So we've got the function, f of x is equal to x squared.
And what I'm concerned with, is finding the area under the curve,
y is equal to f of x, so that's my y axis.
This is my, my, x axis.
Then let me draw my function.
My function looks like this.
At least in the, in the first quadrant.
That's where I'll graph it for now.
I could also graph it, obviously in the second quadrant.
But, what I care about is the area under this curve and above the positive x axis,
between, between x equals 1 and x equals 4, x equals 4.
And I'm tired of approximating areas.
I wanna find the exact area under this curve above the x axis.
And the way we denote the exact area under the
curve, this little brown shaded area, is using the definite integral.
The definite integral from 1 to 4 of f of x, dx.
And the way that, or the way I conceptualize
where this notation comes from, is we imagine a
bunch of infinite, an infinite number of infinitely thin
rectangles that we sum up to find this area.
Let me draw one of those infinitely thin rectangles, maybe not so infinitely thin.
So, let me draw it like this.
So, that would be one of the rectangles, that would be another rectangle.
This should be reminiscent of a Riemann sum.
In fact, that's where the Riemann integral comes from.
Think of a Riemann sum where you have an infinite number
of these rectangles, where the width of each of the rectangles.
This is how I conceptualize it, is dx, and the height of this rectangle is the
function evaluated at an x that's within this interval right over here.
And so, this part right over here is the area of
one of those rectangles, and we were summing them all up.
And this kind of an elongated s, reminiscent of a sigma for summing.
We're summing up the infinite number, of those infinitely thin rectangles,
or the areas of those infinitely thin rectangles between 1 and 4.
So that's where the notation of the definite integral comes from.
But we still haven't done anything.
We've just written some notation that says the exact area of the un-
Between 1 and 4, under the curve f of x, and above the x axis.
In order to actually do anything really productive with this, we have to turn to
the second fundamental theorem of calculus, sometimes called
part two of the fundamental theorem of calculus.
>> Which tells us, that if f has an antiderivative, so if we
have the antiderivative of f, so f of x is derivative,
derivative of some function capital F of x, or another way of saying it is,
f, capital F of x is the antiderivative,
antiderivative of lower case f of x.
Then I can evaluate this thing, and we do
a whole video on conceptually understanding why this makes sense.
We could evaluate this, by evaluating the antiderivative
of f, or an antiderivative of f, at 4.
And from that, subtract the antiderivative evaluated at 1.
So, let's do it for this particular case right over here.
So we are taking, I'll just rewrite this statement.
Instead of writing f of x, I'll write x squared.
So, the definite integral from 1 to 4 of x squared dx.
Well, we're just gonna have to figure out what the antiderivative is.
So if f of x is equal to x squared, what is capital F of x equal to?
What is the antiderivative?
Well, you might remember from your power rule, that if you
take the derivative with respect to x of x to the
third, you are going to get 3x squared, which is pretty
darn close to x squared except for this factor of 3.
So, let's divide both sides by 3.
Let's divide both sides by 3, and you get the derivative
of x to the third divided by 3 is indeed x squared.
Or, you can say this is the same thing as the
derivative with respect to x of, x to the third over 3.
Take the derivative of this.
It'll be 3 times one third.
And then you'll decrement the power, it'll just be x squared.
So, this right over here, once again, is x squared.
It's just equal to, just equal to x squared.
So, in this case, our capital F of x, our antiderivative,
is x to the third, x to the third over 3.
And so we just have to evaluate that at 4 and at 1, and sometimes the way we would,
the, the notation we would use is, we'll say that the antiderivative is x
to the third over 3, and we're going to evaluate it, the one I, I always
just like to write the numbers up here,
at 4 and from that subtracted, evaluated at 1.
Sometimes you'll see people write a little line here too,
we'll say we're evaluate it at 4 and then at 1.
But I'll just do it without the line.
If we're gonna evaluate this thing at 4 and from
that subtract it, subtract it evaluated at 1, so this
going to be equal to 4 to the third power is 64, so it's going to be 64 over 3.
Let me color code it, this is, this right over here, is this, right
over there and then from that, we're
going to subtract this business evaluated at one.
Well, when you evaluate it at 1, you get 1 to the third is one over 3.
You get one third.
So just to be clear, this is this right over there.
And then we are ready to just subtract these fractions.
64 over 3, minus one third, is equal to 63 over 3.
And 3 goes into 63 exactly, exactly 21 times.
So, whatever the units are, the area of this brown area is 21 square units.