Calculus in the AP Physics C Course

Calculus in the AP Physics C Course

The Integral

The Antiderivative

In the simplest case, the antiderivative of a function is found by going backward from the derivative of that function.

Example 1: Suppose the position of an object is described by the position

Then the velocity of the object at any time can be found by taking the derivative of the position, x, with respect to time, t:

Then the acceleration of the object can be found by taking the derivative of the velocity, vinst, with respect to time, t.

If vinst is the derivative of x, then we can say that x is the antiderivative of vinst. Likewise, if a is the derivative of vinst, then vinst is the antiderivative of a.

Example 2: Suppose the velocity of a car is given at any time to be

Find the position x of the car at .

To find the position from the velocity function we need to take the antiderivative, that is, what is the function whose derivative is the given velocity function?

We find that x = antiderivative of vinst =

You can check the answer by simply taking the derivative of x and compare it to the velocity function.

Note that a constant appears in the antiderivative. This is because the derivative of a constant is zero, which means there could be a nonzero constant in the position function, which would represent the initial position of the car at . In order to find out what that constant is we would need more information given to us about the initial conditions of the problem.

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In general, the

antiderivative of

Example 3: For an object undergoing constant acceleration a, find its velocity as a function of time and its position as a function of time.

We begin with constant acceleration, and simply take the antiderivative with respect to time to find the velocity as a function of time:

antiderivative of a =

where the constant in this case must be the initial velocity, v0 . We have then

In the same way, we find the position as a function of time by taking the antiderivative of the velocity function with respect to time:

antiderivative of

where the constant here is the initial position of the object, so.

These are three of the familiar kinematic equations for uniformly accelerated motion.

The Area Under a Curve

Recall that the work done by a force is the scalar product of the force and the displacement through which the force acts. In symbols,

Consider a constant force of 3 N acting on an object and pulling it through a displacement from

to . The work done by the force is the area under the force vs displacement curve, as shown below:

Work = area under curve

No problems here. But what if the force is not constant over the displacement through which it acts? The work done by the force is still the area under the force vs displacement curve, but its not quite as straightforward as finding the area of a rectangle, or is it?

Suppose we have a force that varies over the displacement through which it acts (Figure. A).

Figure A Figure B Figure C

To find the work done (i.e., the area under the force vs displacement curve) by the force from

to we divide the area into rectangles of width and height (Figure B).

The little increment of work done by the force in moving the object through a little displacement is

(Figure C)

So, to approximate the total area under this curve (and thus the work), we can simply sum up the little areas of all the rectangles from a to b.

Total Work =

How can we get a more accurate approximation of the area under the curve (work)?

Let's take more rectangles with smaller widths (smaller 's) so that the "heights of the rectangles will better fit our curve, and then sum the areas of the narrower rectangles as before. This summation method is called Simpson Rule.

But even if we choose a large number of narrow rectangles, Simpson's rule can only give us an approximation of the area under a curve.

We can, however, take the limit of rectangles, as their widths become infinitesimally small, i.e., let , to find the exact area under the curve. We write

Exact Total Work =

This limit of a sum is called a Riemann sum.

The sum becomes an integral; representing the exact area under the F(x) vs x curve.

Note that the becomes a , the integral symbol which takes the shape of an elongated "S", for “summa”, the Latin word for "sum". Also note that the has become a dx , called a differential. The differential dx is an infinitesimally small piece of x, and always accompanies the integral sign. This tells us that we are summing () an infinite number of infinitesimally small elements (dx's) to get the exact whole.

Now we have

and is read "the integral of F of x, dx". The variable in the integrand F(x) and the differential variable dx must match before the integration can be carried out. Remember, in this example, F stands for force.

But how do we carry out summing by integration? In its Simplest form, the integral is an antiderivative.

Example 4: Find the function whose derivative is

Separate the variables:

A quick note on that last step: Although it appeared as though we "multiplied both sides by dx, the derivative is not a quotient, and therefore no multiplication or division took place. However, for our purposes, the rules of separating variables and the rules of multiplication and division are consistent with each other.

Now, integrate both sides, that is, take the antiderivative,

where C is the constant of integration we discussed earlier. Note that on the left side of the equation we summed () all the dy's to get a whole y.

To check our answer, we can take the derivative :

In general, as before with the antiderivative,

This integral is called an indefinite integral, because it has no definite limits to integrate from or to. It gives the antiderivative in terms of the variable x.

Definite Integral

A definite integral is an integral in which the limits a and b are included thus we may evaluate the integral for its actual value.

Recall finding the area under a rectangle:

Work = Area under the curve =

For any curve,

Work = Area under the curve =

Now we need a method for finding the actual value of the integral.

Fundamental Theorem of Calculus: If the function, f, is continuous on [a,b], and F is an antiderivative of f; then

The theorem says that to find , first find an antiderivative F of f. Then evaluate F at and at, and subtract F(a) from F(b). The result is the value of the integral.

Example 5: Find the area under the line y = x from to by integration.

Example 6: Consider a mass that is on a frictionless surface and is attached to a Hookean spring.

The applied force on the mass, which is working against the spring, is not a constant force. Therefore, to find the work this force does against the spring we must integrate.

In order to stretch the spring we must apply a force equal to the magnitude of the Hookean spring force.

The work done in stretching the spring from equilibrium position to is

Since k is a constant, we can factor it out of the integral:

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