Question 1

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?

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?

Accepted Answer

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.

Question 2

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.

Accepted Answer

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.

Question 3

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)

Accepted Answer

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.

Question 4

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?

Accepted Answer

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.

Question 5

Is the definite integral to and from the same point = 0 related at all to the derivative of a constant being = 0

Accepted Answer

Not really. It boils down to something even simpler: x - x = 0.

Question 6

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 .

Accepted Answer

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

Question 7

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!

Accepted Answer

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!

Question 8

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!

Accepted Answer

I think you need to go back and watch a few videos before coming here and learning integrals.....

Calculus, all content (2017 edition)

Course: Calculus, all content (2017 edition) > Unit 4

Definite integral over a single point

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