Integral Calculus

Differential Calculus II

What is the meaning of an integral?

Most textbooks do not begin their discussion of integrals with a formal definition, as they do with limits and derivatives of functions. They typically begin by conceptualizing an indefinite integral as the inverse of a derivative (or antiderivative), and a definite integral as the area under a curve.

Suppose we have a function  defined on , and a function  which is continuous on , differentiable on , and  for all  in . Then,  is known as the antiderivative or indefinite integral of . This is denoted  and is read as, “ is the integral of  with respect to .” Using some of the known differentiation formulas from the previous module, let’s work out a couple of examples.

Find the antiderivative of .

Recall that the derivative of  is . Therefore, the derivative of  is , which is the value of the function inside the integral (known as the integrand). From this, we can see that  is a solution. However, so is  In fact,  where  is any constant.

What would be the answer to this antiderivative if the coefficient  were not there? Prove to yourself that the answer would be . From this, you should be able to derive the following theorem.

A simple approach is to just take the derivative of the right-hand side. You will see that you get the integrand of the left-hand side.

Before we work on the second example, let’s look at a new theorem.

Integral Calculus

Let’s try the second example. Calculate .

Using the above theorem, we rewrite the integral.

To solve for  and , recall that  and . Therefore,  and , and the solution to the indefinite integral (never forget the ) is .

The antiderivative, or indefinite integral, is a function, with infinite possible solutions. Our second conceptualization of the integral is the area under a curve, which is the definite integral of the function. Unlike indefinite integrals, a definite integral is a number rather than a function.

Let’s look at the function  over the interval . Suppose we wish to determine the area bounded by the curve and the x-axis, as depicted below.

We can estimate the area under the curve by drawing rectangles over fixed intervals (in this case, intervals of ), with either the left corner or right corner of the rectangle’s upper width touching the curve, as depicted below.

The area inside the rectangles is easy to compute. The area in the first graph is , while the area in the second graph is . We know the actual area  under the curve is between these two values, .22 and .47. We can narrow the range by taking smaller intervals, such as every .

We then compute the new areas  and .

and

We still have not arrived at a good approximation of the area, but the interval width has narrowed from  to . By continuing to narrow the intervals, both calculations will approximate the true area under the curve (which is ) more and more closely.

This method of measuring the area as a series of rectangles is known as a Riemann sum. More specifically, the first graph is an example of a left Riemann sum, and the second graph is a right Riemann sum. The intervals,  , for the two calculations are examples of partitions. In the first calculation, the right partition contains  subintervals of equal length. In the second calculation, the right partition contains  subintervals. For a function , we represent a Riemann sum using the notation , where  represents the  partition.

As the subintervals become smaller, , and the Riemann sum approaches the true area. In the closed interval , we define the definite integral of the function  as:

We refer to  as the lower limit and  as the upper limit.

To be precise, the area under the curve  in the interval  is not  if the value of  is negative in any part of the open interval. For example, look at the area under the curve  on the interval .

As we will see, if we were to evaluate the definite integral in the interval , we would find , since the shaded area below the x-axis cancels out the symmetrically shaded area above the x-axis. To rectify this, we define the area as . In this example, we split the integral into two portions and find the sum of the positive values. Thus, .

What is the Fundamental Theorem of Calculus?

In order to complete our understanding of definite integrals, we state and prove the Fundamental Theorem of Calculus.

Proof of:

Picture  as the area under the curve  from  to . Then  is the area from  to , and  is the area from  to . If  is small, the area from  to  should be small. Also, since  is continuous, . That area should then approximate a rectangle of height  and width . Therefore,

, or . As , this approximation becomes exact, so .

Proof of ():

Given , and using (i), it follows that , which means  ,or . Therefore,  ,and , so . The second integral is clearly 0 (imagine the area of a rectangle with width 0). Therefore, .

Part (ii) of the theorem states that the value of the definite integral on the interval  is simply the difference of the antiderivative at the endpoints. Our shorthand is . Let’s try a couple of examples solving definite integrals.

Find the actual areas under the curves on the given intervals:

(A) area under  over the interval 

(B) area under  over the interval 

(A)  everywhere in the given interval, so .

(B)  on the interval , and  on the interval , so

Integral Calculus

A brick is dropped from the top of a building that is 500 meters tall. Its velocity after  seconds is given by  meters per second. How long before the brick hits the ground?

You may recall that we found  by taking the first derivative of the position ; that is, . Here, we are given  and need to solve for , which is the antiderivative of the velocity function. From part (ii) of the Fundamental Theorem of Calculus, we know that . In this example, a represents the time corresponding to the start of the fall and b represents the time corresponding to the end of the fall, the time we are looking for. So we will set  seconds,  seconds, and solve for . We know  m and . Therefore, . Also, the total change in the brick’s position is simply its final position minus its initial position, or  meters. Therefore, , or  seconds ().

Before we proceed with the next example, let’s explore a notational detail which may seem strange at first. Recall that the expression  can be read as, “G of x is the integral of f of x, with respect to x.” What if f(x) is a constant function? If f(x) = 5, then  = . This can be read as, “G of x is 5 times the integral of 1, with respect to x.” We now know this is 5x + C, where C is some constant of integration. Typically, this last term is written in the simplified form . If the integrand symbol followed by the dx with nothing in the middle seems strange, imagine a constant of 1 sitting between the integrand symbol and the dx.

Now let’s try the example.

The marginal revenue (rate of change of revenue received per unit change in quantity sold) received by a retailer is given by , where  is the quantity of units sold. Compute the total revenue () as a function of .

According to our definition of marginal revenue, . Therefore, we compute  as the antiderivative of .

To find a complete solution to the problem, we must now solve for C by checking to see what value of C will make our revenue function accurately reflect the conditions set forth in the original problem statement. If we sell a quantity of 0, we make no revenue.

,
or
.

Our final solution is .

How do we find the area bounded by two curves?

We’ve developed a formula for the area under the curve  in : . More specifically, this formula describes the area under the curve  bounded by the lines , and the x-axis. Let’s now consider the area bounded by two curves,  and . Let , and , and look at the area under each curve on .

The curves intersect at  and , and their individual areas are represented by  and .

Now let’s look at both graphs and the area between them on .

The area between the curves is the difference between the areas under each separate curve bounded by the x-axis. On , , and the area is represented by

On , , and the area is

Therefore, to find the area  on  we need to solve the definite integrals.

In some applications, it may be easier to compute the area of the region bounded by two curves by looking at horizontal elements rather than vertical rectangles. For example, look at the region bounded by the parabola  and the line . The area of the region with a vertical rectangle is shown below.

Note that the interval is on , and the line becomes the lower bound on . To compute the area this way, first, we need to solve the first equation for .

These separate equations represent the upper and lower halves of the parabola.

We solve for the total area in two parts. On , the integrand is the difference between the upper and lower halves of the parabola.

On , the integrand is the difference between the upper half of the parabola and the line.

The total area is the sum .

Let’s look at this problem again, but this time, we will compute the area by taking horizontal rectangular elements. Note that the curves intersect at  and .

First, we need to solve both equations for . We get  and . The interval of integration is on the y-range , and the heights of the rectangles are . Therefore, we can compute the total area between the curves with a single integral.

Clearly, this method is easier than the first.

How do we find the volume of a solid of revolution?

Another application of definite integrals is calculating the volume of a solid of revolution. This is the solid created by revolving a portion of a curve about a line, such as the x– or y-axis. Let’s take  on  and see what happens if we revolve that portion of the curve about the x-axis and y-axis.

x-axis	y-axis

For the first graph, imagine a rectangular sliver of width  as it revolves around the x-axis. It forms a circular disk of volume . We can approximate the volume as the sum of rectangles along the interval, just as we have been approximating the areas, but we will skip directly to what happens as . As , the sum of rectangles becomes the definite integral on the interval , and the volume becomes .

For example, solve .

Suppose we wish to revolve this same function on the same interval as in the second graph about the y-axis. First, we need to solve for . (In this case, , as we are only looking at .) Each sliver forms a disk of volume , and we integrate on the y-interval . The volume of the solid revolved around the y-axis is .

Finally, the volume of the solid of revolution bounded by two curves  and  is found by using reasoning similar to that which we have used with areas. If both functions are continuous and  over , then the volume of the solid of revolution of the region bounded by  and  about the x-axis is

Integral Calculus

Another computation it is useful to be able to find is the area of surface of revolution. The area of the surface of revolution obtained by rotating  on  about the x-axis is

The area of the surface of revolution obtained by rotating  on the y-interval  about the y-axis is

Another application of definite integrals is finding the length of a curve, or arc length. You should recall that the distance between point  and point  is . We can approximate the arc length of a curve continuous on  as the sum of straight lines between points  and . Each small segment has a length . As the length intervals become infinitesimally small, the exact length can be computed as the definite integral on . The value  approaches  (the proof of this utilizes the mean value theorem) and . As long as  and  are continuous on , the length of the curve from  to  is given by

If we want to find the length of the curve  from  to  and  and  are continuous on , the length of the curve is

The final application of definite integrals we will examine is the work done by a force acting on an object. If the object is acted on by a force  and is displaced along the x-axis from point  to point , the work done by the force is . Work is usually measured in foot-pounds, or, if the force is in newtons and the displacement is in meters, in joules. Sometimes, force is measured in dynes and displacement in centimeters. Then, work is measured in dyne-cm, or ergs.

Let’s try an example. Consider the curve  from  to . Compute the following:

(A) the area under the curve bounded by the x-axis, using vertical rectangles;

(B) the area under the curve bounded by the x-axis, using horizontal rectangular elements;

(C) the volume of the solid of revolution about the x-axis; and

(D) the volume of the solid of revolution about the y-axis.

(A) 

(B) Solve  for x.  . At , and at . The height of each rectangle is , and its width is . Therefore,

(C) 

(D) 

Let’s try a second example. For the line , find:

(A) the length of arc from  to ; and

(B) the area of surface of revolution about the x-axis over the same interval.

(A) At , ; at , . Also, , so . Therefore,

Since the curve in question is simply a line, we can verify this solution by using the distance formula to find the distance between the points (9, 7) and (18, 10). Indeed, the distance formula gives us the same answer for the length of the arc from  to .

What is integration by substitution?

So far, we have solved definite and indefinite integrals where the integrand is of the form , , and . This section examines integrals of various functions, and techniques for solving integrals of various forms.

The first method we will examine is integration by substitution. We know how to solve for integrals in the form  (). For example, we know how to solve . This method does not work if the power of . However, we can use substitution to solve a similar integral: . Let , so  and substitute  for .

Thus, if we have an integral whose integrand can be placed in the form  , we can use the chain rule and substitution to transform the integral to the form . Then, it can be easily integrated.

For the case , we can determine . Recall from differential calculus that . But . Therefore,

Let’s try a few examples. Compute .

Let . Then . Therefore,

Compute . Let , and . Therefore

What is integration by parts?

Another integration technique is known as integration by parts. If an integral is in the form , but can be solved in the form , use the following.

This formula is derived from the product rule in differential calculus. Recall that . When solving for , we find that . Finally, integrate all three terms to arrive at the formula. The goal of integration by parts is to transform one integral into another integral that is “solvable.” After deciding what to call  and  and solving for  and , one of three things should happen:

(1) is as hard or harder to solve than , and the process is abandoned.

(2)  is easy to solve, and the formula is straightforward.

(3) Solving for  leads to a solution with , which is solved by algebra.

The following examples illustrate each of these scenarios.

Compute  by setting  and .

We find that  Therefore,  However, this new integral is even more difficult to solve than the original, so we abandon this process.

Compute  by setting  and .

We find that . Therefore, . We can easily solve this new integral by substitution. Let , so . The integral is . So,

Integral Calculus

These first two examples illustrate an important point about integration problems and the methods used. For most of our mathematical career, mathematics is generally algorithmic. As long as rules and procedures are memorized and practiced, a solution can usually be reached through careful perseverance. This is not the case when working with integrals. In fact, the integrals of most functions, if they are randomly drawn from the “universe of functions,” are not solvable manually by any of our methods. Even if they are solvable, not every possible method will get us to the solution. Don’t let this frustrate you, and try to remember that with practice, you will begin to recognize patterns in the structure of the functions you are integrating which give clues as to possible methods that might be worth trying.

Compute .

First, let . Set , and . We find that . Therefore,  To solve this new integral, perform integration by parts. Set  and . It is then true that , so  . Now, we have  If we apply simple algebra, we find that  or . Therefore,

What is trigonometric substitution?

If the integrand is in the form , we can solve the integral by trigonometric substitution. Recall the trigonometric identities  and . If we set , the term  simplifies to   If we set , the term  simplifies to  . And if we set , the term  simplifies to 

Integral Calculus

Trigonometric substitution is often easier with definite integrals than with indefinite integrals. With definite integrals, we can leave the answer in terms of  and change the upper and lower limit of the integral into terms of  rather than , where appropriate. With indefinite integrals, we must change  to , which can be tricky. Let’s try an example. Compute . Hint: .

Let , so . Since the 0 and 1 correspond to x-values, not θ-values, we must change the limits of integration. Since , and we are now integrating with respect to θ, we look for a θ value which will make sin θ = 0. This requirement is met when θ = 0. The new upper limit, which corresponds to , is satisfied when , or . Therefore,

The next integration technique is useful when the integrand can be expressed as a rational function, that is, the quotient of two polynomial functions . This method involves breaking up  into the sum of partial fractions. For example, if the integrand is , create partial fractions by setting . By rewriting each term on the right so that the denominators are identical to the denominator on the left and reducing, we have  Factoring and rearranging terms gives  We now have 3 equations and 3 unknowns . The solution you should get is . We can use these terms to rewrite the integrand as the sum of three simple terms and integrate each one separately.

Here is another example. Compute .

Using partial fractions, we rewrite this integral as

When the denominator has factors in the form  for , all factors of  must be included in partial fractions. For example, the quotient  is broken up into partial fractions .

Integral Calculus

Another technique involving rational functions is completing the square. This technique is illustrated in the following example.

Compute  by completing the square.

Set , so . Therefore,

We solve this integral by partial fractions:

.

Therefore, 

The solution is , so

The next form of function requires an understanding of inverse trigonometric functions. For example, the inverse sine function  is satisfied if . We know  is in the interval . In order for each value of x in this interval to have a unique , we restrict  to the interval . The inverse cosine function  exists if and only if  on . The inverse tangent function  exists if and only if  on . The inverse secant function 

Techniques such as substitution and completing the square are often required in order to factor the integrand into one of the above forms.

Using the integration techniques introduced in this module along with other techniques and tricks, we are capable of integrating and, in the case of definite integrals, finding an exact numerical solution. Unfortunately, exact solutions cannot be found for all definite integrals, such as  and . For these circumstances, we rely on methods of numerical or approximate integration. We already examined a method of approximating the value of a definite integral: the Riemann sum. However, the example we looked at demonstrates that this is a slow-converging approximation method. We will now examine two numerical methods that provide better accuracy: the trapezoidal rule and Simpson’s rule.

With Riemann sums, the area of each subinterval is approximated as a rectangle along the partition. The trapezoidal rule more accurately defines each area as a trapezoid, which better represents the change in the function. We illustrate this point by referring back to our original function, . Compare the areas on the interval  of a left Riemann sum, right Riemann sum, and trapezoid with the exact area .

Area under a curve 1	Area under curve 2	Area under curve 3

The area of the first region is . The area of the second region is  . The area of the third region is . The true area is .

Clearly, the trapezoid is closer to the true area than either rectangle. (The trapezoidal area is actually the average of .)

If  is continuous on  and  are regular partitions of , then the trapezoidal rule states:

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Now use the trapezoidal rule to estimate the area under the curve  with .

This is the same interval we examined previously. Recall that we determined  using Riemann sums. First, compute . Therefore,

The trapezoid formula converges to the true area  with 8 intervals more efficiently than either Riemann sum.

The trapezoidal rule, like Riemann sums, approximates the definite integral as the sum of areas formed by straight lines connecting two points. In contrast, Simpson’s rule computes the area bounded by parabolas connecting three points. This gives us a better approximation to the area under the curve. Simpson’s rule states:

Note that Simpson’s rule has  subintervals while the trapezoid rule has  subintervals.

For example, use Simpson’s rule to compute  with .

= 

The reason this value is the exact area is that Simpson’s rule assumes each partition is a parabola, and in this case,  is an actual parabola.

We conclude our analysis of integral calculus with improper integrals. This is a type of definite integral  where either , or  becomes infinite for some value of  on .

If the limit exists and is finite, the improper integral is convergent. If the improper integral does not exist, it is divergent. The integral diverges if it blows up to , or if it has no defined limit. An example of the latter situation is .

Compute the area bounded by the x-axis on  for the following curves:

(A); and

(B).

(A).

(B).

This limit does not exist—it blows up to . This improper integral is divergent.

Another type of improper integral is one that blows up at one or more points on a finite interval. Such a point is represented by a vertical asymptote. If  has a vertical asymptote on  at , and if  exist for  in , and for  in , then

Compute .

The integrand has an infinite discontinuity at , as illustrated below

Applying the above rule to this integral, with , we find

Since neither of these limits exists, the integral is divergent. Had we not recognized the discontinuity and attempted to solve the definite integral, we would have arrived at the wrong answer of . This answer makes no sense, since the integrand is always positive.