Riemann Sums and Definite Integrals - Weeblygourley.weebly.com/uploads/5/0/9/1/5091587/lesson_4.3...4.3 and 4.4 Part 1 filled in Smartboard lesson.notebook December 10, 2013 Definite

4.3 and 4.4 Part 1 filled in Smartboard lesson.notebook December 10, 2013

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Riemann Sums and Definite Integrals


4.3


• Understand the definition of a Riemann sum.

• Evaluate a definite integral using limits.

• Evaluate a definite integral using properties of definite integrals.

Objectives

Riemann SumsIn mathematics, a Riemann sum is a method for approximating the total area underneath a curve on a graph, otherwise known as an integral. It may also be used to define the integration operation. The method was named after German mathematician Bernhard Riemann. Some examples are Upper Sums, Lower Sums, and Midpoint Sums like we learned about in Section 4.2.


Example 1 – A Partition with Subintervals of Unequal Widths

Consider the region bounded by the graph of and the xaxis for 0 ≤ x ≤ 1, as shown in Figure 4.17.

Notice that the rectangles are notthe same width. You don’t have to have equal widths to do a Riemann Sum, (but it is easier to do if the subintervals have equal widths).

Figure 4.17

Riemann Sums


Definite Integrals

Definite Integrals

Basically, as we divide a region into an infinite number of rectangles, each having a width of , we get infinitely close to the actual area of the region. This is called the definite integral and is denoted by

where a and b are upper and lower limits.


Definite Integrals

Figure 4.21

Definite Integrals


As an example of Theorem 4.5, consider the region bounded by the graph of f(x) = 4x – x2 and the xaxis, as shown in Figure 4.22.

Because f is continuous and nonnegative on the closed interval [0, 4], the area of the region is

Figure 4.22

Definite Integrals

You can evaluate a definite integral in two ways—you can use the limit definition or you can check to see whether the definite integral represents the area of a common geometric region such as a rectangle, triangle, or semicircle.

Definite Integrals


Example 3 – Areas of Common Geometric Figures

Sketch the region corresponding to each definite integral. Then evaluate each integral using a geometric formula.

a.

b.

c.

Example 3(a) – Solution

This region is a rectangle of height 4 and width 2.

Figure 4.23(a)


Example 3(b) – SolutionThis region is a trapezoid with an altitude of 3 and parallel bases of lengths 2 and 5. The formula for the area of a trapezoid is h(b1 + b2).

Figure 4.23(b)

cont’d

Example 3(c) – SolutionThis region is a semicircle of radius 2. The formula for the area of a semicircle is

Figure 4.23(c)

cont’d


Because the definite integral in the example below is negative, it does not represent the area of the region shown in Figure 4.20. Definite integrals can be positive, negative, or zero.

For a definite integral to be interpreted as an area, the function f must be continuous and nonnegative on [a, b].

Figure 4.20

Definite Integrals

Properties of Definite Integrals


Properties of Definite IntegralsThe definition of the definite integral of f on the interval [a, b] specifies that a < b.

Now, however, it is convenient to extend the definition to cover cases in which a = b or a > b.

Geometrically, the following two definitions seem reasonable.

For instance, it makes sense to define the area of a region of zero width and finite height to be 0.



Example 4 – Evaluating Definite Integrals

a. Because the sine function is defined at x = π, and the upper and lower limits of integration are equal, you can write

b. The integral has a value of

so you can write

In Figure 4.24, the larger region can be divided at x = c into two subregions whose intersection is a line segment.

Because the line segment has zero area, it follows that the area of the larger region is equalto the sum of the areas of the two smaller regions.

Figure 4.24

cont’dExample 4 – Evaluating Definite Integrals


Example 5 – Using the Additive Interval Property


Note that Property 2 of Theorem 4.7 can be extended to cover any finite number of functions. For example,


Example 6 – Evaluation of a Definite Integral

Evaluate using each of the following values.

Solution:

If f and g are continuous on the closed interval [a, b] and

0 ≤ f(x) ≤ g(x)

for a ≤ x ≤ b, the following properties are true.

First, the area of the region bounded by the graph of f andthe xaxis (between a and b) must be nonnegative.


Second, this area must be less than or equal to thearea of the region bounded by the graph of g and the xaxis (between a and b ), as shown in Figure 4.25. These two properties are generalized in Theorem 4.8.

Figure 4.25



The Fundamental Theorem of Calculus


4.4


• Evaluate a definite integral using the Fundamental Theorem of Calculus.

• Find the average value of a function over a closed interval.• Understand and use the Second Fundamental Theorem of

Calculus.

Objectives




The two major branches of calculus: differential calculus and integral calculus. At this point, these two problems might seem unrelated—but there is a very close connection.

The connection was discovered independently by Isaac Newton and Gottfried Leibniz and is stated in a theorem that is appropriately called the Fundamental Theorem of Calculus.



Example 1 – Solution

The Fundamental Theorem of CalculusThe following guidelines can help you understand the use of the Fundamental Theorem of Calculus.


Example 3(a) – Solution

This region is a rectangle of height 4 and width 2.

Figure 4.23(a)

Example 3(b) – SolutionThis region is a trapezoid with an altitude of 3 and parallel bases of lengths 2 and 5. The formula for the area of a trapezoid is h(b1 + b2).

Figure 4.23(b)

cont’d


Example 3(c) – SolutionThis region is a semicircle of radius 2. The formula for the area of a semicircle is

Figure 4.23(c)

cont’d

Average Value of a Function


In Figure 4.31 the area of the region under the graph of f is equal to the area of the rectangle whose height is the average value.

Average value is like average height.


Figure 4.31


ba is just the total width of the area we are integrating.


Example 4 – Finding the Average Value of a Function

Find the average value of f(x) = 3 x 2 – 2 x on the interval [1, 4].

Solution:The average value is given by

Figure 4.32

The Second Fundamental Theorem of Calculus


The definite integral of f on the interval [a, b] is defined using the constant b as the upper limit of integration and x as the variable of integration.

A slightly different situation may arise in which the variable x is used in the upper limit of integration.

To avoid the confusion of using x in two different ways, t is temporarily used as the variable of integration.





But what if we are doing the derivative of an integral. Then what would happen?

If we are just told to integrate, we evaluate using the First Fundamental Theorem of Calculus:

This result is generalized in the following theorem, called the Second Fundamental Theorem of Calculus.


Remember, this only works if you are taking the derivative of an integral, not the other way around, (integral of a derivative). Also, there must be a constant for the lower limit and x in the upper limit.


Example 7 – Using the Second Fundamental Theorem of Calculus

Evaluate

Solution:Note that is continuous on the entire real line. So, using the Second Fundamental Theorem of Calculus, you can write


Remember we said there must be a constant for the lower limit and an x in the upper limit to use the Second Fundamental Theorem of Calculus. It turns out that you can also use the theorem when the lower limit is a constant and the upper limit is a function of x. The only difference is that we plug in the function of x for t (instead of just the x), and we also multiply by the derivative of the function we plugged in. Here is an example:

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Riemann Sums and Definite Integrals - Weeblygourley.weebly.com/uploads/5/0/9/1/5091587/lesson_4.3...4.3 and 4.4 Part 1 filled in Smartboard lesson.notebook December 10, 2013 Definite