Main content

## Definite integral as area

Current time:0:00Total duration:3:31

# Worked example: Definite integral by thinking about the function's graph

## Video transcript

- [Voiceover] What I want to
do in this video is see if we can evaluate the definite
integral from negative three to three of the square root
of nine minus x squared dx. I encourage you to pause this
video and try it on your own. I'll give you a hint,
you can do this purely by looking at the graph of this function. All right, I'm assuming
you've had a go at it. So let's just think about,
I just told you that you could do this by using
the graph of the function. Let's graph this function. Let's get a y-axis here. This is my y-axis. This is my x-axis. You might be saying, "Oh
well, what is the graph "of this thing?", it
might not jump out at you. It's been a little while
(mumble) of a hint since you've done conic sections, maybe
in your Algebra class. Let's just remind ourselves. If this function, if we said y is equal to some function of x, which
we see is the square root of nine minus x squared. Then we could say, "Well,
that means that y squared must "must be equal to this thing squared." Which is nine minus x squared. Then we could say, "Y squared
plus x squared is equal "to nine." and you might
recognize this as a circle centered at origin with radius equal to three, the square root of nine. So radius is equal to three,
centered at the origin. Now, they graph of this is
not going to be a circle. This is a function. It would be a circle and it would not be a function anymore if
you said the positive and negative square roots of here. When we took the squares
of both sides we got the bottom back I guess you could say. But up here, we're only talking
about the principal root. When you're talking
about the principal root you're really talking about the top. This is the top of a circle centered at the origin with radius three. So this is top of circle
cause it's the positive square root, so let's draw that. It's gonna have a radius of
three, centered at the origin. This is gonna be negative three,
this is going to be three. This is going to be three right over here. So this function is
going to look like this. It's actually only defined
between negative three and three. The absolute value of x
is greater than three, then you're going to get
a negative value in here. Then you can't take the principal root, as if we're defining it over a positive or non-negative
values I should say. So this is the graph. What is the definite integral
from negative three to three. Well it's just the area under the curve and above the x-axis, it's the stuff that I am shading in in green. Well, what's that? We don't need Calculus to figure that out. You can do this with just
traditional geometry. The area of the entire circle, if there were an entire circle,
would just be pi r squared. So it would be pi times three squared, which is equal to nine pi. Now this is only half
of the entire circle. So we're gonna divide that by two. The area is nine pi over two. So this thing is nine pi over two.