# Fundamental Theorem What?

November 24, 2008 1:49 PM Subscribe

Someone explain to me, or show me where I can go to grasp better, the Fundamental Theorem of Calculus.

I'm currently finishing my first semester of Calculus. Going on to take Calc II next semester. So far this year, every single thing in calculus has come to me very naturally, intuitively, and instinctively. I plan on majoring in math in college. Until this.

I simply cannot understand why an indefinite integral of a function evaluated at b minus an indefinite integral of a function evaluated at a equals the definite integral of the function from a to b, or why that's the area under the curve from a to b.

I can use sigma notation and Riemann sums to find the area... but I don't understand how antidifferentiation does the same thing.

I see the first derivative of a function as the slope of that function at a point, that's fine. But I don't see how you can back it up from there and find the area with a reverse operation. It's eluding me.

Now, I understand

I'm looking for either a blog article, youtube video, calc lecture recording, anything that will help me understand this. A visual proof is fine. A formal proof is fine. I've taken a six-week course in mathematical notation and other fun stuff, so even a very formal explanation will suffice, as long as it does so well. I also have a

If anyone wants to just explain it it right here, that's fine, too.

Thanks!

I'm currently finishing my first semester of Calculus. Going on to take Calc II next semester. So far this year, every single thing in calculus has come to me very naturally, intuitively, and instinctively. I plan on majoring in math in college. Until this.

I simply cannot understand why an indefinite integral of a function evaluated at b minus an indefinite integral of a function evaluated at a equals the definite integral of the function from a to b, or why that's the area under the curve from a to b.

I can use sigma notation and Riemann sums to find the area... but I don't understand how antidifferentiation does the same thing.

I see the first derivative of a function as the slope of that function at a point, that's fine. But I don't see how you can back it up from there and find the area with a reverse operation. It's eluding me.

Now, I understand

*that*it works. That is, I've done it dozens and dozens of times, and it always works, so I believe that it works. However, I want to understand why.

I'm looking for either a blog article, youtube video, calc lecture recording, anything that will help me understand this. A visual proof is fine. A formal proof is fine. I've taken a six-week course in mathematical notation and other fun stuff, so even a very formal explanation will suffice, as long as it does so well. I also have a

*very*strong grasp of everything in calc I up to now.

If anyone wants to just explain it it right here, that's fine, too.

Thanks!

My graduate class in computational science and engineering gave me a new insight into the fundamental theorem of calculus. (FToC)

Unfortunately it involves vector calculus and in a kind of hilarious-but-only-to-mathematicians-and-engineers kind of way, you need to have a much deeper understanding of math before you can get this (aptly named) fundamental bit.

The new insight was regarding the FToC as a 1D case of the Divergence Theorem. I'm providing the link for completeness. I don't think it will necessarily help you since as I say, the vector (multivariable) calculus is a little heady for a first year.

One thing that helped me was to think about how the derivative of a function has information about the function. It sort of tells you "where the function is going". The slope is a predictor.

The integral has some information too and we might reasonably expect it to be "the opposite" of what the derivative has. And that information (via the FToC) is that the integral of a function tells us "what a function has been doing". Pardon the license I took with math there. I'm an engineer not a mathematician. An integral is a "history" if you like. All of the function's history from x=a to x=b is buried in the definite integral.

I hope that helps.

posted by KevCed at 2:26 PM on November 24, 2008

Unfortunately it involves vector calculus and in a kind of hilarious-but-only-to-mathematicians-and-engineers kind of way, you need to have a much deeper understanding of math before you can get this (aptly named) fundamental bit.

The new insight was regarding the FToC as a 1D case of the Divergence Theorem. I'm providing the link for completeness. I don't think it will necessarily help you since as I say, the vector (multivariable) calculus is a little heady for a first year.

One thing that helped me was to think about how the derivative of a function has information about the function. It sort of tells you "where the function is going". The slope is a predictor.

The integral has some information too and we might reasonably expect it to be "the opposite" of what the derivative has. And that information (via the FToC) is that the integral of a function tells us "what a function has been doing". Pardon the license I took with math there. I'm an engineer not a mathematician. An integral is a "history" if you like. All of the function's history from x=a to x=b is buried in the definite integral.

I hope that helps.

posted by KevCed at 2:26 PM on November 24, 2008

The b-a stuff is pretty simple, if you allow it to be. The integral, evaluated at a, is the total accumulated area (work with me here) from the starting point of all starting points, i.e. x=0, to a. The integral at b is the total accumulated area from 0 to b. Subtract the two, and you get how much more area accrued while going from a to b.

Now, as to why the integral works at all:

Think of physics. If you've not taken it, you're going to have to trust me. Think of F(t) as the velocity of your car going down a road plotted on a time axis. It's mostly steady, but it ebbs and flows, up and down with traffic. Your speed goes up -you accelerate - and the slope of the curve us positive. You hit the brakes, and the slope is negative, i.e. decelerate. Ok, so that's the derivative of speed.

But say you wanted to find out how far you've gone. You don't have an odometer, but let's say you recorded your speed every minute. You could find the distance by taking the speed*the amount of time at each speed (a bunch of segments). (check your units: s times ft/s=ft) and adding these up.

Now, that's kinda inaccurate - it doesn't account for accelerating and deceleration between each speed, your cruis control sucks, etc. But it would be even worse fi you only recorded the speed every five minutes. Your accuracy would get better, conversely, if you recorded your speed every half a minute. Ok, better yet, every 15 seconds. Etc...you wind up with more rectangles but better accuracy. Now do this continuously - you've made a de facto odometer.

Another more loosey-goosey or holistic explanation looks at dimensions. Zero dimensions is a point. One dimension is a line. Two is area (square), three is volume (cube). In a like way, a zero-order function is a constant - a horizontal line. The next level of function is a line with non-zero slope, which has x^1. Then comes a parabola, more interesting, and using x^2. And up the line to more and more "interesting" functions. Physical measurements, in my example above, are distance, speed, acceleration, and jerk. To go to a higher dimension, you've gotta stretch the dimensions you've already got, and to go to a lower dimension, you've gotta flatten your object. To go up in "interestingness" level of functions, you've got to slap on another power of x, and to find a more boring function, you deduct one power of x. To go down one level on the properties, you take the derivative - which tells you the rate at which you're changing something (speed is the rate you're changing distance, acceleration is the rate you're changing speed, and jerk is the change in acceleration) To go the other way, you take the integral. The distance is the total accumulation of speed. The speed is the total accumulation of acceleration - if all your acceleration and deceleration sum to zero, you're not moving.

posted by notsnot at 2:30 PM on November 24, 2008 [1 favorite]

Now, as to why the integral works at all:

Think of physics. If you've not taken it, you're going to have to trust me. Think of F(t) as the velocity of your car going down a road plotted on a time axis. It's mostly steady, but it ebbs and flows, up and down with traffic. Your speed goes up -you accelerate - and the slope of the curve us positive. You hit the brakes, and the slope is negative, i.e. decelerate. Ok, so that's the derivative of speed.

But say you wanted to find out how far you've gone. You don't have an odometer, but let's say you recorded your speed every minute. You could find the distance by taking the speed*the amount of time at each speed (a bunch of segments). (check your units: s times ft/s=ft) and adding these up.

Now, that's kinda inaccurate - it doesn't account for accelerating and deceleration between each speed, your cruis control sucks, etc. But it would be even worse fi you only recorded the speed every five minutes. Your accuracy would get better, conversely, if you recorded your speed every half a minute. Ok, better yet, every 15 seconds. Etc...you wind up with more rectangles but better accuracy. Now do this continuously - you've made a de facto odometer.

Another more loosey-goosey or holistic explanation looks at dimensions. Zero dimensions is a point. One dimension is a line. Two is area (square), three is volume (cube). In a like way, a zero-order function is a constant - a horizontal line. The next level of function is a line with non-zero slope, which has x^1. Then comes a parabola, more interesting, and using x^2. And up the line to more and more "interesting" functions. Physical measurements, in my example above, are distance, speed, acceleration, and jerk. To go to a higher dimension, you've gotta stretch the dimensions you've already got, and to go to a lower dimension, you've gotta flatten your object. To go up in "interestingness" level of functions, you've got to slap on another power of x, and to find a more boring function, you deduct one power of x. To go down one level on the properties, you take the derivative - which tells you the rate at which you're changing something (speed is the rate you're changing distance, acceleration is the rate you're changing speed, and jerk is the change in acceleration) To go the other way, you take the integral. The distance is the total accumulation of speed. The speed is the total accumulation of acceleration - if all your acceleration and deceleration sum to zero, you're not moving.

posted by notsnot at 2:30 PM on November 24, 2008 [1 favorite]

Seconding the Wikipedia article. The "Geometric intuition" section is quite well explained and might help you get a better hold on it.

Here's a brief rephrasing:

1. Imagine an "area function" A(x) for a given function f(x). When you plug in a value into A(x), it will spit back out the area under f(x) between the the y-axis and the plugged-in value.

2. Now imagine you want to find the area under f(x) between x=a and x=b, where b>a. The area function A(x) tells us the area between x=0 and the inputted value, so the expression A(b)-A(a) will give us the area of the region we're looking for.

3. But wait! Using some cool algebra, we can prove that A(x) is an antiderivative of f(x), using the definition of A(x) and the definition of an antiderivative. And, since all antiderivatives of a given function are just vertical translations of each other, the expression F(b)-F(a) is the same for any antiderivative F(x) you choose.

4. So for any antiderivative F(x), we have that F(b)-F(a) = A(b)-A(a), which is the area under f(x) between x=a and x=b.

Basically, the way to get your head around this is to FIRST construct this magical "area function" and THEN show that it's actually an antiderivative.

posted by DLWM at 2:40 PM on November 24, 2008

Here's a brief rephrasing:

1. Imagine an "area function" A(x) for a given function f(x). When you plug in a value into A(x), it will spit back out the area under f(x) between the the y-axis and the plugged-in value.

2. Now imagine you want to find the area under f(x) between x=a and x=b, where b>a. The area function A(x) tells us the area between x=0 and the inputted value, so the expression A(b)-A(a) will give us the area of the region we're looking for.

3. But wait! Using some cool algebra, we can prove that A(x) is an antiderivative of f(x), using the definition of A(x) and the definition of an antiderivative. And, since all antiderivatives of a given function are just vertical translations of each other, the expression F(b)-F(a) is the same for any antiderivative F(x) you choose.

4. So for any antiderivative F(x), we have that F(b)-F(a) = A(b)-A(a), which is the area under f(x) between x=a and x=b.

Basically, the way to get your head around this is to FIRST construct this magical "area function" and THEN show that it's actually an antiderivative.

posted by DLWM at 2:40 PM on November 24, 2008

notsnot, I think he understands that the integral = 'the area under there curve'. The problem is he don't understand why "the area under there curve" is equal to

Take your speed example. Let's say your car's distance from a start point can be determined by some some function that can be differentiable function distance(t).

The car's speed at any point in time can be found by distance'(t) = speed(t). We know from calculus that the distance the car travels from time t1 to time t2 can be found by taking the absolute integral of speed(t) from time t1 to time t2.

However, common sense also tells us that we can find the difference simply by subtracting distance(t2) - distance(t1).

The fundamental theorem of calculus tells us that the "common sense" answer and the calculus answer are the same.

posted by delmoi at 2:44 PM on November 24, 2008

**both**the integral from A to B**and**the antiderivative at A minus the antiderivative at B.Take your speed example. Let's say your car's distance from a start point can be determined by some some function that can be differentiable function distance(t).

The car's speed at any point in time can be found by distance'(t) = speed(t). We know from calculus that the distance the car travels from time t1 to time t2 can be found by taking the absolute integral of speed(t) from time t1 to time t2.

However, common sense also tells us that we can find the difference simply by subtracting distance(t2) - distance(t1).

The fundamental theorem of calculus tells us that the "common sense" answer and the calculus answer are the same.

posted by delmoi at 2:44 PM on November 24, 2008

The integral from a to b of f'(t) is, intuitively, the area under the curve traced by the graph of f'(t) from a to b. Adding a small value (call it epsilon) to b increases this area by epsilon * f'(b). You can convince yourself of this by drawing a rectangle under the graph at b and letting the width go to 0.

But adding epsilon to b increases the value of f(b) by the very same value, epsilon * f'(b), by the definition of the derivative. The two functions change continuously, so they differ only by a constant. To make them equal, notice that the integral from a to a of f'(t) is 0. So we add the value of the original function, f(a), a constant, to get that (pardon the terrible ascii integral)

posted by panic at 4:47 PM on November 24, 2008

But adding epsilon to b increases the value of f(b) by the very same value, epsilon * f'(b), by the definition of the derivative. The two functions change continuously, so they differ only by a constant. To make them equal, notice that the integral from a to a of f'(t) is 0. So we add the value of the original function, f(a), a constant, to get that (pardon the terrible ascii integral)

/` b | f'(t) dt + f(a) = f(b) `/ aand with a little algebra, the fundamental theorem of calculus comes out:

/` b | f'(t) dt = f(b) - f(a). `/ a

posted by panic at 4:47 PM on November 24, 2008

*I see the first derivative of a function as the slope of that function at a point, that's fine. But I don't see how you can back it up from there and find the area with a reverse operation. It's eluding me.*

The short answer: mean value theorem.

The long answer: Here goes the formal explanation to complement the excellent solutions other people have given (Where were you guys for my single-variable calculus class anyway?), which is a simplified version of the Wikipedia proof done in reverse:

Let f be a function. We're looking for a function F such that f = derivative(F) = F'. Suppose that one exists. (This is the anti-derivative.)

We know from Riemann sums that

area(f(x) | from a to b) = A = sum(f(c_i)(delta x_i)).

area(f(x) | from a to b) = A = sum(f(c_i)(delta x_i)).

We can replace f(c_i) with F'(c_i) since f = F'. So

A = sum(F'(c_i)(delta x_i)). [Equation 1]

A = sum(F'(c_i)(x_i - x_{i-1}).

A = sum(F'(c_i)(delta x_i)). [Equation 1]

A = sum(F'(c_i)(x_i - x_{i-1}).

**The mean value theorem**

We also know the mean value theorem: If a function G is continuous on an interval [a, b], then there exists some k such that a < k < b and also

G'(c) = (F(b) - F(a))/(b - a).

G'(c) = (F(b) - F(a))/(b - a).

We want to figure out how to rejigger Equation 1 so that we can use the mean value theorem. From Riemann sums, we know that we can choose any c_i we feel like. Usually, calculus classes teach you to choose the left point of all those itty-bitty rectangles. Or the right point. Or the center point. Or do a quadratic approximation (Simpson's Rule).

**Cleverly choosing our Riemann sum**

We can instead choose something clever: We know each triangle interval is something like [x_{i-1}, x_i]. We know, by the mean value theorem, that there exists a k such that x_{i-1} < k < x_i. Why not let c_i be k?

Then we get this:

A = sum(F'(k_i)(delta x_i)). [Equation 1]

A = sum(F'(k_i)(delta x_i)). [Equation 1]

But, according to the mean value theorem:

A = sum((F(x_i) - F(x_{i-1}))/(x_i - x_{i-1}) * (x_i - x_{i-1})).

A = sum((F(x_i) - F(x_{i-1}))/(x_i - x_{i-1}) * (x_i - x_{i-1})).

This cancels out into

A = sum((F(x_i) - F(x_{i-1})).

A = sum((F(x_i) - F(x_{i-1})).

**Telescoping**

It seems like we can't simplify this any further. But, if we telescope the sums out, we get this:

A = [F(x_2) - F(x_1)] + [F(x_3) - F(x_2)] + ... + [F(x_n) - F(x_{n-1})].

A = F(x_n) - F(x_1).

A = F(b) - F(a).

A = [F(x_2) - F(x_1)] + [F(x_3) - F(x_2)] + ... + [F(x_n) - F(x_{n-1})].

A = F(x_n) - F(x_1).

A = F(b) - F(a).

The last line follows because the first rectangle's left end is "a" and the last rectangle's right end is "b", assuming we have n rectangles.

Of course, calculus doesn't use the letter A. It uses the indefinite integral. Therefore,

A = area(f(x) | a to b) = integral(f(x) dx | a to b) = F(b) - F(a)

A = area(f(x) | a to b) = integral(f(x) dx | a to b) = F(b) - F(a)

where F'(x) = f(x).

**Loose ends and bric-a-brac**

Here's a list of things we've glossed over:

- Which functions have to be continuous? Is it OK for f or F to have discontinuities? Or even weirder? What are the assumptions? (Consult a textbook.)
- How do you prove the mean value theorem? (Wikipedia has more information.)
- Can we generalize this to 3-D functions? 3-D vector fields? Arbitrary dimensions? (Yes. Multi-variable calculus and then multi-variable analysis.)
- Can we do this without Riemann sums at all? (Yes. See the Lebesgue integral.)

posted by shadytrees at 4:50 PM on November 24, 2008 [1 favorite]

Setup:

Let f(x) be a differentiable real-valued function on the interval [a,b].

The derivative f'(x) is defined to be the limit, as h goes to 0, of the quantity (f(x+h) -f(x))/h.

(*) In particular, f'(x)h is approximately f(x+h) - f(x), for small values of h.

The integral from a to b of f'(x) is defined as follows. Take an n-partition of the interval [a,b], i.e. points (x_0, x_1, ..., x_n) such that x_0 =a <>

Taking h = x_i - x_{i-1}, and using (*), we see that

f'(x_0)(x_1-x_0) + f'(x_1)(x_2-x_1) + ... + f'(x_1)(x_n-x_{n-1}) is approximately (f(x_1)-f(x_0)) + (f(x_2)-f(x_1))+ ... + (f(x_n) -f(x_{n-1}).

All the terms cancel except the first and last, so S_n is approximately equal to f(x_n) - f(x_0). One can show the limit of S_n as n goes to infinity equals f(x_n) - f(x_0).

Moral:

f'(x)((x + h) - x) is approximately the area of the rectangle under the curve f'(x) between x and x+h, and it also approximately equal to the f(x+h) - f(x) (by definition of derivative).

posted by metastability at 5:08 PM on November 24, 2008

Let f(x) be a differentiable real-valued function on the interval [a,b].

The derivative f'(x) is defined to be the limit, as h goes to 0, of the quantity (f(x+h) -f(x))/h.

(*) In particular, f'(x)h is approximately f(x+h) - f(x), for small values of h.

The integral from a to b of f'(x) is defined as follows. Take an n-partition of the interval [a,b], i.e. points (x_0, x_1, ..., x_n) such that x_0 =a <>

Taking h = x_i - x_{i-1}, and using (*), we see that

f'(x_0)(x_1-x_0) + f'(x_1)(x_2-x_1) + ... + f'(x_1)(x_n-x_{n-1}) is approximately (f(x_1)-f(x_0)) + (f(x_2)-f(x_1))+ ... + (f(x_n) -f(x_{n-1}).

All the terms cancel except the first and last, so S_n is approximately equal to f(x_n) - f(x_0). One can show the limit of S_n as n goes to infinity equals f(x_n) - f(x_0).

Moral:

f'(x)((x + h) - x) is approximately the area of the rectangle under the curve f'(x) between x and x+h, and it also approximately equal to the f(x+h) - f(x) (by definition of derivative).

posted by metastability at 5:08 PM on November 24, 2008

on preview: ... such that x_0 = a < x_1 < x_2 ... < x_n = b.

posted by metastability at 5:10 PM on November 24, 2008

posted by metastability at 5:10 PM on November 24, 2008

Some of you need to learn about the

posted by delmoi at 5:31 PM on November 24, 2008 [1 favorite]

_{<sub>}tag.posted by delmoi at 5:31 PM on November 24, 2008 [1 favorite]

This book should be useful, and your local library probably has it:

posted by parudox at 6:44 PM on November 24, 2008

*Calculus Made Easy: Being a Very-Simplest Introduction to Those Beautiful Methods of Reckoning Which Are Generally Called by the Terrifying Names of the Differential Calculus and the Integral Calculus*posted by parudox at 6:44 PM on November 24, 2008

Alright, one more question, and I think I get it. The wikipedia intuitive proof REALLY helped. Now I need explained to me how the magic Area function is found... is it any of the indefinite integrals? A specific one? If I can get that, I think I'll really have it.

And I worded that part in my question wrong about "majoring in math. Until now." The "Until now." should have been after the part about grasping everything in calculus immediately. I'm definitely majoring in math.

posted by Precision at 9:10 PM on November 24, 2008

And I worded that part in my question wrong about "majoring in math. Until now." The "Until now." should have been after the part about grasping everything in calculus immediately. I'm definitely majoring in math.

posted by Precision at 9:10 PM on November 24, 2008

An indefinite integral can't be a function just of x, because it doesn't have a value depending solely on x, because it's an expression describing the

The area function is likewise not a straight function of x. It too is a function of another function f(x) and

Now, it's possible to show that the area function A(f(x), a, b) will always boil down to F(b) - F(a) where f(x) is the derivative of F(x). In general you won't have F(x), but you

posted by flabdablet at 12:32 AM on November 25, 2008

*general*antiderivative of f(x), and as such will always include an*arbitrary*constant term. In fact it's a function of another function f(x), and two independent variables: x, and the arbitrary constant c. It has the convenient form F(f(x), c) = F(x) + c such that f(x) is the derivative of F(x).The area function is likewise not a straight function of x. It too is a function of another function f(x) and

*two*independent variables, a and b; A(f(x), a, b) gives you the area bounded by the straight line x=a to the left, the x axis below, the straight line x=b to the right, and the curve f(x) above (with reasonable extensions to the idea of area to allow areas to be negative if left/right or below/above get swapped).Now, it's possible to show that the area function A(f(x), a, b) will always boil down to F(b) - F(a) where f(x) is the derivative of F(x). In general you won't have F(x), but you

*will*have the indefinite integral F(f(x), x, c) = F(x) + c, and you can simply plug that in instead, with*any assumed value you like*for c, as long as you make it the*same*value in both substitutions so it cancels out in the subtraction. If you don't do that, you get a nonzero result for A(f(x), a, a), which (barring odd things like Dirac functions) is nonsensical.posted by flabdablet at 12:32 AM on November 25, 2008

Think about composing the area function as the sum of n vertical rectangular slivers, each one being h wide and approximately f(x) high.

The area of the sliver at x=a is exactly f(a) * h, and this is an

Next,

That gives f(a) * h (which is the area of our sliver) = F(a+h) - F(a).

Now, the area function A(f(x), a, b), where b=a+nh, is going to be approximately the sum of the areas of all the h-width slivers between x=a and x=b. Which means it's going to be

F(a+h) - F(a) +

F(a+2h) - F(a+h) +

F(a+3h) - F(a+2h) +

...

F(a+(n-1)h) - F(a+(n-2)h) +

F(a+nh) - F(a+(n-1)h)

All the guts cancels out and the result is F(a+nh) - F(a).

The approximation gets better as h shrinks and n grows; in the limit, it's exact. And in the limit, we get

f(x) = lim

which is the same thing as saying that f(x) is the derivative of F(x).

posted by flabdablet at 1:02 AM on November 25, 2008

The area of the sliver at x=a is exactly f(a) * h, and this is an

*approximation*to the area function A(f(x), a, a+h).Next,

*define*F(x) such that f(x) is (F(x+h) - F(x)) / h.That gives f(a) * h (which is the area of our sliver) = F(a+h) - F(a).

Now, the area function A(f(x), a, b), where b=a+nh, is going to be approximately the sum of the areas of all the h-width slivers between x=a and x=b. Which means it's going to be

F(a+h) - F(a) +

F(a+2h) - F(a+h) +

F(a+3h) - F(a+2h) +

...

F(a+(n-1)h) - F(a+(n-2)h) +

F(a+nh) - F(a+(n-1)h)

All the guts cancels out and the result is F(a+nh) - F(a).

The approximation gets better as h shrinks and n grows; in the limit, it's exact. And in the limit, we get

f(x) = lim

_{h->0}(F(x+h) - F(x)) / hwhich is the same thing as saying that f(x) is the derivative of F(x).

posted by flabdablet at 1:02 AM on November 25, 2008

...which is the same thing as saying that F(x) is any

posted by flabdablet at 4:40 PM on November 25, 2008

*particular*antiderivative of f(x); that is, F(x) is the indefinite integral of f(x) with a definite though arbitrary value substituted for c.posted by flabdablet at 4:40 PM on November 25, 2008

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