- #1

- 53

- 0

Suppose something is too hard to integrate. Can i use the riemann sum to estimate the area of an interval as closely as possible, and then use that approximation to help me integrate? Im sure that it must be possible, but i dont know how.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter okkvlt
- Start date

- #1

- 53

- 0

Suppose something is too hard to integrate. Can i use the riemann sum to estimate the area of an interval as closely as possible, and then use that approximation to help me integrate? Im sure that it must be possible, but i dont know how.

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

In the years before Newton and Leibniz published their work on the calculus, most mathematicians were interested in 2 difficult questions:Can somebody explain to me, geometrically and intuitively, the fundamental theorem of calculus? I understand that i can find the area between the graph of f'(x) and the x axis where b>x>a by finding f(b)-f(a), but i dont understand why.

1) find the equation of a tangent to a curve. (Or equivalently, find a linear local approximation to a function.)

2) find the area of a general area.

The second problem goes back to Archimedes who broke the interior of a parabola into smaller and smaller portions. Pascal and DesCartes worked on the first problem. The "fundamental theorem of calculus" (and the reason Newton and Leibniz are considered the "founders" of Calculus and not Archimedes, Pascal, or DesCartes) says that those are basically "inverse" problems.

The proof is found in any Calculus book. It's too long to give here but basically involves looking at the area under a curve from, say, a to x and then from a to x+h. The difference between those areas is the area from x to x+h and that leads to the formula for the derivative.

Yes, that's a simple way to give an approximate integral. If by "help me integrate" you mean then use that to find anSuppose something is too hard to integrate. Can i use the riemann sum to estimate the area of an interval as closely as possible, and then use that approximation to help me integrate? Im sure that it must be possible, but i dont know how.

An important reason to learn about Riemann sums, other than historical, is that the basic idea of dividing into pieces and then adding will help see how to set up integrals in applications.

- #3

- 2,111

- 18

[tex]

[a,b] = [a,a+\Delta x]\;\cup\; [a+\Delta x, a + 2\Delta x]\;\cup\;\cdots\;\cup\;[a + n \Delta x, b],

[/tex]

[tex]

\Delta x = \frac{b-a}{n+1},

[/tex]

the integral of f' is approximately

[tex]

\int\limits_a^b f'(x) dx \;\approx\; \sum_{k=0}^N f'(a + k\Delta x) \Delta x.

[/tex]

On the other hand the difference f(b)-f(a) is

[tex]

f(b) - f(a) = \big(f(b) - f(b-\Delta x)\big)\; +\; \big(f(b-\Delta x) - f(b-2\Delta x)\big) \;+\;\cdots \;+\; \big( f(a+\Delta x) - f(a)\big) \;=\; \sum_{k=0}^N \big( f(a + (k+1)\Delta x) \;-\; f(a + k\Delta x)\big).

[/tex]

Substitute

[tex]

f(a + (k+1)\Delta x) - f(a + k\Delta x) \approx f'(a + k\Delta x) \Delta x

[/tex]

in and you get

[tex]

\int\limits_a^b f'(x) dx \;\approx\; f(b) - f(a),

[/tex]

which is close to the fundamental theorem of calculus. The rigor proof of what happens on the limit [tex]\Delta x\to 0 [/tex], [tex]N\to\infty[/tex], is of course more difficult, but I hope this answered your original question somehow. I would have preferred explaining it with picture, though. You should draw something yourself, to see what is happening.

- #4

- 655

- 3

There is a method of navigation known as "dead reckoning" used by sailing ships and wilderness explorers who don't have GPS or other modern technology. It is still used in inertial navigation technology used by missiles and other such things.

The idea is, if you kept track of where you started, what direction you went, and how fast and for how long, you can figure out your position.

Here is an example of how it might work: suppose you are walking in the wilderness, and you know your walking speed is about 2 miles/hr. Every hour you look at your compass and write down which direction you are going. After 8 hours of hiking, you can estimate your position compared to where you started by drawing a series of connected line segments on a map 2 miles long each, in the directions you wrote down. Here is a picture to explain it better:

http://img408.imageshack.us/img408/909/deadreckoninggu7.png [Broken]

The more frequently you take write down what direction you are going, the more accurate your approximation will be. If you wrote down what direction you are going every second, the approximate path would look basically exactly the same as the actual path. You can imagine that if you took an infinite amount of measurements of direction, the actual and estimated path would be the same.

This is actually the fundamental theorem of calculus in disguise. If the path you actually take is f(x), then "direction" you write down is the tangent vector, (1,f'(x)), and the dead reckoning estimation process of drawing line segments is precisely the reimann sum of f'(xi)delta x.

The idea is, if you kept track of where you started, what direction you went, and how fast and for how long, you can figure out your position.

Here is an example of how it might work: suppose you are walking in the wilderness, and you know your walking speed is about 2 miles/hr. Every hour you look at your compass and write down which direction you are going. After 8 hours of hiking, you can estimate your position compared to where you started by drawing a series of connected line segments on a map 2 miles long each, in the directions you wrote down. Here is a picture to explain it better:

http://img408.imageshack.us/img408/909/deadreckoninggu7.png [Broken]

The more frequently you take write down what direction you are going, the more accurate your approximation will be. If you wrote down what direction you are going every second, the approximate path would look basically exactly the same as the actual path. You can imagine that if you took an infinite amount of measurements of direction, the actual and estimated path would be the same.

This is actually the fundamental theorem of calculus in disguise. If the path you actually take is f(x), then "direction" you write down is the tangent vector, (1,f'(x)), and the dead reckoning estimation process of drawing line segments is precisely the reimann sum of f'(xi)delta x.

Last edited by a moderator:

- #5

- 10

- 0

APPLICATION so you can USE it:

In the most concise wording, since it's nicer to show it with an actual graph; Use the FTC usually when given a graph of the derivative (f '(x) OR (d/dx) where the integral of that would give you the original graph. When the equation for d/dx is NOT given, just use the graph and (x,y) coordinates to find points on f(x).

i.e. if you see on d/dx that points (2,4) {f '(2)=4} and (3,8) {f '(3)=8} exist,

and the problem might tell you that f(2) = 20 (where f(x) is the original function),

then you can use FTC to say:

S(2,3) f '(x)dx = f(3) - f(2)

(where S(2,3) is an integral (a,b)

so looking on d/dx again, since it's the derivative, and let's say it looks relatively close to the graph Maze gave us http://img408.imageshack.us/img408/909/deadreckoninggu7.png [Broken] ;

you can find the area using geometry from f '(2) to f '(3), where A= 1/2*b*h:

1/2 * (3-2) * (8-4) = 2

NOW we can find F(3)

S(2,3) f '(x)dx = 2 = f(3) - f(2)

2 = f(3) - 20

20 + 2 = f(3)

22 = f(3)

I am so sorry that I can't articulate. I hope this helps somewhat with understanding what it's used for. In Calc II, knowing the FTC helped my understanding of Integration By Parts, allowing you to take the integral of any polynomial / function.

In the most concise wording, since it's nicer to show it with an actual graph; Use the FTC usually when given a graph of the derivative (f '(x) OR (d/dx) where the integral of that would give you the original graph. When the equation for d/dx is NOT given, just use the graph and (x,y) coordinates to find points on f(x).

i.e. if you see on d/dx that points (2,4) {f '(2)=4} and (3,8) {f '(3)=8} exist,

and the problem might tell you that f(2) = 20 (where f(x) is the original function),

then you can use FTC to say:

S(2,3) f '(x)dx = f(3) - f(2)

(where S(2,3) is an integral (a,b)

so looking on d/dx again, since it's the derivative, and let's say it looks relatively close to the graph Maze gave us http://img408.imageshack.us/img408/909/deadreckoninggu7.png [Broken] ;

you can find the area using geometry from f '(2) to f '(3), where A= 1/2*b*h:

1/2 * (3-2) * (8-4) = 2

NOW we can find F(3)

S(2,3) f '(x)dx = 2 = f(3) - f(2)

2 = f(3) - 20

20 + 2 = f(3)

22 = f(3)

I am so sorry that I can't articulate. I hope this helps somewhat with understanding what it's used for. In Calc II, knowing the FTC helped my understanding of Integration By Parts, allowing you to take the integral of any polynomial / function.

Last edited by a moderator:

- #6

mathwonk

Science Advisor

Homework Helper

2020 Award

- 11,125

- 1,323

the proof is that when you multiply the height by the base, you get the area.

then you have to take some limits. but this is basically it.

if your graph is not a rectangle, the area is the base times some intermediate height.

thus when you divide the area by the base you get some intermediate height.

but (assuming your function is continuous) as the length of the base of a region under a graph,

bounded by two vertical sides, goes to zero, that intermediate height approaches

the height at the left vertical side of the region,

i.e. the derivative at c, of the area function under a graph, with respect to the x variable, is the height at x=c.

- #7

Gib Z

Homework Helper

- 3,346

- 5

Assume f(t) is a continuous strictly increasing positive function over some closed interval. The integral [tex]A(x)=\int^x_a f(t) dt[/tex], where a and x are inside the closed interval, and x>a, gives the area between t=a, t=b, f(t) and the t axis.

By the definition of the derivative, [tex]A'(x) = \lim_{h\to 0} \frac{ \int^{x+h}_a f(t) dt - \int^x_a f(t) dt }{h} = \lim_{h\to 0} \frac{1}{h} \int^{x+h}_x f(t) dt[/tex].

Since f(t) is strictly increasing and positive, that last integral can not be greater than the product of the maximum value in the interval of integration, f(x+h), and the width of the interval, h. Similarly, it can not be less than the minimum value, f(x), multiplied by the width, h.

Summarizing: [tex] hf(x) < \int^{x+h}_x f(t) dt < hf(x+h)[/tex]. Dividing through by h, taking limits limits as h approaches zero, we see that [tex]A'(x) = f(x)[/tex], or [tex]\frac{d}{dx} \int^x_a f(t) dt = f(x) [/tex], one statement of the fundamental theorem.

Now: [tex]A(x) - A(a) = \int^x_a f(t) dt - \int^a_a f(t) dt = \int^x_a f(t) dt[/tex]. And as we have already seen, A'(x) = f(x), as required.

- #8

dx

Homework Helper

Gold Member

- 2,011

- 18

[tex] \int_{a}^{x} f(x) dx [/tex]

with respect to x is just

[tex] \frac{ f(x) \Delta x}{\Delta x} = f(x) [/tex].

Share: