- 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
Based on the fundamental theorem of calculus, we can use antiderivatives to compute integrals. Created by Sal Khan.
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- I am trying to jive in my head how the mandatory +c (constant) after antiderivatives of functions doesn't matter for definite integrals.
- I'm seven years late, but it's because it cancels out.
s stands for the long s
s f(x) = (F(a) + C) - (F(b) + C)
Note that while we sometimes treat C as an arbitrary constant, here we're using a specific antiderivative of f(x), and we're using the same one; so the C's are the same, and can cancel out.(36 votes)
- What is the point of Riemann approximation when we have the Second Fundamental Theorem of Calculus?(43 votes)
- Also, if you ever go into computer programming, it is much easier to use a really precise Riemann sum instead of actually integrating. This is because we can't make the programs flexible or intelligent enough to handle all forms of integrals. (As a side-note, this is why most graphing calculators can only do definite integrals; they can't do the actual integral, only find the area)(23 votes)
- So basically, the antiderivative of a function at "x" tells you the area from 0 to x under the curve?(26 votes)
- In all of my life experience, I have only ever been concerned with f(x) (the curve itself). What exactly is the value in knowing the area under a curve?(4 votes)
- Think of it this way
You're an engineer/architect. I'm your boss. I tell you: "I want an arch bridge that looks awesome. Give me one."
So you create blueprints for a bridge. But how would you know how much cement to use? Let's say the equation for your curve is y = -x²/1000 + 10. You can use integral calculus to find the amount of cement you will need.
If you are a statistician, you will need to find the area of a Gaussian curve more than once. Its equation: ƒ(x) = ae^((x-b)²/-2c²).
If you are counting an infinite series (which comes up a lot), the area under the curve is almost exactly the answer.
If anyone else wants to add a couple other reasons, they can.
Hope this helps!(31 votes)
- why it is called "definite" integral?(1 vote)
- In a definite integral the region of ingratiation is defined. The integral from a to b means that we are integrating from a to b which is a definite length.
In an indefinite integral the region of integration is not defined.(8 votes)
- I don't get the explanation from the "Area between a curve and an axis" exercises. https://puu.sh/mQCj2/e83d91d76e.png Can someone explain it in a not-too-rigorous way?(2 votes)
- Make sure you fully understand the differential calculus lessons or learning integration will be more difficult.
Here is a great introduction to integration and area under the curve.
If you still have any specific questions after going through the math is fun page, write them in the comments and I will reply.
Keep Studying and Have Fun!(5 votes)
- what is the difference between definite integral and indefinite integral?(1 vote)
- Another thing that might help. The answer to an indefinite integral is a function. The answer to a definite integral is a value, a number. For example, in the problem for this video, the indefinite integral is (1/3)x^3 + c. The definite integral, evaluated from 1 to 4 is 21. You use the indefinite integral to find the definite integral evaluated between two values.(5 votes)
- at3:47why does sal take the derivative of x^3?(3 votes)
- It s true that it is easy if you have also watched the videos of how to take anti-derivatives. But if one is only watcing the videos of definite integrals, he just might not know how to take the anti-derivate of a function.(2 votes)
- At3:45, why doesn't Sal add the constant C as we generally do while using the reverse power rule?
Is what Sal did theoretically valid?
Thank you in advance :)(2 votes)
- The + C is only for when taking the indefinite integral of something. Sal is finding the definite integral here(2 votes)
- Is there an intuitive reason for why the integral of f from a->b = F(a)-F(b)?(1 vote)
- Begin by thinking of f as the rate of change and F as the cumulative change in something we're measuring. If we're measuring distance traveled by an automobile, f would be given by the speedometer and F would be given by the odometer. We can see that F, the odometer, is the integral of f, the speedometer, by imagining a situation where we don't have an odometer but need to know how far we've traveled. We can get this answer by constantly checking our speed, say, every second, and multiplying that speed by 1/3600 (the fraction of an hour given by one second) to get the distance traveled in one second, and adding all these increments.
Now, suppose you want to know how far you traveled between time (a) and time (b). You would simply notice the reading of your odometer at time (a) and subtract that number from the reading at time (b). That's the same as F(b) - F(a), which is what the fundamental theorem of calculus tells us.
Make sure you keep this straight: it's F(b) - F(a), not the other way around as you wrote.(3 votes)
So we've got the function, f of x is equal to x squared. And what I'm concerned with, is finding the area under the curve, y is equal to f of x, so that's my y axis. This is my, my, x axis. Then let me draw my function. My function looks like this. At least in the, in the first quadrant. That's where I'll graph it for now. I could also graph it, obviously in the second quadrant. But, what I care about is the area under this curve and above the positive x axis, between, between x equals 1 and x equals 4, x equals 4. And I'm tired of approximating areas. I wanna find the exact area under this curve above the x axis. And the way we denote the exact area under the curve, this little brown shaded area, is using the definite integral. The definite integral from 1 to 4 of f of x, dx. And the way that, or the way I conceptualize where this notation comes from, is we imagine a bunch of infinite, an infinite number of infinitely thin rectangles that we sum up to find this area. Let me draw one of those infinitely thin rectangles, maybe not so infinitely thin. So, let me draw it like this. So, that would be one of the rectangles, that would be another rectangle. This should be reminiscent of a Riemann sum. In fact, that's where the Riemann integral comes from. Think of a Riemann sum where you have an infinite number of these rectangles, where the width of each of the rectangles. This is how I conceptualize it, is dx, and the height of this rectangle is the function evaluated at an x that's within this interval right over here. And so, this part right over here is the area of one of those rectangles, and we were summing them all up. And this kind of an elongated s, reminiscent of a sigma for summing. We're summing up the infinite number, of those infinitely thin rectangles, or the areas of those infinitely thin rectangles between 1 and 4. So that's where the notation of the definite integral comes from. But we still haven't done anything. We've just written some notation that says the exact area of the un- Between 1 and 4, under the curve f of x, and above the x axis. In order to actually do anything really productive with this, we have to turn to the second fundamental theorem of calculus, sometimes called part two of the fundamental theorem of calculus. >> Which tells us, that if f has an antiderivative, so if we have the antiderivative of f, so f of x is derivative, derivative of some function capital F of x, or another way of saying it is, f, capital F of x is the antiderivative, antiderivative of lower case f of x. Then I can evaluate this thing, and we do a whole video on conceptually understanding why this makes sense. We could evaluate this, by evaluating the antiderivative of f, or an antiderivative of f, at 4. And from that, subtract the antiderivative evaluated at 1. So, let's do it for this particular case right over here. So we are taking, I'll just rewrite this statement. Instead of writing f of x, I'll write x squared. So, the definite integral from 1 to 4 of x squared dx. Well, we're just gonna have to figure out what the antiderivative is. So if f of x is equal to x squared, what is capital F of x equal to? What is the antiderivative? Well, you might remember from your power rule, that if you take the derivative with respect to x of x to the third, you are going to get 3x squared, which is pretty darn close to x squared except for this factor of 3. So, let's divide both sides by 3. Let's divide both sides by 3, and you get the derivative of x to the third divided by 3 is indeed x squared. Or, you can say this is the same thing as the derivative with respect to x of, x to the third over 3. Take the derivative of this. It'll be 3 times one third. And then you'll decrement the power, it'll just be x squared. So, this right over here, once again, is x squared. It's just equal to, just equal to x squared. So, in this case, our capital F of x, our antiderivative, is x to the third, x to the third over 3. And so we just have to evaluate that at 4 and at 1, and sometimes the way we would, the, the notation we would use is, we'll say that the antiderivative is x to the third over 3, and we're going to evaluate it, the one I, I always just like to write the numbers up here, at 4 and from that subtracted, evaluated at 1. Sometimes you'll see people write a little line here too, we'll say we're evaluate it at 4 and then at 1. But I'll just do it without the line. If we're gonna evaluate this thing at 4 and from that subtract it, subtract it evaluated at 1, so this going to be equal to 4 to the third power is 64, so it's going to be 64 over 3. Let me color code it, this is, this right over here, is this, right over there and then from that, we're going to subtract this business evaluated at one. Well, when you evaluate it at 1, you get 1 to the third is one over 3. You get one third. So just to be clear, this is this right over there. And then we are ready to just subtract these fractions. 64 over 3, minus one third, is equal to 63 over 3. And 3 goes into 63 exactly, exactly 21 times. So, whatever the units are, the area of this brown area is 21 square units.