Hello everyone. Tom from come everystep calculus.com and
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.
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
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
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