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# Evaluating simple definite integral

Based on the fundamental theorem of calculus, we can use antiderivatives to compute integrals. Created by Sal Khan.

Video transcript

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.