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7.1 Indefinite Integrals Calculus
Learning Objectives
A student will be able to:
Find antiderivatives of functions. Represent antiderivatives. Interpret the constant of integration graphically. Solve differential equations. Use basic antidifferentiation techniques. Use basic integration rules.
Introduction
In this lesson we will introduce the idea of the antiderivative of a function and formalize as indefinite integrals. We will derive a set of rules that will aid our computations as we solve problems.
Antiderivatives
Definition
A function is called an antiderivative of a function if for all in the domain of
Example 1:
Consider the function Can you think of a function such that? (Answer:
many other examples.)
Since we differentiate to get we see that will work for any constant Graphically, we can
think the set of all antiderivatives as vertical transformations of the graph of The figure shows two such transformations.
With our definition and initial example, we now look to formalize the definition and develop some useful rules for computational purposes, and begin to see some applications.
Notation and Introduction to Indefinite Integrals
The process of finding antiderivatives is called antidifferentiation , more commonly referred to as integration. We have a particular sign and set of symbols we use to indicate integration:
We refer to the left side of the equation as “the indefinite integral of with respect to " The function is called the integrand and the constant is called the constant of integration. Finally the symbol indicates that we are to integrate with respect to
Using this notation, we would summarize the last example as follows:
Using Derivatives to Derive Basic Rules of Integration
As with differentiation, there are several useful rules that we can derive to aid our computations as we solve problems.
The first of these is a rule for integrating power functions, and is stated as follows:
We can easily prove this rule. Let. We differentiate with respect to and we have:
The rule holds for What happens in the case where we have a power function to integrate with
say. We can see that the rule does not work since it would result in division by. However,
if we pose the problem as finding such that , we recall that the derivative of logarithm functions had this
form. In particular,. Hence
In addition to logarithm functions, we recall that the basic exponentional function, was special in that its derivative was equal to itself. Hence we have
Again we could easily prove this result by differentiating the right side of the equation above. The actual proof is left as an exercise to the student.
As with differentiation, we can develop several rules for dealing with a finite number of integrable functions. They are stated as follows:
Differential Equations
We conclude this lesson with some observations about integration of functions. First, recall that the integration process
allows us to start with function from which we find another function such that This latter equation is called a differential equation. This characterization of the basic situation for which integration applies gives rise to a set of equations that will be the focus of the Lesson on The Initial Value Problem.
Example 4:
Solve the general differential equation
Solution:
We solve the equation by integrating the right side of the equation and have
We can integrate both terms using the power rule, first noting that and have
Lesson Summary
- We learned to find antiderivatives of functions.
- We learned to represent antiderivatives.
- We interpreted constant of integration graphically.
- We solved general differential equations.
- We used basic antidifferentiation techniques to find integration rules.
- We used basic integration rules to solve problems.
Multimedia Link
The following applet shows a graph, and its derivative,. This is similar to other applets we've explored with a
function and its derivative graphed side-by-side, but this time is on the right, and is on the left. If you edit the
definition of , you will see the graph of change as well. The parameter adds a constant to. Notice that
you can change the value of without affecting. Why is this? Antiderivative Applet.
Review Questions
In problems #1–3, find an antiderivative of the function
In #4–7, find the indefinite integral
4
x dx
x x
^
- Solve the differential equation.
- Find the antiderivative of the function that satisfies
- Evaluate the indefinite integral (^) x dx. (Hint: Examine the graph of f ( ) x x .)
Review Answers
2
4 4
x
x dx C
x x x
^ ^ ^ ^
7.2 The Initial Value Problem
Learning Objectives
Find general solutions of differential equations Use initial conditions to find particular solutions of differential equations
Introduction
In the Lesson on Indefinite Integrals Calculus we discussed how finding antiderivatives can be thought of as finding
solutions to differential equations: We now look to extend this discussion by looking at how we can designate and find particular solutions to differential equations.
Let’s recall that a general differential equation will have an infinite number of solutions. We will look at one such equation and see how we can impose conditions that will specify exactly one particular solution.
Example 1:
Suppose we wish to solve the following equation:
Solution:
We can solve the equation by integration and we have
We note that there are an infinite number of solutions. In some applications, we would like to designate exactly one solution. In order to do so, we need to impose a condition on the function We can do this by specifying the value of
for a particular value of In this problem, suppose that add the condition that This will specify exactly one value of and thus one particular solution of the original equation:
Substituting into our general solution gives or
Hence the solution is the particular solution of the original equation satisfying the initial condition
We now can think of other problems that can be stated as differential equations with initial conditions. Consider the following example.
Example 2:
Suppose the graph of includes the point and that the slope of the tangent line to at any point is given by the
expression Find
Solution:
We can re-state the problem in terms of a differential equation that satisfies an initial condition.
with.
By integrating the right side of the differential equation we have
Hence is the particular solution of the original equation satisfying the initial
condition
Finally, since we are interested in the value , we put into our expression for and obtain:
Lesson Summary
- We found general solutions of differential equations.
- We used initial conditions to find particular solutions of differential equations.
Multimedia Link
The following applet allows you to set the initial equation for and then the slope field for that equation is displayed.
In magenta you'll see one possible solution for. If you move the magenta point to the initial value, then you will see the graph of the solution to the initial value problem. Follow the directions on the page with the applet to explore this idea, and then try redoing the examples from this section on the applet. Slope Fields Applet.
Review Questions
In problems #1–3, solve the differential equation for
In problems #4–7, solve the differential equation for given the initial condition.
- and.
- and
- and
7. and f ( )^ 3 3 ^12
- Suppose the graph of f includes the point (-2, 4) and that the slope of the tangent line to f at x is -2x+4. Find f(5).
Initial Condition & Integration of Trig Functions Practice
- Find the particular solution (^) y f ( ) x that satisfies the differential equation and initial condition.
a. f '( ) x 3 x 3, f (1) 4 b. f '( ) x 6 ( x x 1), f (10) 10
c. 3
2 3 '( ) , 0, (2) 4
x f x x f x
d. 2 '( ) sec , 2 3 3
f x x f
(^)
- Find the equation of the function f whose graph passes through the point.
f '( ) x 6 x 10, (^) 4,2
- Find the function f that satisfies the given conditions.
a. f "( ) x 2, f '(2) 5, f (2) 10 b. 2 3 f "( ) x x , f '(8) 6, f (0) 0
4. Integrate.
a. (^) (2sin x 3cos x dx ) b. 1 csc cot t t dt
c. 2 (^) csc cos d d. 2 t^ sin t dt
Answers:
1. a.
3 f ( ) x 2 x^2 3 x 1 b. f ( ) x 2 x^3 3 x^2 1710
c. 2
1 1 1 ( ) 2
f x x x
d. f ( ) x tan x 3
3 f ( ) x 4 x^2 10 x 10
3. a.
2 f ( ) x x x 4 b.
4 ( ) 9 3 4
f x x
- a. 2cos x 3sin x C b. t csc t C
c. cot sin C d.
3 cos 3
t t C
7.3 The Area Problem
Learning Objectives
Use sigma notation to evaluate sums of rectangular areas Find limits of upper and lower sums Use the limit definition of area to solve problems
Introduction
In The Lesson The Calculus we introduced the area problem that we consider in integral calculus. The basic problem was this:
Suppose we are interested in finding the area between the axis and the curve of from to
We approximated the area by constructing four rectangles, with the height of each rectangle equal to the maximum value of the function in the sub-interval.
We then summed the areas of the rectangles as follows:
and
We call this the upper sum since it is based on taking the maximum value of the function within each sub-interval. We noted that as we used more rectangles, our area approximation became more accurate.
We would like to formalize this approach for both upper and lower sums. First we note that the lower sums of the area
of the rectangles results in Our intuition tells us that the true area lies
Another Look at Upper and Lower Sums
We are now ready to formalize our initial ideas about upper and lower sums.
Let be a bounded function in a closed interval and the partition of into subintervals.
We can then define the lower and upper sums, respectively, over partition , by
where is the minimum value of in the interval of length and is the maximum value of in the interval of length
The following example shows how we can use these to find the area.
Example 2:
Show that the upper and lower sums for the function from to approach the value
Solution:
Let be a partition of equal sub intervals over We will show the result for the upper sums. By our definition we have
We note that each rectangle will have width indicated:
We can re-write this result as:
We observe that as
We now are able to define the area under a curve as a limit.
Definition
Let be a continuous function on a closed interval Let be a partition of equal sub intervals over Then the area under the curve of is the limit of the upper and lower sums, that is
Example 3:
Use the limit definition of area to find the area under the function from to
- from to
- from to
- from to
In problems #9–10, state whether the function is integrable in the given interval. Give a reason for your answer.
- on the interval
- on the interval
Review Answers
(note that we have included areas under the x-axis as negative values.)
- Yes, since is continuous on
- No, since ;
7.4 Definite Integrals
Learning Objectives
Use Riemann Sums to approximate areas under curves Evaluate definite integrals as limits of Riemann Sums
Introduction
In the Lesson The Area Problem we defined the area under a curve in terms of a limit of sums.
where
and were examples of Riemann Sums. In general, Riemann Sums are of form where each is the value we use to find the length of the rectangle in the sub-interval. For example, we used the maximum function value in each sub-interval to find the upper sums and the minimum function in each sub-interval to find the lower sums. But since the function is continuous, we could have used any points within the sub-intervals to find the limit. Hence we can define the most general situation as follows:
Definition
If is continuous on we divide the interval into sub-intervals of equal width with. We let be the endpoints of these sub-intervals and let beany sample points in these sub-intervals. Then the definite integral of from to is
Example 1:
Evaluate the Riemann Sum for from to using sub-intervals and taking the sample points to be the midpoints of the sub-intervals.
Solution:
If we partition the interval into equal sub-intervals, then each sub-interval will have length So we
have and
Lesson Summary
- We used Riemann Sums to approximate areas under curves.
- We evaluated definite integrals as limits of Riemann Sums.
Multimedia Link
For video presentations on calculating definite integrals using Riemann Sums (13.0) , see Riemann Sums, Part 1
(6:15) and Riemann Sums, Part 2 (8:32).
The following applet lets you explore Riemann Sums of any function. You can change the bounds and the number of partitions. Follow the examples given on the page, and then use the applet to explore on your own. Riemann Sums
Applet. Note: On this page the author uses Left- and Right- hand sums. These are similar to the sums and that you have learned, particularly in the case of an increasing (or decreasing) function. Left-hand and Right-hand sums are frequently used in calculations of numerical integrals because it is easy to find the left and right endpoints of each interval, and much more difficult to find the max/min of the function on each interval. The difference is not always important from a numerical approximation standpoint; as you increase the number of partitions, you should see the Left- hand and Right-hand sums converging to the same value. Try this in the applet to see for yourself.
Review Questions
In problems #1–7 , use Riemann Sums to approximate the areas under the curves.
- Consider from to Use Riemann Sums with four subintervals of equal lengths. Choose the midpoints of each subinterval as the sample points.
- Repeat problem #1 using geometry to calculate the exact area of the region under the graph of from to (Hint: Sketch a graph of the region and see if you can compute its area using area measurement formulas from geometry.)
- Repeat problem #1 using the definition of the definite integral to calculate the exact area of the region under the graph of from to
- from to Use Riemann Sums with five subintervals of equal lengths. Choose the left endpoint of each subinterval as the sample points, or use trapezoids.
- Repeat problem #4 using the definition of the definite integral to calculate the exact area of the region under the graph of from to
- Consider Compute the Riemann Sum of f on [0, 1] under each of the following situations. In each case, use the right endpoint as the sample points, or use trapezoids. a. Two sub-intervals of equal length. b. Five sub-intervals of equal length. c. Ten sub-intervals of equal length. d. Based on your answers above, try to guess the exact area under the graph of f on [0, 1].
- Consider. Compute the Riemann Sum of f on [0, 1] under each of the following situations. In each case, use the right endpoint as the sample points. a. Two sub-intervals of equal length. b. Five sub-intervals of equal length.
c. Ten sub-intervals of equal length. d. Based on your answers above, try to guess the exact area under the graph of f on [0, 1].
- Find the net area under the graph of ; to (Hint: Sketch the graph and check for symmetry.)
- Find the total area bounded by the graph of and the x-axis, from to
- Use your knowledge of geometry to evaluate the following definite integral: (Hint: set 2
y 9 x^ and square both sides to see if you can recognize the region from geometry.)
Review Answers
- using the Left sum, Area = 13.68 using the Trapezoid Sum
- a. using the Right sum, Area = 1.125 using the Trapezoid Sum b. using the Right sum, Area = 1.02 using the Trapezoid Sum c. using the Right sum, Area = 1.005 using the Trapezoid Sum d. is a good guess
- a. b. c. d.
- The graph is symmetric about the origin; hence net Area = 0
- Area = 1/
- The graph is that of a quarter circle of radius 3; hence Area =
7.5 Evaluating Definite Integrals
Learning Objectives
Use antiderivatives to evaluate definite integrals Use the Mean Value Theorem for integrals to solve problems Use general rules of integrals to solve problems
Introduction
In the Lesson on Definite Integrals, we evaluated definite integrals using the limit definition. This process was long and tedious. In this lesson we will learn some practical ways to evaluate definite integrals. We begin with a theorem that provides an easier method for evaluating definite integrals. Newton discovered this method that uses antiderivatives to calculate definite integrals.
Theorem:
If is continuous on the closed interval then
where is any antiderivative of
