INTRODUCTION TO POLYNOMIAL CALCULUS

1. Straight Lines

Given two distinct points in the plane, there is exactly one straight line that contains them both. This is one of the important principles of Plane Geometry. If the plane is equipped with a Cartesian coordinate system, it should be possible to write down an equation for such a line in terms of the x and y coordinates. In this section we shall show how to do this. The notion of slope will be very useful in this task.

The slope of a line is defined to be the change in the y coordinate (the rise) divided by the change in the x coordinate (the run) as we move from one point on the line to a second point on the line (see Figure 1.1). That is, if (x_{1}, y_{1}) and (x_{2}, y_{2}) are two points on the line, then the slope of the line is the number m, where

m = \frac{y_{2} − y_{1}}{x_{2} − x_{1}}

This number does not depend on which two points on the line are chosen. In fact, if two other points (x_{3}, y_{3}) and (x_{4}, y_{4}) are chosen, then it follows from the similar triangles in Figure 1.2 that

\frac{y_{2} − y_{1}}{x_{2} − x_{1}} = \frac{y_{4} − y_{3}}{x_{4} − x_{3}}

and so the points (x_{3}, y_{3}) and (x_{4}, y_{4}) give the same value for the slope m as do the points (x_{1}, y_{1}) and (x_{2}, y_{2}).

The slope measures whether a line rises or falls as we move to the right and how steeply it does so. Positive slope means the line rises to the right, while negative slope means it falls to the right. Slope zero means the line is horizontal, since it means that the y coordinate does not change at all from point to point on the line. A vertical line has no slope (some would say it has infinite slope). This is because the x coordinates of points on such a line are all the same and so the denominator is zero in the equation defining slope.

Example 1. Find the slope of the line which contains the points (1, 2) and (3, 5). Solution: Here the rise is 5 − 2 = 3 and the run is 3 − 1 = 2 and so the slope is

\frac{rise}{run} = 3/2.

The point-slope form of the equation of a line. Using slope we can easily write down an equation for any line that is not vertical. Let m be the slope of the line and let (x_{0}, y_{0}) be some fixed point on the line. If (x, y) is a variable point on the line, then

m = \frac{y − y_{0}}{x − x_{0}}

and so

y − y_{0} = m(x − x_{0})

This is called the point-slope form of the equation of a straight line. It gives the equation of the line in terms of the slope m of the line and a point (x_{0}, y_{0}) on the line.

Example 2. Find the equation of the line which passes through the point (1, 3) and has slope 2.

Solution: Here the slope m is 2 and the point (x_{0}, y_{0}) is (1, 3) and so the equation of the line is

y − 3 = 2(x − 1) or y = 2x + 1

Example 3. Find the equation of the line which contains the points (−1, 2) and (0, 5) Solution: The slope of this line is

m = \frac{5 − 2}{0 − (−1)} = 3

and the line contains the point (0, 5); so the equation of the line is

y − 5 = 3(x − 0) or y = 3x + 5

The slope-intercept form of the equation of a line. If a line crosses the y-axis at the point (0, b), then the number b is called the y-intercept of the line. If the line has slope m, then the point-slope form of its equation (using the point (0, b)) is y − b = m(x − 0) or

y = mx + b

This is the point-intercept form of the equation of the line. Note that every line except a vertical line does cross the y-axis and, thus, has a slope-intercept form for its equation.

Example 4. What is the equation of the line with slope −5 and y-intercept 4 ?

Solution:

y = −5x + 4

The general equation of a line. Every equation of the form

Ax + By + C = 0

is the equation of a line and, conversely, every line has an equation that can be put in this form. A non-vertical line has an equation of the form y = mx + b, which can be written

mx − y + b = 0

which is of the form Ax + By + C with A = m, B = −1 and C = b. On the other hand, a vertical line has the form

x = k

where k is a constant. This can be re-written as

x − k = 0

which is of the form Ax + By + C = 0 with A = 1, B = 0, and C = −k.

Example 5. Find the slope and y-intercept for the line which is described by the equation 2x + 3y − 6 = 0.

Solution: If we solve for y in this equation, we get the equation

y = −\frac{2}{3}x + 2

The equation of the line is now in slope-intercept form and we can just read off the slope (−\frac{2}{3}) and the y-intercept (2).

Parallel and perpendicular lines. If the two lines in Figure 1.3 are parallel, then the two triangles are similar, from which it follows that the two lines have the same slope. On the other hand, if the two lines have the same slope then the two triangles in Figure three must be similar, since they contain right angles with proportional adjacent sides. This implies that the two lines are parallel since they make the same angle with a horizontal line. Thus, two lines are parallel if and only if they have the same slope.

What about perpendicular lines? There is an old carpenter’s trick for checking that an angle is a right angle. One measures off three units along one leg and makes a mark and then measures four units along the other leg and makes another mark. The angle is then a right angle if and only if the distance between the two marks is five. This works because 3^{2} + 4^{2} = 5^{2} and an angle in a triangle is a right angle if and only if

a^{2} + b^{2} = c^{2}

where a and b are the lengths of the two agacent sides of the angle and c is the length of the opposite side (the Pythagorean Theorem).

Let us see how we can apply this to tell when two lines are perpendicular. We may as well assume that neither line is horizontal, since it is easy to see when a line is perpendicular to a horizontal line (it must be vertical). Then the first line intersects the x-axis in a point (x_{1}, 0) and the second line intersects the x-axis in a point (x_{2}, 0). Let’s suppose the two lines are not parallel and so they intersect each other in a point (x_{0}, y_{0}) (see Figure 1.4). Then they intersect at a right angle if and only if a^{2} + b^{2} = c^{2}, where a is the distance from (x_{0}, y_{0}) to (x_{1}, 0), b is the distance from (x_{0}, y_{0}) to (x_{2}, 0) and c is the distance from (x_{1}, 0) to (x_{2}, 0). This leads to the equation

(x_{1} − x_{0})^{2} + y_{0}^{2} + (x_{2} − x_{0})^{2} + y_{0}^{2} = (x_{2} − x_{1})^{2}

which simplifies to

2x_{0}^{2} + 2y_{0}^{2} − 2x_{0}x_{1} − 2x_{0}x_{2} = −2x_{1}x_{2}

On dividing by 2 and rearranging terms this becomes

y_{0}^{2} = −x_{0}^{2} + x_{0}x_{1} + x_{0}x_{2} − x_{1}x_{2}

or, after factoring,

y_{0}^{2} = −(x_{0} − x_{2})(x_{0} − x_{1})

or

\frac{y_{0}}{x_{0} − x_{2}} = −\frac{x_{0} − x_{1}}{y_{0}}

which says that

m_{2} = −\frac{1}{m_{1}}

where m_{1} is the slope of the first line and m_{2} is the slope of the second line. Thus, two non horizontal lines are perpendicular if and only if the slope of the second is the negative reciprocal of the slope of the first.

Example 6. Find the equation of the line which is parallel to the line with equation 2x + 3y = 5 and passes through the point (1, 2).

Solution: We solve for y in the equation of the first line in order to put its equation in the form

y = −\frac{2}{3}x + \frac{5}{3}

This shows that the line has slope −\frac{2}{3}. The line which has the same slope and passes through the point (1, 2) has the equation

y − 2 = −\frac{2}{3}(x − 1) or y = −\frac{2}{3}x + \frac{8}{3}x or 2x + 3y = 8

Example 7. Find the equation of the line which is perpendicular to the line 2y + x = 3 and meets it at the point (1, 1).

Solution: If we solve the first equation for y it becomes

y = −\frac{1}{2}x + \frac{3}{2}

which tells us that its slope is −\frac{1}{2}. A line perpendicular to this line will, therefore, have slope 2. Since the line we seek must pass through the point (1, 1), its equation is

y − 1 = 2(x − 1) or y = 2x − 1

2. Slope of a Curve

We know about the slope of a straight line. It is the change in the y-coordinate divided by the change in the x-coordinate (rise divided by run) as we move from a given point on the line to any other point on the line. The law of similar triangles says that this ratio is independent of the two points on the line that are chosen.

What about the slope of a curve that is not a straight line? Does this make sense? If so, how do we calculate it? Let’s look at an example, say the curve

y = x^{2}

The graph of this curve (Figure 2.1) makes it clear that, if its slope makes sense, it cannot be a fixed number. To the left of x = 0 the curve slopes downward (negative slope) while to the right of x = 0 the curve slopes upward (positive slope) and it rises more steeply the further to the right we go. Thus, if it makes sense at all, the slope must depend on where we are on the curve.

In fact, the slope of the curve y = x^{2} does make sense at each point (x, y) on the curve (but it changes as the point changes). This is suggested by the fact that if a microscopically tiny piece of the curve is magnified enough to be visible then it looks like a straight line. This is illustrated by the graphs in Figure 2.1 which show the curve near the point (1, 1) magnified by various factors. In other words, the smaller the segment of the curve we look at near (1, 1), the more the curve looks like a straight line. The slope of this straight line should be what we mean by the slope of the curve at the point (1, 1).

How can we calculate this slope? We do the same thing we would do if the curve were a straight line. We calculate the change in y divided by the change in x as we move from one point on the curve to another. However, now we choose special points. We let the first one be (1, 1) itself and we choose the second one to be near (1, 1). The nearer to (1, 1) we make it, the more the curve between these two points looks like a straight line and the closer our ratio will be to the slope of this line. Let’s try this for some choices of points on the curve y = x^{2} (see Figure 2.2). In each case we will be calculating the rise divided by the run between a first point, (1, 1), and some nearby second point on the curve. In other words we will be calculating the slope of the line joining these two points.

With second point (3, 9) we get

slope = \frac{9 − 1}{3 − 1} = \frac{8}{2} = 4

With second point (2, 4) we get

slope = \frac{4 − 1}{2 − 1} = \frac{3}{1} = 3

With second point (1.5, 2.25) we get

slope = \frac{2.25 − 1}{1.5 − 1} = \frac{1.25}{.5} = 2.5

With second point (1.1, 1.21) we get

slope = \frac{1.21 − 1}{1.1 − 1} = \frac{.21}{.1} = 2.1

With second point (1.01, 1.0201) we get

slope = \frac{1.0201 − 1}{1.01 − 1} = \frac{.0201}{.01} = 2.01

A graphic description of what is happening here is shown in Fig. 2.

One can now guess that if we continue doing this for second points that are closer and closer to (1, 1), the slope we are calculating will simply get closer and closer to the number 2. This would mean that the slope of the curve y = x^{2} at the point (1, 1) is 2.

Let us now take a perfectly general second point – one that is obtained by changing the x coordinate of the first point by an amount h so that the second point becomes (1 + h, (1 + h)^{2}). The smaller h is chosen the closer the second point is to (1, 1). The slope we get for the line joining (1, 1) to this second point is then

slope = \frac{(1 + h)^{2} − 1}{h} = \frac{2h + h^{2}}{h} = 2 + h

Now if h is chosen very small then this number will be very close to 2. In other words, as h approaches 0 our slope approaches 2. This should convince us that the slope of the curve y = x^{2} at the point (1, 1) is 2.

The fact that 2 + h approaches 2 as h appoaches 0 is commonly written

lim_{h→0}(2 + h) = 2

Here lim_{h→0}(2+h) is shorthand for “the limit of 2+h as h approaches 0” and it simply means the number that 2 + h approaches as h approaches 0.

The method used above works for other curves as well.

Example 1. Find the slope of the curve y = x^{2} − 3x at the point where x = 2.

Solution: For the function f(x) = x^{2} − 3x we have

f(2) = −2
f(2 + h) = (2 + h)^{2} − 3(2 + h) = −2 + h + h^{2}

Thus, the slope of the curve when x = 2 is

slope = lim_{h→0} \frac{f(2 + h) − f(2)}{h} = lim_{h→0} \frac{h + h^{2}}{h} = lim_{h→0}(1 + h) = 1

Returning to the curve y = x^{2}, we would like a formula for the slope of this curve at any point of the curve, not just at the point (1, 1). We use the same technique. Given a point (x, x^{2}) on the curve, we move to a nearby second point (x + h, (x + h)^{2}), obtained by changing the x coordinate of the first point by an amount h. Then we calculate the slope of the line joining these two points. It is

slope = \frac{(x + h)^{2} − x^{2}}{h} = \frac{2xh + h^{2}}{h} = 2x + h

As h approaches zero this slope approaches the number 2x. In other words

lim_{h→0}(2x + h) = 2x

We conclude that the slope of the curve y = x^{2} at the point (x, x^{2}) is 2x.

The preceding discussion leads to the following definition of the slope of a curve which is given as the graph of a function f:

Definition A. The graph of a function f is said to have slope m at a point (x, f(x)) on the graph provided

lim_{h→0} \frac{f(x + h) − f(x)}{h} = m

That is, provided the expression \frac{f(x+h)−f(x)}{h} approaches m as h approaches 0. In this case, we call the number m the derivative of f at x and denote it by f′(x).

Thus, the derivative of a function f is another function f′ of the same variable x. Its value at x is the slope of the graph of f at the point (x, f(x)). In other words, it is the instantaneous rate of change of y with respect to x as we pass through the point (x, f(x)) while moving along the graph of f.

For now we will take the statement “\frac{f(x+h)−f(x)}{h} approaches m as h approaches 0” as intuitively understood but later in the course we will need to make this idea more precise. We will do this when we study limits. As we shall see in the next section, when f is a polynomial it is quite easy to see what happens to the expression \frac{f(x+h)−f(x)}{h} as h approaches 0 and so to study derivatives of polynomials we do not need a sophisticated study of limits.

The discussion preceding the above definition shows that the derivative of x^{2} is 2x. We give two other examples of the calculation of a derivative.

Example 2. Find the derivative of a constant function f(x) = c.

Solution:

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h} = lim_{h→0} \frac{c − c}{h} = lim_{h→0} 0 = 0

Thus, the derivative of a constant function is 0. This just reflects the fact that the graph of a constant function is a horizontal line and, thus, has slope 0.

Example 3. Find the derivative of the function f(x) = 3x − 2.

Solution:

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h} = lim_{h→0} \frac{3(x + h) − 2 − (3x − 2)}{h} = lim_{h→0} 3 = 3

Thus, the derivative of the function f(x) = 3x − 2 is the constant 3. This is not surprising, since the graph of the function f(x) = 3x − 2 is a straight line with slope 3.

3. Derivative of a Polynomial

In the last section we defined the derivative f′(x) of a function f(x) to be the slope of the curve y = f(x) at the point (x, f(x)). This, in turn, is defined to be the number that the expression \frac{f(x+h)−f(x)}{h} approaches as h approaches 0. In other words

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h}

We will now use this definition to calculate the derivative of any polynomial. We begin by calculating the derivative of the monomial x^{n} for each value of n. For n = 0, 1, 2, 3 this was done in the previous section and its problem set. However, we will repeat these calculations here in order to demonstrate the pattern that emerges.

If f(x) = x^{0} = 1 then

f′(x) = lim_{h→0} \frac{1 − 1}{h} = lim_{h→0} 0 = 0

If f(x) = x^{1} = x then

f′(x) = lim_{h→0} \frac{x + h − x}{h} = lim_{h→0} 1 = 1

If f(x) = x^{2} then

f′(x) = lim_{h→0} \frac{(x + h)^{2} − x^{2}}{h} = lim_{h→0}(2x + h) = 2x

If f(x) = x^{3} then

f′(x) = lim_{h→0} \frac{(x + h)^{3} − x^{3}}{h} = lim_{h→0}(3x^{2} + 3xh + h^{2}) = 3x^{2}

The pattern here suggests that, for any natural number n, the derivative of x^{n} should be nx^{n−1}. This is, in fact, true. The proof is not difficult. From the definition of derivative, we have that

(x^{n})′ = lim_{h→0} \frac{(x + h)^{n} − x^{n}}{h}

If we expand (x + h)^{n} using the binomial theorem we obtain

(x + h)^{n} = x^{n} + nx^{n−1}h + \frac{n(n − 1)}{2}x^{n−2}h^{2} + ⋯ + nxh^{n−1} + h^{n}

It is not important here to know exactly each term in this expansion. It is important to know the first two terms and the fact that every term except the first two has a factor of h raised to a power of at least two. When we subtact x^{n} from this expansion and divide by h we obtain

\frac{(x + h)^{n} − x^{n}}{h} = nx^{n−1} + \frac{n(n − 1)}{2}x^{n−2}h + ⋯ + nxh^{n−2} + h^{n−1}

Here, every term except the first one has a factor of h. Thus, when we take the limit as h appoaches 0, all terms except the first one will vanish. We conclude that

(x^{n})′ = lim_{h→0} \frac{(x + h)^{n} − x^{n}}{h} = nx^{n−1}

This proves the following theorem:

Theorem A. If n is any non-negative integer, then

(x^{n})′ = nx^{n−1}

We now know the derivatives of a large number of functions. For example:

(x^{4})′ = 4x^{3}
(x^{10})′ = 10x^{9}
(x^{95})′ = 95x^{94}

Next we would like to be able to find the derivative of a polynomial like x^{3} + 4x^{2} − x + 5 which is a linear combination of monomials. We need the following theorem:

Theorem B. If f and g are functions and c is a constant, then

(1)  (cf)′ = cf′
(2)  (f + g)′ = f′ + g′

That is, the derivative of a constant times a function is that constant times the derivative of the function and the derivative of the sum of two functions is the sum of their derivatives.

We won’t prove this theorem now. Its proof is in the textbook and will be done later in the course. Using this theorem and the preceding one, we can now find the derivative of any polynomial.

Example 1. Find the derivative of 3x^{2} + 2x + 5. Solution:

(3x^{2} + 2x + 5)′ = (3x^{2})′ + (2x)′ + (5)′  Th B(2)
= 3(x^{2})′ + 2(x)′ + (5)′  Th B(1)
= 3 · 2x + 2 · 1 + 0  Th A
= 6x + 2

Example 2. Find the derivative of x^{5} − 11x^{3} + 9x + 2. Solution:

(x^{5} − 11x^{3} + 9x + 2)′ = (x^{5})′ + (−11x^{3})′ + (9x)′ + (2)′
= 5x^{4} − 11 · 3x^{2} + 9 · 1 + 0
= 5x^{4} − 33x^{2} + 9

Example 3. Find the slope of the curve y = x^{4} − 3x^{2} + 2 at the point (1, 0). Solution: We first find the derivative of x^{4} − 3x^{2} + 2:

(x^{4} − 3x^{2} + 2)′ = 4x^{3} − 3 · 2x + 0 = 4x^{3} − 6x

We then evaluate the derivative at x = 1 to get the slope of the curve at the point (1, 0). Thus, the answer is

4 − 6 = −2

Example 4. Find the rate of change of the function f(x) = x^{3} − 4x with respect to x when x = 2.

Solution: We first find the derivative of f(x):

f′(x) = (x^{3} − 4x)′ = 3x^{2} − 4

We then evaluate the derivative at x = 2 to find the rate of change of f(x) with respect to x at x = 2. Thus, the answer is

3 · 2^{2} − 4 = 8

Velocity and Acceleration

Much of Physics and Engineering is concerned with the mathematics of moving bodies. Suppose a body moves along a straight line. If we fix an origin for this line and units for measuring distance on the line, then the position of the body at any time t is described by its coordinate on the line, often denoted by s(t). The velocity v(t) of the body is then defined to be the rate of change of s(t) with respect to t - that is, the derivative s′(t) of s(t). Similarly, the acceleration a(t) of the body is defined to be the rate of change of v(t) with respect to t - that is, the derivative v′(t) of v(t). In summary,

v(t) = s′(t)
a(t) = v′(t)

Example 5. If a ball if dropped off the top of a 64 foot high building, then its height s(t) above the ground t seconds later is described by the formula

s(t) = −16t^{2} + 64

What is its velocity when it hits the ground? What is its acceleration at any time? Solution: The ball hits the ground when

−16t^{2} + 64 = 0.

This happens when t^{2} = 4; that is, when t = 2. Thus, we need to know the velocity of the ball when t = 2. But

v(t) = s′(t) = −16 · 2t + 0 = −32t

At time t = 2 this is −64. Thus, the ball hits the ground with velocity −64 ft/sec. The acceleration at any time t is

a(t) = v′(t) = (−32t)′ = −32

Thus, the acceleration is a constant -32 ft/sec/sec.

4. Antiderivatives of Polynomials

We have introduced the operation of differentiation (finding the derivative of a function) and shown how to differentiate any polynomial. In this section we will study the reverse operation.

Definition A. If f(x) is a function, then an antiderivative for f is a function having f(x) as its derivative.

Antidifferentiation undoes or reverses differentiation. If f is the derivative of g, then g is an antiderivative of f. Notice that we said “an antiderivative”, not “the antiderivative”. This is because a function which has an antiderivative always has infinitely many. In fact, if g is an antiderivative of f then g + c is also an antiderivative of f for any constant c. This follows from the fact that the derivative of a constant is zero.

Given a function f, does f have an antiderivative? This is a difficult question in general and will be studied in some detail later in the course. However, for polynomials it is quite easy to see that the answer is yes. We begin by looking at the monomial x^{n}, where n is a non-negative integer. Is this the derivative of some other polynomial? If we differentiate x^{n+1} we get

(x^{n+1})′ = (n + 1)x^{n}

If we divide by n + 1 this leads to the equation

(\frac{x^{n+1}}{n + 1})′ = x^{n}

Thus, we have found an antiderivative for x^{n}, namely \frac{x^{n+1}}{n+1} . This is not the only antiderivative for x^{n}, since we can add any constant to a function without changing its derivative. Thus, any function of the form

\frac{x^{n+1}}{n + 1} + c

where c is a constant, will have derivative x^{n}.

Definition B. If f(x) is a function, the set of all antiderivatives for f(x) is denoted

∫ f(x) dx

and is called the indefinite integral of f(x).

We have shown that every function of the form \frac{x^{n+1}}{n+1} + c is an antiderivative for x^{n}. Conversely, every antiderivative for x^{n} is of this form (as long as we consider the domain for x^{n} to be the whole real line or, at least, an interval on the real line). However, to prove this requires one of the big theorems of calculus - the Mean Value Theorem - which will not be discussed until later in the course. For now we will simply assume that this is true. Then we can state the following theorem:

Theorem A. If n is a non-negative integer, then

∫ x^{n} dx = \frac{x^{n+1}}{n + 1} + c

where c ranges over all constants.

Remember, this theorem simply says that the set of antiderivatives for x^{n} is the set of functions of the form \frac{x^{n+1}}{n+1} + c. If we apply this for n = 0, 1, 2 we get:

∫ 1 dx = ∫ x^{0} dx = x + c

∫ x dx = ∫ x^{1} dx = \frac{x^{2}}{2} + c

∫ x^{2} dx = \frac{x^{3}}{3} + c

If h is an antiderivative for f and a is a constant, then

(ah)′ = ah′ = af

and so ah is an antiderivative for af. Similarly, if h is an antiderivative for f and k is an antiderivative for g, then

(h + k)′ = h′ + k′ = f + g

and so h + k is an antiderivative for f + g. Thus, we have proved the following theorem

Theorem B. If f and g are functions and a is a constant, then

(1)  ∫ af(x) dx = a ∫ f(x) dx
(2)  ∫ (f(x) + g(x)) dx = ∫ f(x) dx + ∫ g(x) dx

In other words, the integral of a constant times a function is the constant times the integral of the function and the integral of the sum of two functions is the sum of the integrals of the two functions.

Theorems A and B combined allow us to integrate (find the indefinite integral of) any polynomial.

Example 1. Integrate x^{3} + 3x − 6.

Solution:

∫ (x^{3} + 3x − 6) dx = ∫ x^{3} dx + ∫ 3x dx + ∫ −6 dx  Th B(2)
= ∫ x^{3} dx + 3 ∫ x dx − 6 ∫ 1 dx  Th B(1)
= \frac{x^{4}}{4} + \frac{3x^{2}}{2} − 6x + c  Th A

Note that we can always check our answer to an integration problem by differentiating it to see if we get back the original function. For example, let’s differentiate the answer in the preceding example:

(\frac{x^{4}}{4} + \frac{3x^{2}}{2} − 6x + c)′ = \frac{4x^{3}}{4} + \frac{3 · 2x}{2} − 6 + 0 = x^{3} + 3x − 6

Indeed, we do get back our original function and this verifies that our answer was correct.

Example 2. Find the antiderivative of x^{3} − 3x^{2} that has value 0 when x = 1.

Solution: All antiderivatives of x^{3} − 3x^{2} are given by

∫ (x^{3} − 3x^{2}) dx = \frac{x^{4}}{4} − x^{3} + c

When x = 1 this becomes

1/4 − 1 + c = −3/4 + c

Thus, if we want the antiderivative that has value 0 when x = 1, we should choose c = 3/4. The answer is then

\frac{x^{4}}{4} − x^{3} + \frac{3}{4}

Example 3. The acceleration experienced by an object due to the force of gravity is −32ft/sec^{2}. A projectile is fired straight up with an initial velocity of 128 ft/sec, after which the only force acting on it is gravity. What is its velocity t seconds later? When does it reach its maximum height?

Solution: Acceleration is the rate of change of velocity with respect to time - that is, it is the derivative of velocity as a function of time. We know the acceleration is −32ft/sec^{2}. Thus, the velocity v(t) is an antiderivative of the constant function −32. So

v(t) = ∫ −32 dt = −32t + c

When t = 0 the velocity is the initial velocity 128 ft/sec. Thus, the constant c must be 128. Therefore, the velocity at time t is

v(t) = −32t + 128

The object reaches its maximum height when its velocity drops to zero (it is not going up anymore). This happens when

−32t + 128 = 0

that is, when t = 4 seconds.

Example 4. For the preceding example, assume that the object was fired from an initial height of 10 feet off the ground and then find the height above the ground of the object at any time t. Find the maximal height achieved by the object.

Solution: Velocity is the rate of change of distance (in this case height) with respect to time. That is, it is the derivative of height with respect to time. Thus, the height s(t) we are seeking is an antiderivative of of the velocity v(t) = −32t + 128. This means

s(t) = ∫ (−32t + 128) dt = −16t^{2} + 128t + c

When t = 0 the height is the initial height 10 feet. Thus, c = 10 and

s(t) = −16t^{2} + 128t + 10

The maximum height occurs when t = 4 by the previous example. So the maximal height achieved is −16 · 16 + 128 · 4 + 10 = 266 feet.

5. Definite integrals

There are two matters which we will take as intuitively and operationally acceptable for the time being. Later in the course we will study these issues at greater depth.

1. If two differentiable functions f and g have the same derivative in an interval I, then they differ by a constant. This is the same as saying that if a function h has derivative equal to 0 everywhere in I, then it is constant (take h = f − g). Now this is intuitively obvious, if we paraphase it as saying that if a function never changes anywhere in the interval I, then it is is constant. However, that paraphrase has problems: the derivative is the instantaneous rate of change, and the assertion: “ a function which appears not to change at any given instant in fact never changes” , when made precise loses its intuitive clarity.

2. Area is well-defined; regions we shall consider in this course have area. Again, calculation of areas has been an ongoing problem since the beginning of recorded history, and the Calculus provides a most effective way of doing this, if we know that it makes sense. As we shall see below, area of a curvilinear figure is found by approximating the figure by collections of rectangles. That the approximations lead to a good concept of area is a subtle problem which for the time being, we shall ignore.

From time to time in this course we will touch on the difficulties with these assumptions, but for the present, we shall assume them as true statements.

Suppose that y = f(x) is a nonnegative function on the interval [a, b]. The area under the curve y = f(x) from a to b is denoted

∫_{a}^{b} f(x)dx.

Theorem. The area under y = f(x) is equal to F (b) − F (a), for any indefinite integral F (x) of f(x) .

Demonstration. For x any point between a and b , let A(x) be the area under the curve from a to x . Let’s calculate A′(x). For a small increment h the area, A(x + h) up to x + h is approximated by A(x) + hf(x), where hf(x) is the area of the rectangle of base h and height f(x). Thus

(1)  [A(x + h) − A(x)] = f(x)h + error,

where “error” is bounded by [f(x + h) − f(x)]h, if, for example, f is an increasing function. Thus

\frac{A(x + h) − A(x)}{h} − f(x)

is bounded by f(x + h) − f(x) which will go to zero if f is at all a reasonable function (we say, f is continuous). We conclude

lim_{x−>0} \frac{A(x + h) − A(x)}{h} = f(x).

Thus A(x) is an indefinite integral of f . If F is any other indefinite integral,

(2)  A(x) − F (x) = C,

where C is some constant which we can easily find since A(a) = 0. Substituting in (2) we obtain C = −F (a). Thus, the area under the curve from a to b is A(b) = F (b) − F (a) .

Examples 1. Find the area under y = x^{n} from 0 to 1.

An indefinite integral is F (x) = \frac{1}{n+1}x^{n+1}, so the area is F (1) − F (0) = \frac{1}{n+1}.

2. Find the area under the curve y = x^{2} + 2x^{3} from 2 to 4.

∫_{2}^{4}(x^{2} + 2x^{3})dx = \frac{1}{3}x^{3} + 2\frac{1}{4}x^{4}|_{2}^{4} =

= \frac{1}{3}64 + \frac{1}{2}256 − \frac{1}{3}8 − \frac{1}{2}16 = 416/3.

The definite integral will be defined later in this course as the limit of a sum, as follows. Suppose that y = f(x) is defined over the interval [a, b]. Let us subdivide the interval by a sequence of partition points {x_{i}}:

x_{0} = a < x_{1} < x_{2} < ⋯ < x_{n−1} < x_{n} = b

Consider the sum

f(x_{1})(x_{1} − x_{0}) + f(x_{1})(x_{2} − x_{1}) + ⋯ + f(x_{n})(x_{n} − x_{n−1})

which can be written, using the summation notation

Σ_{1}^{n}f(x_{i})(x_{i} − x_{i−1}).

If, for example, we were calculating the area under the graph of a positive function, this would be the area of a polygonal figure approximating the curvilinear figure, and would thus give an approximate value for the area. If we divide the figure more finely, using more points {x_{i}}, we would expect to get a better approximation to the area. Under suitable conditions (the continuity of the function f), these sums do approach a limit which is the definite integral of the function f from a to b.

The Fundamental Theorem of the Calculus will tell us that the definite integral can be computed as F (b) − F (a), where F is any indefinite integral of f.

Example. Suppose a particle moves in a straight horizontal line so that its velocity directed to the right at time t is v(t) = t^{2} − t^{3} meters per minute. How far to the left or right of the initial position of the particle will it be after 2 minutes?

Answer:

∫_{0}^{2} v(t))dt = ∫_{0}^{2}(t^{2} − t^{3})dt = \frac{1}{3}t^{3} − \frac{1}{4}t^{4}|_{0}^{2} = \frac{8}{3} − \frac{16}{4} − 0 = −\frac{4}{3} :

the particle is 4/3 of a meter to the left of its initial position.