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# Area between curves

By integrating the difference of two functions, you can find the area between them. Created by Sal Khan.

Video transcript

What I want to do
in this video is find the area of the region
in the first quadrant right over here that's
below the graph of y is equals to square root of
x, but above the graph of y is equal to x squared. So we're talking about this
region right over here. And we know what the
two endpoints are. The two endpoints are when these
two functions equal each other. So x squared is
equal to square root of x at x is equal to 0
and at x is equal to 1. So how do we think about this? Well, one way to do it
is just to think about, well, what's the
area between y is equal to square root
of x and the x-axis? So that's going to be, and
we're going from 0 to 1, so this is the area under
square root of x from 0 to 1 dx. So this is literally denoting
this entire region, right over here, all the way up
to this boundary point. And then from that,
subtract out this region. The area underneath
y is equal to x squared, but above the x-axis. So from that, we could
subtract from 0 to 1. That's the interval
we care about. The area under x squared dx. And this would be
completely legitimate. And you might say, hey, look,
I have the same boundaries of integration, I have
dx right over here. Couldn't I have also written
this as the definite integral from 0 to 1 dx of square
root of x minus x squared? And if you asked that,
I would say absolutely. And this is actually a different
way of conceptualizing it. Here, you're taking the area. And instead of, if
you conceptualize dx as the width of each
of your rectangles, now the height of the
rectangle isn't just a function between the x-axis
and the function itself. The height is the difference
between the two functions. So in this case, this would
be the width of a rectangle, and then the height would be
the difference between the two functions. And then you would have another
rectangle right over there. All of them have width
dx, and then the height is the difference
between the functions. And then as you take the
limit as you have super thin, as these rectangles get thinner
and thinner and thinner, and you have more and
more and more them, you are essentially
approximating the area. So this is one way of
conceptualizing it. Or just using, I guess
the tools we already had, the area under the
curve of square root of x. Subtracting from that the area
under the curve of x squared. These are completely legitimate
ways of conceptualizing it. But now let's evaluate it. So what's the antiderivative? We know from the second
fundamental theorem of calculus, we can say that
this is right over here is just going to be the antiderivative
of the square root of x. Well, square root of
x is the same thing as x to the 1/2 power,
so we just increment it. So we say x to the 3/2 power,
and then divide by 3/2. So that's the same thing
as multiplying by 2/3. And we're going to
evaluate that from 0 to 1. And from that, we subtract the
antiderivative of x squared, is x to the third over 3. And we're going to evaluate
that from 0 to 1 as well. Now this first expression,
if we evaluated it at 1. 2/3 times 1 and 3/2. That's just going to be 2/3. And then from that,
we're going to subtract this thing evaluated
at 0, which is just 0. So we're just left with 2/3. And then we're going to
subtract this thing evaluated at 1, minus 1/3 third. And then, also, we're going
to subtract this thing. Well, just to be
clear, we're going to subtract this thing evaluated
at 1, and then from that, it evaluated at 0,
which is just 0. But this whole thing simplifies
to just subtracting 1/3. And there you have it. We get our answer. 2/3 minus 1/3 is equal to 1/3. So this area right over here. Whatever units we're using
is 1/3 square units of area.