If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC>Unit 8

Lesson 4: Finding the area between curves expressed as functions of x

# 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.

## Want to join the conversation?

• 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?
• 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.)
• 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.
• 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!
• 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.
• how is this even possible if we don't know for sure if there is a constant at the end of that antiderivative?
• 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!
• I thought the derivative of cos was equal to -sin, not just normal sin?
• 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.
• 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?
• 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?
• What is the proper example, in any field of knowledge, for the negative area? What is it intuitively?
• 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
• 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?
• 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.
• 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..
• how can we find the definite integral of a modulus function say |x+2| with limits -5 to 5
• ∫ |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:

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.