Vectors and Vector Operations - University of Michigan
4. Integrals
As we mentioned earlier, calculus revolves around two main concepts, derivatives and integrals.
1. Derivatives: Derivatives correspond physically to an instantaneous rate of change of one variable y which is related to another variable x by a formula of the form y = f(x). This is the limit of average rates of change as the changes approach zero. Thus we divide the change in y by the change in x and take the limit as the changes approach zero.
2. Integrals: There are two important aspects to integrals. These are called indefinite integrals and definite integrals.
Indefinite integrals (or anti-derivatives): This is the process of recovering a formula from its derivative. For example, finding an object's position when one is given the velocity v as a function of time t, i.e. v = f(t). We look at this process first in section 4.1.
Definite integrals: It turns out that one way find how far you have traveled in certain period of time is to divide the time period into small segments and sum up your velocity during each segment times the length of the time segment. In the limit as the length of the segments approach zero, you get the distance traveled. This illustrates the second aspect of integrals, namely as a limit of sums. This type of integral is called a definite integral. We look at that in sections 4.2 and 4.3.
4.1 Indefinite Integrals
In order to appreciate why we are interested in integrals, let’s look at some examples of derivatives
Function (((((((((((((((( Derivative
position = f(time) ( velocity
s = f(t) v =
s = t2 v = 2t
velocity = f(time) ( accelaeration
v = f(t) a =
v = sin 2t a = 2 cos 2t
mass = f(position) ( linear density
M = f(x) ( =
M = 2x + x2 ( = 2 +
charge = f(time) ( current
Q = f(t) i =
Q = cos 2t i = - 2 sin t
cost = f(number produced) ( marginal cost
C = f(x) MC =
C = 4x + MC = 4 -
In each case we have some physical quantity which is a function of another physical quantity and the derivative is another interesting quantity. However, often we are given a formula for the quantity in the right hand column and from that we want to compute the formula for the quantity in the left hand column. For example, we might be given a formula for the current as a function of time and from that we want to compute a formula for the net charge that has gone past as a function of time.
To do this we want to find a function if we are given its derivative. The process of doing this is called integration.
To be more precise, an integral (or an indefinite integral) of a function v = f(t) is another function x = F(t) whose derivative is f(t), i.e.
= f(t)
or
= f(t)
We write f(t) dt to indicate that we want to find a function x = F(t) whose derivative is f(t) and we write
f(t) dt = F(t)
to indicate that = f(t).
Example 1. Suppose you are driving along the road and
s = distance from home (mi)
t = time since noon (hr)
Then = v the velocity. Now, suppose that instead of being given s as a function of t, you are given the velocity as a function of t. In particular, suppose v = t2. Now we want to find our position s as a function of t. Since = v = t2, we want to find an integral of v = t2, i.e. an integral of f(t) = t2.
Solution. We are looking for a function s = F(t) such that
= t2
or
= t2
We express the fact that we are looking for a function whose derivative is t2 by saying that we want to find t2 dt.
It is important to note, that we are not taking the derivative of t2. Rather, we want something whose derivative is t2. If we think about it, s = t3 is close, since
= 3t2
We need to do something to get rid of the 3. The solution is to multiply by . If we let
s = t3 then
= ( 3t2 = t2
So an integral of v = t2 is s = t3. We express this fact by writing t2 dt = t3.
Since the derivative of a constant is zero, we can add any constant to t3 and it will also be an integral of v = t2. For example s = t3 + 4 and s = t3 – 6 are also integrals of v = t2. We can summarize the situation by saying the integrals of v = t2 are s = t3 + C for any constant C and we would write t2 dt = t3 + C. These are all the integrals of v = t2.
Let's return to the problem of finding the formula for our position. At this point we know s = t3 + C. In order to find C we need to be given some additional piece of information. For example, suppose at 12 noon we are 2 miles from home. So s = 2 when t = 0. Plugging this into s = t3 + C we get 2 = 03 + C, so C = 2 and s = t3 + 2.
Example 2. Find t dt.
Solution. We are looking for a function s = F(t) such that
= t
or
= t
If we think about it, s = t2 is close, since = 2t. To get rid of the 2, we multiply by . If s = t2 then = ( 2t = t. So an integral of v = t is s = t2 and t dt = t2. To get all the functions whose derivative is t we add a constant, so t dt = t2 + C.
Example 3. Find t4 dt.
Solution. As in the preceding examples, s = t5 is close, since = 5t4. To get rid of the 5, we multiply by . If s = t5 then = ( 5t4 = t4. So t4 dt = t5 or t4 dt = t5 + C if we want to get all the functions whose derivative is t4.
We can generalize the preceding three examples by saying
tn dt = tn+1
This formula holds if n is a negative and/or a fraction as well as if n is a positive integer. For example,
dt = t1/2 dt = t1/2 + 1 = t3/2
dt = t-2 dt = t-2 + 1 = - t-1 = -
The sum, difference and constant multiple rules are true for integrals because they are true for derivatives. They say
[ f(t) + g(t) ] dt = f(t) dt + g(t) dt
[ f(t) - g(t) ] dt = f(t) dt - g(t) dt
[c f(t)] dt = c f(t) dt
For example,
[ t3 + t-4 ] dt = t3 dt + t-4 dt = t4 + t-3 = t4 -
[ t3 - t-4 ] dt = t3 dt - t-4 dt = t4 - t-3 = t4 +
5t3 dt = 5 t3 dt = 5 ( t4 = t4
Another useful rule is that the integral of a constant is just the constant time t, i.e.
c dt = ct
For example,
8 dt = 8t
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