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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 6

Lesson 6: Applying 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
- Definite integrals properties review

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# Definite integral over a single point

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

What happens when the bounds of your integral are the same?

## Want to join the conversation?

- Alright, I'm a little curious where my gap in knowledge is with this one. It is understandable that the resultant area = 0, but if I say that an integral is the anti-derivative, whereas the function being integrated is the anti-derivative, shouldn't that c signify a portion of the original function?

Say like this -

The derivative of x^2 is 2x

At point 2, c, the rate of change for x^2 is 4 because 2*x = 4.

Thus shouldn't the anti-derivative be x^2=2^2=4, not 0?(16 votes)- The antiderivative of 2x is x^2+C.

The indefinite integral is the same as the antiderivative, but the definite integral is not.

If the indefinite integral of f(x) is F(x), then the definite integral from a to b is F(b) - F(a). We can choose the C in the antiderivative to be anything, but it has to be the same for both. C = 0 is the most convenient.

So the definite integral of 2x from c to c is c^2 - c^2 which equals 0.(8 votes)

- Im confused, how am I supposed to solve a integral? Antiderivatives, Reimann Sum? I am just getting patches of different things about integrals but I am not really getting the whole picture.(11 votes)
- Solving integrals is far more difficult that derivatives. So, the methods for solving them will gradually be introduced. But, you need to understand what integrals represent first.

Antiderivatives are a special type of integral. So, if you can take some of the basic antiderivatives then you are on your way to using integration in a more general way.(16 votes)

- Is the definite integral to and from the same point = 0 related at all to the derivative of a constant being = 0(3 votes)
- Not really. It boils down to something even simpler: x - x = 0.(14 votes)

- What is the difference between F(c) and the definite integral from c to c of f(x) dx?

For example, the definite integral from c to c of 2x dx is 0, but F(c) is c^2?

What is the difference in meanings of the two, like the definite integral is supposed to represent the total change in x over the interval, but what does F(c) mean? (I'm asking this in AP context, because they usually ask you to explain the meaning of things in the FRQs)(5 votes)- I'll take issue kindly with one thing you said and hope it might bring some focus. "the definite integral is supposed to represent the total change in x over the interval"

Actually the definite integral is the SUM of the TINY changes in _y_ over the x-interval. (The total change in x over an interval, Dx, isn't defined by the function, it is defined by the interval itself since you are speaking about a function of x.)

Whenever you think of a definite integral it's good to begin with a picture in your mind of a Reiman sum rectangle configuration where you add the areas, and imagine the rectangles becoming infinitely many and infinitely thin. If you think of it this way you realize that an integral over an interval with the same start and endpoint MUST be zero no matter the function. There is no interval to calculate the area over.

One more thing - I made a lot of progress in my understanding of the integral both definite and indefinite by doing a lot of d-r-t physics/calculus type problems - The kind that has you integrate - initial value problems and such.(6 votes)

- If we take all points between
`a`

and`b`

, the integral from and to the same point will be 0, but we all know together they create a definite area. What am I missing?(4 votes)- What you might be missing is we are talking about the situation when the a and b of the interval over which we will integrate [a,b] are equal to the same value, say c.

If c is a point on the line and a=c and b=c then a and b are both equal to the same point.

Does a point have any width?

Does a line have any width?

A point is a zero dimensional object with no width or breadth.

A line is the extension of a point into a 1 dimensional object, which has length, but no width.

If we go from a=c to b=c, we are just staying on c, a point, whose extension up to the point on the graph f(c) produces a line, which has no area at all.

Now if a≠b, the the distance between a and b is a line segment and when we extend that 1 dimensional line segment up to the function values it produces , we get a 2 dimensional figure (a section of a plane) which does have area.(4 votes)

- how is it possible that the line has no width ?

it does occupy space .

what if we divide the area between two points into little lines you are saying that it will be equal zero .(3 votes)- it is the definition of a one dimensional object that it has no width. it doesen't really exist but in this case we are trying to represent a one dimensional object with a three dimensional drawing but the concept is all that matters(3 votes)

- In some of the previous videos, the integral of f(x) would be F(x), where f(x) = F'(x). But in this video the integral of f(x) over a single point is 0. I know there is a difference between taking antiderivatives and taking the area under a curve, but the mathematical notation seems to be the same. Is there a simple explanation for this? Thanks!(2 votes)
- The reason an integral works, in essence, is because we're using the idea of an infinitesimal (which is sometimes a controversy in math), that is, an infinitely small part, basically the "opposite" of infinity. The dx, in the integral, represents this infinitesimal, it is an incredibly small width (change in x), such that as we take the limit and all that it is basically 0. This is why it doesn't really work for a single point. When we're taking more than 1 dx we get something, when we only take 1 dx, and the limit of dx approaches 0, we're getting 0. There are better proofs for this by using the actual Sigma notation (Riemann sum) for an integral.

Another good proof would be the fundamental theorem of calculus, which basically relates the anti-derivate to the integral and derivatives, it states:

Given the integral F(x) and it's antiderivative f(x) such that f'(x) = F(x), and b is the upper bound of integration, and a is the lower bound, we have:

F(x) = f(b) - f(a)

As you can see when a = b (the upper bound is equal to the lower bound), we get x - x = 0, we get one value, and subtract that same value from it, resulting in 0. This is why the integral of a single point is simply 0.

You could relate it to the thing strips as well (the tiny tiny Reimann rectangles), as we make this rectangle smaller and smaller, it's width gets closer and closer to 0... well, what's the area of a rectangle when the width is 0? It's 0. Then you might be wondering why calculus and the Reimann sum and integrals even work at all! Yep... therein lies the slight controversy over the infinitesimal. However, it works, and the ideas and proofs behind it all are sound, as we can all see since calculus indeed works. The real issue is the controversy over infinity, which is present everywhere in maths, but despite being a controversy it doesn't affect the validity of the maths present!(4 votes)

- If the area on the same point is 0, then reinmann sum won't work as it slices the region below the graph into infinitesimally small segments as the change in X approaches zero....can someone please explain why it works?(2 votes)
- WHY ARE WE USING NOTATION FOR DEFINIATE INTEGRALS and the phrase definite integral but not graphing or calculating them! We're just playing with graphs of the derivation of an integral. I get that the derivative of a function is also the integral of its derivative, by why this obsession with f(x)?? There's an FX hair design in Charlottesville. I'll give your their number if you want. USE THE PROPER NOTATION AND STANDARD FORM. This is the section on integrals not derivatives!(1 vote)
- I think you need to go back and watch a few videos before coming here and learning integrals.....(3 votes)

- Would you please provide an specific example of a problem simplified by using this concept? I'm having trouble visualizing how it could be useful.(4 votes)

## Video transcript

- [Voiceover] We've already
taken definite integrals and we've seen how they
represent or denote the area under a function
between two points and above the X axis. But let's do something interesting. Let's think about a definite
integral of F of X DX, It's the area under the curve,
F of X, but instead of it being mean between two different X values, say A and B like we see in multiple times, let's say it's between the same one. Let's say it's between C and C. Let's say C is right over here. What do you think this
thing right over here is going to be equal to? What does this represent? What is this equal to? I encourage you to pause the video and try to think about it. Well if you try to visualize
it, you're thinking, well the area under the curve F of X, above the X axis, from X
equals C to X equals C. So this region, I guess we could call it, that we think about
it, does have a height. The height here is F of C. What's the width? Well there is no width,
we're just at a single point. We're not going from C
to C plus some delta X or C plus some even very small change in X or C plus some other very small a value. We're just, saying at the point C. When we're thinking
about area we're thinking about how much two-dimensional
space you're taking up. But this idea, this is
just a one-dimensional, I think you could think
of it as a line segment. What's the area of a line segment? Well a line segment has no area. So this thing right over here is going to be equal to zero. Now you might say, I get that. I see why that could make sense, why that makes intuitive sense. I'm trying to find the area of a rectangle where I know it's height,
but it's width is zero. So that areas going to be zero is one way to think about it. But Sal, why are you even
pointing this out to me? As well see, especially
when we do more complex definite integration
problems and solving things sometimes recognizing this will help you simplify an integration
problem dramatically. Or you could work to be able
to get to a point like this so that you can cancel things out. Or you can say, hey that
thing right over there is just going to be equal to zero.