Midterm Review
3-1: One Type of Integral of a Function
Deblyn Lawrence
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 (ba), even if f(x) is not a
constant.
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)
Sample Problems
1. Estimate the definite integral by counting
squares on a graph for the following
function.
f(x) = -0.1x^2+7
a. x = 0 to x = 5
b. x = -1 to x = 6
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)
a. Draw the graph of function v in the domain [0, 10].
b. What is the range of the velocity function?
c. Approximately how many seconds will it take the car to reach 60
ft/second?
d. Approximately how far will the car have traveled when it reaches 60
ft/sec?
e. At approximately what rate is the velocity changing when t = 5?
f.
What special mane is given to the rate of change of velocity?
Problem 1A: To make graph, first graph on grapher and find coordinate
points by using the table function. The definite integral is 30.775, or
30.8
Problem 1B: The definite integral is 41.7
Problem 2A: Again, to make graph, first graph on grapher and find
coordinate points by using the table function.
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.
Problem 2C: It will take approximately 8.7 seconds for the car to reach
a velocity of 60 ft/sec.
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)
Problem 2E: Rate of velocity change at t = 5 is
6.24 (ft/sec)/sec.
1. We only want to pay attention to the ycoordinate.
2. 40.95 – 34.39 = 6.56 (distance from 4 sec
to 5 sec)
3. 46.86 – 40.95 = 5.91 (distance from 5 sec
to six sec)
4. (6.56 + 5.91) = 6.24 (take the average of
the distances)
Problem 2E, Con’t: Notice the red points.
Problem 2F
Acceleration.