1. Midterm Review 3-1: One Type of Integral of a Function
Deblyn Lawrence

2. Definite Integral Text Book Definition:
Definite Integral- the definite integral of the function f from x = a to x = b gives a way to find the product of (b-a), even if f(x) is not a constant.

3. Definite Integral…Again, In English
Definite Integral- finding the area under the curve and above the x-axis. The problem may show you where you should end the counting by saying “count from x = 1 to x = 7”.
You can find this by counting squares (make sure you know the scale of the boxes)

4. Sample Problems
1. Estimate the definite integral by counting squares on a graph for the following function.
f(x) = -0.1x^2+7
x = 0 to x = 5
x = -1 to x = 6

5. Sample Problems
2. You have been hired by an automobile manufacturer to analyze the predicted motion of a new sports car they are building. When accelerated hard from a standing start, the velocity of the car, v(t) ft/sec, is expected to vary exponentially with time, t seconds, according to the equation.
v(t) = 100(1-0.9^t)
Draw the graph of function v in the domain [0, 10].
What is the range of the velocity function?
Approximately how many seconds will it take the car to reach 60 ft/second?
Approximately how far will the car have traveled when it reaches 60 ft/sec?
At approximately what rate is the velocity changing when t = 5?
What special mane is given to the rate of change of velocity?

6. Problem 1A: To make graph, first graph on grapher and find coordinate points by using the table function. The definite integral is , or 30.8

7. Problem 1B: The definite integral is 41.7

8. Problem 2A: Again, to make graph, first graph on grapher and find coordinate points by using the table function.

9. Problem 2B
Range: 0 ≤ y ≤ 100. This is the range because the limit of the car’s velocity is 100, as indicated by the original equation.

10. Problem 2C: It will take approximately 8
Problem 2C: It will take approximately 8.7 seconds for the car to reach a velocity of 60 ft/sec.

11. Problem 2D: The car travels approximately 295 ft when it reaches 60 ft/sec. ** I divided 118 by .4 because of my scale (4 boxes = 10 ft)

12. Problem 2E: Rate of velocity change at t = 5 is 6.24 (ft/sec)/sec.
1. We only want to pay attention to the y-coordinate.
– = 6.56 (distance from 4 sec to 5 sec)
– = 5.91 (distance from 5 sec to six sec)
4. ( ) = 6.24 (take the average of the distances)

13. Problem 2E, Con’t: Notice the red points.

14. Problem 2F
Acceleration.