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## Integral Calculus

### Unit 1: Lesson 8

Properties of definite integrals- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Finding definite integrals using algebraic properties
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Definite integrals over adjacent intervals
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Functions defined by integrals: challenge problem
- Definite integrals properties review

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# Finding definite integrals using area formulas

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

Since definite integrals are the net area between a curve and the x-axis, we can sometimes use geometric area formulas to find definite integrals. See how it's done.

## Video transcript

- [Instructor] We're told to
find the following integrals, and we're given the graph
of f right over here. So this first one is the definite integral from negative six to
negative two of f of x dx. Pause this video and see if
you can figure this one out from this graph. All right we're going
from x equals negative six to x equals negative two, and the definite integral
is going to be the area below our graph and above the x-axis. So it's going to be this
area right over here. And how do we figure that out? Well this is a semicircle, and we know how to find
the area of a circle if we know its radius. And this circle has radius two, has a radius of two. No matter what direction
we go in from the center, it has a radius of two. And so the area of a
circle is pi r squared. So it would be pi times our
radius which is two squared, but this is a semicircle, so I'm gonna divide by two. It's only 1/2 the area of the full circle. So this is going to be four pi over two, which is equal to two pi. All right let's do another one. So here we have the definite integral from negative two to one of f of x dx. Pause the video and see if
you can figure that out. All right let's do it together. So we're going from negative two to one, and so we have to be a
little bit careful here. So the definite integral, you could view it as the area below the, below the function and above the x-axis. But here the function is below the x-axis. And so what we can do is, we can figure out this area, just knowing what we know about geometry, and then we have to
realize that this is going to be a negative value
for the definite integral because our function is below the x-axis. So what's the area here? Well there's a couple of
ways to think about it. We could split it up into a few shapes. So you could just view it as a trapezoid or you can just split
it up into a rectangle and two triangles. So if you split it up like this, this triangle right over here has an area of one times two times 1/2. So this has an area of one. This rectangle right over here has an area of two times one, so it has an area of two. And then this triangle
right over here is the same area as the first one. It's going to have a base
of one, a height of two, so it's one times two times 1/2. Remember the area of a triangle
is 1/2 base times height. So it's one. So if you add up those areas,
one plus two plus one is four, and so you might be tempted
to say oh is this going to be equal to four? But remember our function
is below the x-axis here, and so this is going
to be a negative four. All right let's do another one. So now we're gonna go from
one to four of f of x dx. So pause the video and see
if you can figure that out. So we're gonna go from here to here, and so it's gonna be this
area right over there. So how do we figure that out? Well it's just the formula
for the area of a triangle, base times height times 1/2. So or you could say 1/2 times our base, which is a length of, see we have a base of
three right over here, go from one to four, so 1/2 times three times our height, which is one, two, three, four, times four. Well this is just going to get us six. All right last but not least, if we are going from
four to six of f of x dx. So that's going to be
this area right over here, but we have to be careful. Our function is below the x-axis, so we'll figure out this
area and then it's going to be negative. So this is a half of a
circle of radius one. And so the area of a circle
is pi times r squared, so it's pi times one squared. That would be the area if
we went all the way around like that, but this is
only half of the circle, so divided by two. And since this area is
above the function and below the x-axis, it's going to be negative. So this is going to be equal
to negative pi over two. And we are done.