Main content

## Integral Calculus

### Course: 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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.

## Want to join the conversation?

- Did sal forget to also calculate the other triangle in the interval [1,4]? .5(3*4)= 6 and .5(1*4)=2 then 2 + 6 = 8 I believe(10 votes)
- This is actually 6. 4-1 = 3 for the base, and then the height is 4. This gives us a (b*h)/2 of (3*4)/2 = 6.

The error was that you need to take the distance between 1 and 4, not from 0.(6 votes)

- So what if he combined the last two problems into one - sum1-6 f(x)dx? Is it just the sum of the areas of the triangle and semicircle, 6 + (-pi/2)?(5 votes)
- I thought area under the curve meant that there couldn't be negative numbers(3 votes)
- When we work with definite integrals, we use
**signed**areas. Areas above the x-axis are "signed" positive and areas below the x-axis are "signed" negative.

The magnitude of the area itself is positive but depending on its orientation, it is assigned a sign.(4 votes)

- This is very convenient to us with or knowledge so far, but what would we do if the equation was shifted down 1 unit (or less)? How would we do the first problem, using our current knowledge, without using trapezoidal sums or Riemann sums, since we're just dealing with a semi-circle?(3 votes)
- Interesting question - I think you could still solve it geometrically, it's just a significantly tougher problem. Check out this resource: https://www.mathopenref.com/segmentarea.html

You could compute the central angle given the radius of the circle and the height offset (however much you wanted to "shift the function down"). Computing the area of the circle segment that's greater than 0 gives the positive part of the area for x in`[-6, -2]`

. The negative part could then be found by computing the area that's within the semicircle but outside of the positive segment, then subtracting that from the rectangle formed by`y=0`

and`y=heightOffset`

to get the negative area for x in`[-6, -2]`

. (Disclaimer: I'm just sort of mulling through this in my head, so I'm not sure if that's exactly right, but it seems like the correct approach.)

That being said, the point of this problem is to be relatively convenient and provide a good way to grasp the concept at hand (integrals as areas under curves). The alternatives you listed are designed to solve more complex problems; not everything has to be done using the least available knowledge.(3 votes)

- If we were to switch places with the numbers for each integral, for example, if we switched the 4 by the 6 and vice-versa in the 4th example shown in the video, would the area of the semi-circle be considered positive despite being below the x-axis? (I'm new to integration so I don't know much about all the properties yet, but I think I saw that somewhere, so I am not so sure if what I said is true)(1 vote)
- Area is always positive.

However any area underneath the x-axis is negative when perform the integration. If you remember the explanation Sal gave using rectangles to approximate an area you realise if f(x)<0 then area of the rectangle will come out as negative f(x) dx is less than 0.

If switch the bounds of the integrand then the result will switch signs.

Try integrating from some function f(x) from a to b will lead result of

F(b)-F(a)

while swapping the bounds gets you

F (a)-F (b) = -( F (a) - F (b) ) which is opposite the above example(3 votes)

- For the third question, wouldnt it be 8 bc 6+ 2 from the second triangle on x=3 to x=4(1 vote)
- Even if you do it like that, you'll get (1/2)(2)(4) + (1/2)(1)(4) = 4 + 2 = 6. The area of the triangle from 1 to 3 is 4, not 6(1 vote)

## 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.