-substitution Definite Integralsap Calculus Integrals

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