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Evaluating Definite Integrals

Evaluating Definite Integrals. This power point presentation is created and written by Dr. Julia Arnold Using the 5 th edition of S. T. Tan’s text Applied Calculus. Properties of the Definite Integral. Rule 1:. Rule 2:. Rule 3: c, a constant. Rule 4:.

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Evaluating Definite Integrals

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  1. Evaluating Definite Integrals This power point presentation is created and written by Dr. Julia Arnold Using the 5th edition of S. T. Tan’s text Applied Calculus

  2. Properties of the Definite Integral Rule 1: Rule 2: Rule 3: c, a constant Rule 4: Rule 5:

  3. Example 1: Evaluate Solution:

  4. 2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2 First of all evaluating the integral above is different (sometimes from finding the area under the curve). Consider these next two problems: 1. Evaluate

  5. 2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2 First of all evaluating the integral above is different (sometimes from finding the area under the curve). Consider these next two problems: 1. Evaluate Solution

  6. 2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2 First let’s look at the graph. Part of the area lies below the x-axis When x = 1, f(1)= 0 Let’s find the area below the x axis by integrating from 0 to 1. (1,0)

  7. 2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2 As you can see from the result, areas below the x-axis are negative, although area itself is a positive quantity. Thus the area below the x- axis is 2/3. (1,0)

  8. 2. Find the area under the curve bounded by the curve, the x axis and the vertical lines x = 0 and x = 2 (1,0) Now let’s find the area from 1 to 2. The area is then 4/3 + 2/3 = 6/3 or 2.

  9. Average Value of a Function Suppose f is integrable on [a,b]. Then the average value of f over [a,b] is Find the average value of over the interval [0,4]

  10. Average Value of a Function Find the average value of over the interval [0,4] Y= 4/3

  11. Now that you’ve seen some examples, see if you can answer the following questions. The area under the curve is the same as the definite integral? True or False (Run your mouse over the answer you think is correct True False

  12. Now that you’ve seen some examples, see if you can answer the following questions. The area under the curve is the same as the definite integral? True or False (Run your mouse over the answer you think is correct True As seen in an earlier example. Go back if you don’t recall it. The definite integral was equal to 2/3, while the area under the curve was 6/3

  13. To find out the area under a curve, you should • View the graph • Separate the areas below the x axis from the areas above the x-axis • Both of the above • D. None of the above A B C D

  14. To find out the area under a curve, you should • View the graph • Separate the areas below the x axis from the areas above the x-axis • Both of the above • D. None of the above A B C You should probably do both A, and B

  15. Evaluate the definite integral and find the area under the curve for: Which of the following is the definite integral?

  16. Evaluate the definite integral and find the area under the curve for: Which of the following is the definite integral? Remember you must multiply out polynomials before integrating them.

  17. Evaluate the definite integral and find the area under the curve for: Using the graph below the area under the curve is which of the following?

  18. Evaluate the definite integral and find the area under the curve for: Using the graph below the area under the curve is which of the following? If you don’t change anything, you will get the definite integral value, which is what you would get for the first answer. Notice the limits of integration have been reversed in the first integral. This causes the integral value which is negative from 0 to 1 to become positive for 1 to 0.

  19. Now let’s compute the integrals: The area under the curve from [0,2] is 3/2. While the definite integral from [0,2] is 4/3.

  20. In a real life application of this material, one can find average yearly sales, concentration of a drug in the bloodstream, flow of blood in an artery, depreciation using the double declining-balance method. However, we will leave it to the business professional , or biologist to apply the material. Go to the homework.

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