Class 12 math (India)
- The fundamental theorem of calculus and definite integrals
- Intuition for second part of fundamental theorem of calculus
- Area between a curve and the x-axis
- Area between a curve and the x-axis: negative area
- Definite integrals: reverse power rule
- Definite integral of rational function
- Definite integral of radical function
- Definite integral of trig function
- Definite integral involving natural log
- Definite integrals: common functions
- Area using definite integrals
Area between a curve and the x-axis: negative area
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.)(36 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!(27 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:
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)
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.