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Area and the Definite Integral Tidewater Community College Mr. Joyner, Dr. Julia Arnold and Ms. Shirley Brown using Tans 5th edition Applied Calculus for.

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Presentation on theme: "Area and the Definite Integral Tidewater Community College Mr. Joyner, Dr. Julia Arnold and Ms. Shirley Brown using Tans 5th edition Applied Calculus for."— Presentation transcript:

1 Area and the Definite Integral Tidewater Community College Mr. Joyner, Dr. Julia Arnold and Ms. Shirley Brown using Tans 5th edition Applied Calculus for the managerial, life, and social sciences text

2 Area and the Definite Integral Suppose a states annual rate of oil consumption each year over a 4-year period is a constant function, such as f(t) = 2, where t is measured in years and f(t) is measured in millions of barrels. We would graph such a function as:

3 Area and the Definite Integral How could you use the graph to answer the question, How much oil was used during the four years? Answer: 2(4-0) = 8 million barrels It seems reasonable that the answer is 8 million barrels and this is the same as the area of the rectangle…the area under the curve. (Yes, even a straight line is considered a curve in mathematics!)

4 Area and the Definite Integral How would you find the area above the x-axis and under the curve f(x) = x from x = 0 to x = 4? Does it seem reasonable to find the area of the triangle… the area under the curve.

5 Area and the Definite Integral Imagine now that our function, f(x) = x, represents the oil consumption of the same state over the same four years. What was the oil consumption over the four years? Lets explore.

6 The average rate of change over the first year is (1 – 0)/2 = 0.5 million barrels per year. Area and the Definite Integral For the first year, the state started off consuming no oil (zero barrels per year), and ended the year consuming oil at the rate of 1 million barrels per year. Since the rate of increased consumption is linear, we can use the average rate of change over the year to represent the entire first year. For one year, (the first year) the amount of oil consumed is: 1(0.5) = 0.5 million barrels

7 The average rate of change over the 2-year period is (2 – 0)/2 = 1 million barrels per year. Area and the Definite Integral For the first two years, the state started off consuming no oil and ended the two-year year period consuming oil at the rate of 2 million barrels per year. Again, since the rate of increased consumption is linear, we can use the average rate of change over the 2 years to represent the entire two-year period. For two years. the amount of oil consumed is: 2(1) = 2 million barrels.

8 Area and the Definite Integral In a similar fashion we can reason that the total oil consumption is as follows: Total Number of Years Average Rate of Consumption Total oil consumption 1.51(.5) =.5 212(1) = 2 31.53(1.5) = 4.5 424(2) = 8 Lets compare these values to… the areas of the right triangles under the curve f(x).

9 Area and the Definite Integral Triangle areas: Base (total number of years) Height (average rate of consumption) Area = ½(bh) (total oil consumption) 11½(1)(1) = 0.5 22½(2)(2) = 2 33½(3)(3) = 4.5 44½(4)(4) = 8 We get the same values!

10 Area and the Definite Integral How would we find the area under the following curve and above the x-axis? The work is a little more complex, but the approach is the same… area under the curve! f(x) x

11 We could use rectangles to approximate the area. First, divide the horizontal base with a partition. We will start with a partition that has two equal subintervals: [0,1] and [1,2] f(x) x Area and the Definite Integral

12 In this slide, the subinterval rectangles have been drawn using the right-hand endpoints of each subinterval. We could have used the left endpoints, the midpoints of the intervals, or any point in each subinterval. 0 1 2 1unit 2 2 +1 = 5 1 2 + 1 = 2 Area and the Definite Integral

13 0 1 2 1 5 2 1 The area of the first rectangle is 2 and is greater than the area under the curve. The area of the second rectangle is 5 and is also greater than the area under the curve. The total area of the rectangles is 2 + 5 = 7 which is greater than the desired area under the curve. Area and the Definite Integral

14 In this slide, lets draw the rectangles using the left-hand endpoints. 0 1 2 1unit 1 2 +1 = 2 0 2 + 1 = 1 Area and the Definite Integral

15 The area of the first rectangle is 1 x 1 = 1 and is less than the area under the curve. 0 1 2 1unit 2 1 The area of the second rectangle is 1 x 2 = 2 and is also less than the area under the curve. The total area of the rectangles using the left-hand endpoints is 1 + 2 = 3 which is less than the desired area under the curve. Now, lets calculate the areas of these new rectangles. Area and the Definite Integral

16 Now we know that the area under the curve must be between 3 and 7. Can we get a more accurate answer? Yes! And to do so, we need to construct more subintervals using a finer partition. Area and the Definite Integral

17 Area Under A Curve The area under a curve on an interval can be approximated by summing the areas of individual rectangles on the interval. If the rectangles are circumscribed above the curve, the area will be greater than the desired area. (In our example, this area was 7.) If the rectangles are inscribed under the curve, the approximated area will be less than the desired area. (In our example, this area was 3.) Area and the Definite Integral

18 By using a finer partition (one with more subintervals), we make each rectangle narrower (increasing their number), and we get an area value that is closer to the true area under the curve. By taking the limit as n (the number of rectangles) approaches infinity, the actual area is approached. Area Under A Curve Area and the Definite Integral

19 Definition: A sum such as the one below is called a Riemann sum: Area Under A Curve Area and the Definite Integral

20 Let f(x) be a nonnegative, continuous function on the closed interval [a, b]. Then, the area of the region under the graph of f(x) is given by where x 1, x 2, x 3, … x n are arbitrary points in the n subintervals of [a,b] of equal width. Area Under A Curve The area under a curve can be approached by taking an infinite Riemann sum. Area and the Definite Integral

21 On the following six slides, you will see the number of rectangles increasing for the graph of y = x 3 on the interval [-1, 2]. Observe the way that the area of rectangles more closely approximates the area under the curve as the number of rectangles increases. Area and the Definite Integral

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28 When there are 10 rectangles, Area 2.4675 20 rectangles, Area 3.091875 40 rectangles, Area 3.416719 100 rectangles, Area 3.615675 1000 rectangles, Area 3.736507 5000 rectangles, Area 3.7473 As you can see, the sum of the areas of the rectangles increases as you increase the number of rectangles. As the number of rectangles increases, they begin to better fit the curve giving a closer and closer approximation of the area (3.75) under the curve and above the x-axis. Area and the Definite Integral

29 We are almost done with this lesson in that we can now turn our attention to the definition and discussion of the definite integral for all functions, not just nonnegative functions. Area and the Definite Integral

30 Let f(x) be defined on [a,b]. If exists for all choices of representative points x 1, x 2, x 3, … x n in the n subintervals of [a,b] of equal width, then this limit is called the definite integral of f(x) from a to b and is denoted by. The number a is the lower limit of integration, and the number b is the upper limit of integration. Definition: The Definite Integral Area and the Definite Integral

31 Definition: The Definite Integral Thus As long as f(x) is a continuous function on a closed interval, it has a definite integral on that interval. f(x) is said to be integrable when its integral exists. Area and the Definite Integral

32 The definite integral of a function is a number. example: The indefinite integral of a function is another function. example: The Definite Integral Please make this important distinction between the indefinite integral of a function and the definite integral of a function: You will learn how to calculate this number in the next section of your textbook. Area and the Definite Integral

33 If f(x) is a nonnegative, continuous function on [a, b], then is equal to the area of the region under the graph of f(x) on [a, b]. The Definite Integral Lets take a quick look at the geometric interpretation of the definite integral, and well be done. Area and the Definite Integral

34 If f(x) is a nonnegative, continuous function on [a, b], then is equal to the area of the region under the graph of f(x) on [a, b]. The Definite Integral f(x) y x a b What happens if f(x) is not always nonnegative? Area and the Definite Integral

35 If f(x) is simply a continuous function on [a, b], then is equal to the area of the region below the graph of f(x) and above the x-axis minus the area of the region above the graph of f(x) and below the x-axis on [a, b]. The Definite Integral f(x) y x a b Area of region 1 – Area of region 2 + Area of region 3 Area and the Definite Integral

36 Lets look at a problem similar to #3 of the section 6.3 Exercises (page 475) in the textbook. We shall use right endpoints. Let f(x) = 3x. Part a: Sketch the region R under the graph of f(x) on the interval [0,2] and find its exact area using geometry. The Definite Integral Area and the Definite Integral

37 The Definite Integral Solution - Part a: Since the area under the curve from [0,2] forms a right triangle, we can calculate the exact area using the geometry formula A= ½ bh ; b = 2 and h = 6, so ½ (2)(6) = 6 Area = 6 f(x) = 3x 0 2 1 6 4 2 Area and the Definite Integral

38 Let f(x) = 3x. Part b: Use a Riemann sum with four subintervals (n = 4) of equal length to approximate the area of region R. Choose the representative points to be the right endpoints of the subintervals. The Definite Integral Area and the Definite Integral

39 1 2 1/2 3/2 f(x) = 3x Solution – Part b: Divide the length of the interval (2) by the number of subintervals (4). Each subinterval has a length equal to ½ unit;. This produces the 4 subintervals: [0, 1/2], [1/2, 1], [1, 3/2], and [3/2, 2]. The Riemann Sum is A = [f(1/2) + f(1) + f(3/2) + f(2)](1/2) Area and the Definite Integral

40 Let f(x) = 3x. Part c: Repeat part (b) with eight subintervals (n = 8) of equal length. The Definite Integral Area and the Definite Integral

41 Solution – Part c: Divide the length of the interval (2) by the number of subintervals (8). Each subinterval has a length equal to 1/4 unit;. This produces the 8 subintervals: The Riemann Sum is A = [f(1/4) + f(1/2) + f(3/4) + f(1) + f(5/4) + f(3/2) + f(7/4) + f(2)](1/4) [0, 1/4], [1/4, 1/2], [1/2, 3/4], [3/4, 1]. [1, 5/4], [5/4, 3/2], [3/2, 7/4], and [7/4, 2] Area and the Definite Integral

42 Let f(x) = 3x. Part d: Compare the approximations obtained in parts (b) and (c) with the exact area found in part (a). Do the approximations improve with larger n ? The Definite Integral Solution – Part d: Compare the values. What is you conclusion? Hopefully, you see that the approximation does improve with a finer partition (and thus, larger values of n). Area and the Definite Integral


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