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AP®︎/College Calculus AB
Unit 8: Lesson 4
Finding the area between curves expressed as functions of x- Area between a curve and the x-axis
- Area between a curve and the x-axis: negative area
- Area between a curve and the x-axis
- Area between curves
- Worked example: area between curves
- Area between two curves given end points
- Area between two curves
- Composite area between curves
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Area between a curve and the x-axis: negative area
AP.CALC:
CHA‑5 (EU)
, CHA‑5.A (LO)
, CHA‑5.A.1 (EK)
With integrals, it can be helpful to introduce the notion of "negative area". See why! Created by Sal Khan.
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It broke my mind. I always thouhgt that S(circle) = Pr^2.
if r = 1, then S must be eaqual to P.But, in this video, it seems to be 4. Can somebody explain where is mistake in my thoughts?(12 votes)
You are right that the area of a circle with radius of 1 would be equal to pi. What Sal is showing here, though, is how to find the area between the curve described by y = f(x) = cos x and the x-axis, which is not quite circular. (For instance, the circumference of a circle with a radius of 1 would be 2pi, while the variable curve of cos x is somewhat longer, as shown in Sal's graph.)(35 votes)
When should we use delta x as opposed to dx in format? I've researched it, but no one gives me a really decent answer.(7 votes)
Δx is a term you can use for any generic change in x. For example, your final position minus your starting position gives you your total change in position (Δx). dx is only used in calculus, and it is important that you understand the conceptual difference. Whereas Δx represents an arbitrary, but still measurable, difference, dx represents an infinitesimal difference, and refers to a very specific quantity within a formula - a differential. So, you should use Δx when you are just referring to a general change in x ( S = x + Δx ), and use dx only when dealing derivatives and integrals ( dy/dx = 3x^2 ). Happy mathing!(26 votes)
Sal! could you explore negative areas a little more?
How come area from x = 0 to x = 2pi is 0? shouldn't it be 4? or are we not measuring space?(1 vote)- "Areas" measured by integration are actually signed areas, meaning they can be positive or negative. Areas below the x-axis are negative and those above the x-axis are positive.
If you are integrating from 0 to 2*pi and getting a result of 0, then half of the area is positive and half of the area is negative; they are, in a sense, canceling each other out.
This will happen if you integrate sin(x) from 0 to 2*pi. If you want the total area, in the geometric sense of the word, then there are two options:
First, you can integrate from 0 to Pi and then from Pi to 2*pi. The "area" from Pi to 2*Pi will be negative. Make this number positive, and add it to the area from 0 to Pi.
Or, if you think of this problem graphically, you will see that the region from 0 to Pi and the region from Pi to 2*Pi are congruent. So find the positive area and then multiply by 2. Thus, 2*integral( sin(x),0,Pi) will be the total area bound by the graph of sin(x) from 0 to 2*pi.(19 votes)
how is this even possible if we don't know for sure if there is a constant at the end of that antiderivative?(3 votes)
From what I understand of your question, you are asking why we don't keep the +c at the end of the antiderivatives to signify that they could be shifted I up or down by any constant c when we solve for the integral. The simple reason is that in the process of evaluating the integral, the constants will always cancel out each other when we subtract the antiderivatives. For example, if we keep the c while evaluating integral(2x+1) from 1 to 2, we will see how this happens.
Integral(2x+1)= (x^2+x+c) evaluated at 2 - (x^2+x+c) evaluated at 1
=(4+2+c)-(1+1+c)= (6+c)-(2+c)=6-2+c-c Here the c's cancel out.
=4. We are left with no c's in the final answer.
This cancellation occurs when evaluating any integral, so there is no need to include them at all when you solve for one; it is already known that they will cancel. Hope this helped!(7 votes)
- I thought the derivative of cos was equal to -sin, not just normal sin?(0 votes)
- It is but Sal is not taking the derivative of cos. He's taking the anti-derivative of cos. Meaning what value or function, when you take the derivative of it, you get cos? The answer is sin.(13 votes)
- If asked 'what is the area below the curve' or something to that affect which would result in Sal's workings, what exactly would the answer be; would it be 4 or 0?(2 votes)
- One person said 0, one person said 4.
He asks for the area 'below the curve' here. If he asked for the area 'between' the curve and the x axis, would the answer be 4? The negative area is above the curve.
If the teacher asked for the shaded area, wouldn't we consider all areas as positive in the end?(5 votes)
What is the proper example, in any field of knowledge, for the negative area? What is it intuitively?(3 votes)
Its got tremendous applications in physics especially. Consider a Force vs. displacement curve, the integral F.dx will give you work done. If area is +ve so is work and vice versa. You can judge whether a force is conservative or not only by looking at the plot. PS the limits must be chosen wisely though, from left to right the area is conventionally +ve and from right to left conventionally -ve(2 votes)
If you are finding the area of the sin function do you use radians or pi as units? and if you have a function in terms of x are the units 1x by 1y square units?(0 votes)- Pi is not a unit, it's a constant. Generally you will use radians because they are easier to work with than degrees. If the units are not given you can indeed just say the answer is A units^2.(7 votes)
- In which case would we consider a negative sign outside the integral?
(not particularly pertaining to the question above,but in general)
Ive seen this done a couple of times and im confused..(3 votes) - how can we find the definite integral of a modulus function say |x+2| with limits -5 to 5(2 votes)
∫ |x+2| dx from -5 to 5
You can remove the absolute value sign by multiplying by -1 when (x + 2) is negative.
(x + 2) is negative when it is less than -2.
So then you have 2 functions (x+2) and (-x - 2)
Remember you can always split up the interval and add the integrals.
If you need to review this concept watch this:
https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/properties-definite-integral/v/breaking-integral-interval
Splitting up the integral we have:
∫ -x-2 dx (from -5 to -2) + ∫ x+2 dx (from -2 to 5)
= [-x²/2 - 2x] (-5 to -2) + [x²/2 + 2x] (-2 to 5)
= [-2 + 4 + 25/2 - 10] + [25/2 + 10 - 2 + 4]
= 9/2 + 49/2 = 29
Hope that helps.(2 votes)
Video transcript
What I've got here is the graph
of y is equal to cosine of x. What I want to do is
figure out the area under the curve y is equal to
f of x, and above the x-axis. I'm going to do it
over various intervals. So first, let's think about
the area under the curve, between x is equal to 0 and
x is equal to pi over 2. So we're talking about
this area right over here. Well, the way we denote it is
the definite integral from 0 to pi over 2 of cosine of x dx. And remember, all
this is, is kind of, it's reminiscent of
taking a sum of a bunch of super thin rectangles
with width dx and height f of x for each of
those rectangles. And then you take
an infinite number of those infinitely
thin rectangles. And that's kind of what this
notation is trying to depict. But we know how to
do this already. The second fundamental
theorem of calculus helps us. We just have to figure out what
the antiderivative of cosine of x is. Or what an antiderivative
of cosine of x is. Evaluate it at pi
over 2, and from that subtract it evaluated at 0. So what's the antiderivative
of cosine of x? Or what's an antiderivative? Well, we know if we take
the derivative-- let me write this up here. We know that if we take the
derivative of sine of x, we get cosine of x. So the antiderivative of
cosine of x is sine of x Now, why do I keep saying sine
of x is an antiderivative? It's not just the
antiderivative. Well, I could also
take the derivative of sine of x plus any
arbitrary constant, and still get cosine of x. Because the derivative
of a constant is 0. This could be pi, this could
be 5, this could be a million, this could be a googol, this
could be any crazy number. But the derivative of this is
still going to be cosine x. So when I say that we just have
to find an antiderivative, I'm just saying, look, we just have
to find one of the derivatives. Sine of x is probably the
simplest, because in this case the constant is 0. So let's evaluate. So one way that we
can denote this, the antiderivative cosine, or
an antiderivative of cosine of x is sine of x. And we're going to
evaluate at pi over 2. And from that, subtract
and evaluate it at 0. So this is going to
be equal to-- let me do it right over here. This is going to be equal
to sine of pi over 2 minus sine of 0, which is equal
to sine of pi over 2 is 1. Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region
right over here, this area is equal to 1. Now let's do
something interesting. Let's think about the area. Let's think about the area
under the curve between, let's say pi over
2 and 3 pi over 2. So between here and here. So we're talking
about this area. We're talking about that
area right over here. This is 3 pi over 2. So once again, the
way we denote the area is the definite integral from pi
over 2 to 3 pi over 2 of cosine of x dx. The antiderivative-- or an
antiderivative of cosine of x is sine of x. Evaluated at 3 pi
over 2 and pi over 2. So this is going
to be equal to sine of 3 pi over 2 minus
sine of pi over 2. What's sine of 3 pi over 2? If we visualize the unit
circle really fast, 3 pi over 2 is going all the way 3/4
around the unit circle. So it's right over there. So sine is the y-coordinate
on that unit circle. So it's negative 1. So this right over
here is negative 1. This right over here,
sine of pi over 2 is just going
straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1,
which is equal to negative 2. We've got a negative area here. We got a negative area. Now how does that make sense? We know in the real world,
areas are always positive. But what is negative 2
really trying to depict? Well, it's trying to
sign it based on the idea that now our function
is below the x-axis. So we could kind of think
we have an area of 2, but it's all below the
x-axis in this case. And so it is signed
as negative 2. The actual area is 2, but
since it's below the x-axis, we get a negative
right over here. Now let's do one
more interesting one. Let's find the definite
integral from 0 to 3 pi over 2 of
cosine of x dx. Now this is just denoting
this entire area. Going from 0 all
the way 3 pi over 2. And what do you think
is going to happen? Well, let's evaluate it. It's going to be sine of 3
pi over 2 minus sine of 0. Which is equal to
negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under, all this orange
area that I just outlined, is clearly not negative, or any
area isn't a negative number. And the area isn't even 1. But what just happened here? Well, we saw in the first case
that this first area was 1. The area of this first
region right over here is 1. And then the area of the second
region, we got negative 2. And so one way to
interpret it is that your net area above
the x-axis is negative 1, or another way to say it,
your net area is negative 1. So it's taking the 1
region above the x-axis and subtracting the 2 below it. So what the definite
integral is doing when we evaluate it using the
second fundamental theorem of calculus, it's
essentially finding the net area above the x-axis. And if we get a
negative number, that means that the net area is act--
that actually most of the area is below the x-axis. If we get 0, then that
means it all nets out. And if you want to
see a case that's 0, take the integral from 0 all
the way all the way 2 pi. And this will evaluate to 0
because you have an area of 1 and another area of
1, but it nets out with this area of negative 2. Let's try that out. So if we go from 0 to
2 pi of cosine of x dx, this is going to be equal to
sine of 2 pi minus sine of 0, which is equal to 0 minus 0,
which is indeed equal to 0. Now clearly there was
area here, but all of the area that
was above the x-axis netted out with all the area
that was below the x-axis.