BAM Mr. Tarrou. In this video we're going to be taking a look at one small concept
within the umbrella within the introduction of what it is to have and evaluate a definite
integral and there's many many applications of definite integrals it's one of the favorite
sections within calculus that ap likes to use for writing questions for their standardized tests
or whatever calculus class you are in in the description below you will find
links to my past lessons that i've worked all the different types of applications mean value
theorem average value theorem word problems net change theorem if you will total displacement
many applications it makes sense because we're just starting into our journey of the second
major branch of calculus which is moving from differential calculus into integral calculus so
we're just started looking at maybe estimating area between the curve and the axis using
three or four or five maybe rectangles left and right endpoint rectangles maybe using the midpoint
rectangle riemann sum or estimating that area with say like trapezoids and then we moved into the
limit process and worked with that full-blown definition of a definite integral which says
that we have the limit as n approaches infinity of the summation where i starts at 1 and goes to
n now let's not forget that n is the number of rectangles trapezoids whatever it is
whatever common geometric region you're using to estimate the area say under a curve we like to
in calculus kind of just use those rectangles as a very basic shape because we have in the
integrand sort of that delta x the width of those rectangles and f of c sub i the height of each of
those rectangles and we're not introducing the definition of a definite integral but
i'm trying to now through my now sixth year of teaching calculus and making more videos
remind ourselves of what is that definition of a definite integral that we're kind of
just using and maybe forgetting what it actually represents it's basically a limit of sums so that
expression is then equal to the definite integral from a to b of f of x dx
we through the process of evaluating this definite integral
find the antiderivative capital f of x of the integrand or lowercase f of x dx
and we carry those low and upper limits along with us and then ultimately we say okay we've evaluated
the antiderivative of the integrand or the capital f to the lowercase f let's not forget by the way
that that is f lowercase f is really f prime of x if you take the derivative of your antiderivative
you get well the original expression that you are integrating with that little plus c constant of
integration so then we take that upper and lower limit we plug it into our our antiderivative
our capital f and we get a numerical value which really can take on many applications
but when we're first just rolling into that idea of a definition of definite integral
your textbook is more than likely talking about area under a curve and the x-axis or between the
curve and the x-axis or the equation y is equal to zero though there's many many applications
we're going to focus though not on word problems not in applications just with the geometry of
what's going on here and talk about this idea of an accumulation function we're going to evaluate
the definite integrals from one to one of x squared plus three dx the integral from one
to two of x squared plus three dx the integral from one to three of x squared plus 3 dx and then
the integral of from 1 to 4 of well x squared plus 3 dx and you're thinking i'm not watching
a video of some dude evaluating four definite integrals when they're all based off of the same
well function as we have here just a simple parabola shifted up three units we're not really
going to work out and long form the value of all four of those examples we're going to recognize
that each of these four examples has the exact same integrand and they also have the exact same
lower limit though with accumulation functions you might be able to argue that the lower limit um
maybe doesn't need to be a constant that's how we're setting this one up
and what i would like to do with this and and to get across the idea of what is an accumulation 4
it's it's almost like i want to create an equation that i could then drop into an excel
spreadsheet and allow someone to in this case find the area under this parabola
off of the initial value of one now we could go i'm moving to the right just to kind of
for simplicity of our examples we can go to the left or the right from x equals one and
have this accumulation function be such that i don't actually have to start all four of these
problems from scratch i would be able to just have a function which then would tell
me what's the area under the curve as we move off of this initial value of x of 1.
so that if you were making some kind of excel spreadsheet and somebody wanted
the area under this curve at least if they want the air into the curve
some distance off of the initial value of one they could just plop in that two that three that four
or any number indeed that they would like and out comes a number they can calculate the area without
actually having to know all of the calculus so you can think of an accumulation function as a way of
avoiding maybe repetitive type calculations on the same function so how do we do that
well first of all let's not remember or let let's remember one of our properties of definite
integrals and how it is at least in fundamental geometric representation area how are you going to
accumulate in the area along the well this may be you can you can integrate along the horizontal or
vertical axis but how are you going to accumulate in the area along the x-axis because we have an
independent variable of x here and a dx if you don't actually move left to right
like the first example number one is like what's the area of this line segment between 1 and
1 the area of a line that's a one-dimensional object the area is equal to zero whenever you
have a definite integral where the lower and upper limits are the same the answer is equal
to zero well now let's actually focus on these last three examples of one to two one two three
one to four indeed one to any numerical value that i that i want how would we evaluate that
well instead of doing three separate problems how about we represent we recognize that each of these
integrands are the same each of these lower limits are the same and just say the only thing that's
actually changing is that upper limit so let's shortcut this process and instead of talking about
all three of these each individually independently let's say that we have the definite integral from
1 to x of x squared plus 3 dx let's just drop in a variable in that upper limit now we do have a
little bit of a problem a little bit of confusion that you can see here maybe potentially developing
in this definite integral which is going to become an accumulation function that function that's
going to allow me to just simply plug in a 2 a 3 or 4 or any number that i would like actually
and tell me the area under this curve off of the initial value of 1 which is the variables
in my integrand are the same as the variable that i've introduced in my either it could be
lower could be upper or maybe potentially both but whenever you have a variable expression as
one or both of the limits in a definite integral it is a little bit confusing so when i write x
am i talking about the variable x which is my upper limit or am i talking about variable x
or a variable that is in my integrand now also you have know that when you evaluate a definite
integral as long as the upper and lower limits are constants you're going to get a numerical value
if you do an indefinite integral well then you get some function that such that if you
take its derivative you get the integrand so indefinite integral gives you function
a definite integral with lower and upper limits which are constants gives you a numerical value
in the beginning it's area under a curve but it really depending on the
word problem the application of your to which your calculations apply to could be represent anything
well anyway it's confusing to have the variables in the integrand be the same as the variable in
the upper limit and the reason why i came back to this formula here the variable that's in the
integrand is really going to be replaced by whatever values or variables that you plug in
from the lower and upper limit so really in terms of the variable of our final answer we are kind of
getting into the idea as a definite integral as a function as opposed to a definite integral as
an numerical value this is the variable which is going to be the independent variable of our final
answer thus the variables in the integrand well it's confusing to have them be the same and i
saw i know i said a lot there just to kind of say something simple what we need to do is introduce
a sort of a dummy variable in the integrand just anything that's not the same as the variable
that you may have in one or both of your limits because it's going to go away anyway and then
you have some clarity when you write variables out are you talking about some kind of variable in the
integrand or about that upper and or lower limit so working through now the integration process
using that algebraic power rule because we don't have any trig going on here c to the second power
right you've got a variable base fixed exponent we're just going to raise that exponent by 1.
2 plus 1 is 3 and then divide by that increased variable just kind of like the opposite if you
or inverse process you went through through doing the derivative using the power rule the integral
of 3 is equal with respect to c is equal to 3c and then we have that upper lower limit of x and 1.
now we can go ahead and plug in these values you know kind of continue that definite derivative
process so we're going to plug in the x and get x cubed over 3 plus 3 times x and then we're going
to bring in our lower limit as the definition of a definite integral says that we're going to be
doing we plug in the upper limit we plug in the lower limits we're plugging in that lower limit
of one and we have one cubed over three plus three times one well that's going to give us
x cubed over three plus three x and then what do we have here we have a minus
one third plus three or three and one-third and now what this is going to tell us is the area
under
the curve the curve being y is equal to x squared plus 3 to the right
of 1 because remember we need to have this be our lower limit
and as we move you know so as we have that lower limit of 1 as we move to the right those upper
limit values are going to maintain that idea of being larger and that is now okay so now how do
we use this we're going to take this value and we're going to be able to plug in the value of 2
and get the answers well to these 3 problems so we have
2 cubed over 3 plus 3 times 2 minus well 10 3 and i just realized that i don't have a
calculator in this room that's going to be 16 over 3 and now we can just simply take
again that value of 3 plug it into this equation or you know actually go through that process
again and that would give you an answer of 44 over 3 and we can put in a value of 4
and find out the area as we move off to the right of 1 just as we continue to move off to the right
and accumulating that area as we go if we go from 1 to 4 then this total area
under the parabola between the parabola and the x axis that total area is going to be equal to well
at least in terms of area it would be 30 square units now i am saying that you are going to
get the area under the curve to the right of 1. that that's what this accumulation function does
if you wanted to put a number that's indeed well less than 1 this accumulation function
is still going to work but again the only application to definite integrals the only
application is definitely not finding area it can take all kinds of values it can take values
of positive or negative that negative value may have some kind of you know real life meaning in an
application problem but it doesn't make any sense to say that there is like a negative area but
let's think about that if we plug in if we want to find the definite integral from 1 to negative 2 of
x squared plus 3 dx well the lower limit of a definite integral needs to indeed be the lower
limit so we would rewrite this using another property of definite integrals as being the
opposite of the definite integral from negative 2 to 1 of x squared plus 3 dx now if you evaluate
this definite integral you are going to get an answer of negative 12. if you plug negative 2 into
this expression or this accumulation function you're going to get an answer of negative 12.
now what does that mean is it really the area under the curve to the right of 0. well
actually 12 if you cover up that 12 is the area under the curve between 1 and negative 2
and at least in terms of this scenario you can think of the negative not as a negative area
but area to the left of that value of one so like this is area to the left of one
and um these other answers of what 16 over 3 4 over 3 and 30 are areas to the right but again you
might be talking about this function representing say the rate of travel and if you integrate a rate
of change you get an amount of change so like this might be like a net change theorem where you find
the net change in position from let's say second one to second two it could be all kinds of things
so it's not just area under curve but that is the process and sort of like what the purpose is
of an accumulation function in pure geometry instead of doing repetitive um
work with constant upper and lower limits you can just sort of come up with a an equation or
expression if you will that depends how you write with with or without an equal sign that gives you
the answer to just a whole bunch of definite integrals if there's just a slight change in
the limits so you get a nice convenient formula so i'm mr tarrou bam go to your homework
it