Integrals with no limits. (Backwards differientation)

[QuantumTheory]

Alright, I understand clearly what a integral with no limits is, what it does, etc.

It is simply backwards differentation (differentation deals with the instanteous slope of a parabola)

And has no limit (no a to b), thus there must be a constant of integration since two (or more) of the same functions can look the same.

It looks like:
$$\int$$


I understand that these integrals are backwards differentation, thus backwards slopes?

This doesn't make any sense, what exactly is their use then?

I see a lot of integrals without limits (They do not deal with area like the other type of integrals) being written down about the paradoxes of space (like on the discovery channel/science channel)

This makes me even more interested on their use of them! Because I am hopefully going to be a professor in astrophysics! So far I'm 16, and I have a great start! I even tutor kids after school in math, its great.

I understand the use of integrals with limits , from a to b like used in area, it makes sense! However I don't see the use of integrals without limits (thus with the C constant!)

Help?

 

[agro]

Consider this problem: given a function f(x), we wish to find a function F(x) such that F'(x) = f(x). The process is called antidifferentiation and F(x) is called an antiderivative of f(x). This "general" problem arises in various applications.

Consider this problem:

I have a particle moving along an x-axis such that its velocity at time t is v(t) = 2t. What is its position function, x(t)? We know from physics that x'(t) = v(t). So we want to find a function x(t) such that x'(t) = 2t. This is an antidifferentiation problem.

How do we solve it? We use educated guesswork. If you're familiar with differentiation then you'll quickly realize that if x(t) = t², x'(t) = 2t.

So, we're finished right?

Not really. Note that x(t) = t² + 2, x(t) = t² - 1, and even x(t) = t² +1000 qualifies as a solution (since in all of those cases, x'(t) = v(t) ).

Since we have no additional information about x(t), we write the general solution x(t) = t² + C.

Now that took a lot of time! The notation to describe what we did is

$$x(t) = \int v(t)\;dt = \int 2t\;dt = t^2 + C$$

The notation $\int 2t\;dt$ is used to denote all functions x(t) such that x'(t) = 2t.

Now, if we have additional information about x(t) we can find the value of C. For example, if we have the information that the particle is at x = 5 when t = 0, then we can quickly conclude that x(t) = t² + 5.

Other uses of indefinite integral (the one with upper and lower limit is called definite integral):

- Suppose we know r(t), the rate of change of a city population at time t. Find the function p(t) that describe the population of the city at time t! (note that p'(t) = r(t) )
- Suppose we know a(t), the acceleration of a particle moving along x-axis at time t. Find its position function x(t)! (note that x'(t) = v(t) and v'(t) = a(t))

As a comparison: a definite integral (the one you use to find area) yields a number (in that case the area), while an indefinite integral yields a family of functions (as shown earliear, this family has the form of F(x) + C).

Oh, and btw, the definite and indefinite itegrals are nicely connected to each other. Let f(x) be continuous on [a, b]. Then it can be proved that:

$$\int_a^b f(x)\;dx = \left[\int f(x)\;dx\right]_a^b$$

I hope that clears things up :)...

 

[QuantumTheory]

So basically, $$\int_a^b f(x)\;dx = \left[\int f(x)\;dx\right]_a^b$$

Is saying that F(x) dx = $$\int_a^b f(x)\;dx$$?

The first integral with limits, is the area under a curve of $$f(x)$$ of each piece of the area of the curve ($$dx$$).

But, you can solve the area under the curve with the antiderivative of f(x) dx?

Thanks.

Is this right?

 

[Hurkyl]

Yep. This connection between antidifferentiation and integration is so important, it's called the fundamental theorem of calculus.

$$\int_a^b f'(t) \, dt = f(b) - f(a)$$

and if f is continuous,

$$\frac{d}{dx} \int_a^x f(t) \, dt = f(x)$$