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Anti-differentiation (graphically and numerically) Quiz

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Web resources available

This quiz tests the work covered in the lecture on graphical interpretation of the anti-derivative and corresponds to Section 6.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There are more web quizzes at Wiley, select Section 1.

There are not many relevant web resources but it may be useful to review the derivative function.

Visual Calculus, as usual, has links to an applet at http://archives.math.utk.edu/visual.calculus/4/ftc.9/ and some useful explanations.

Question 1

Suppose ′ F (x) = f (x) and F (2) = 4, F (3) = 8 and F(4) = 11.
Which of the following statements are correct?

a)
∫ 3 f(x) = 7 2
b)
∫ 4 f(x ) = 7 2
c)
∫ 4 f(x) = 3 3
d)
∫ 3 f (x) = - 4 2

 

There is at least one mistake.
For example, choice (a) should be false.
Note that ∫ 3 f(x) = F(3)- F(2) = 8 - 4 = 4 2
There is at least one mistake.
For example, choice (b) should be true.
∫ 4 f(x) = F(4)- F(2) = 11 - 4 = 7 2
There is at least one mistake.
For example, choice (c) should be true.
∫ 4 f(x) = F (4)- F(3) = 11- 8 = 3 3
There is at least one mistake.
For example, choice (d) should be false.
Note that ∫ 3 2 f(x) = F(3)- F(2) = 8 - 4 = 4
Your answers are correct
  1. False. Note that ∫ 3 f(x) = F(3)- F(2) = 8 - 4 = 4 2
  2. True. ∫ 4 f(x) = F(4)- F(2) = 11 - 4 = 7 2
  3. True. ∫ 4 f(x) = F (4)- F(3) = 11- 8 = 3 3
  4. False. Note that ∫ 3 2 f(x) = F(3)- F(2) = 8 - 4 = 4

Question 2

Consider the graph of f(x) below.
PIC
Let F′(x ) = f(x).
Which one of the following statements is true.

a)
F has a local minimum at x = - 1 and at x = 3 and a local maximum at x = 1.
b)
F has a local maximum at x = - 1 and at x = 3 and a local minimum at x = 1.
c)
F has a local maximum at x = - 0.4, a local minimum at x = 2.4 and a point of inflection at x = 1.
d)
F has a local minimum at x = - 1, a point of inflection at x = 1 and a local maximum at x = 3.

 

Your answer is correct.
The derivative of F changes sign from negative to positive at both x = - 1 and at x = 3 so there are minima at these points. The derivative changes sign from positive to negative at x = 1 so there is a maximum at this point.
Not correct. Choice (b) is false.
Try again, look at the way the derivative changes sign at these points.
Not correct. Choice (c) is false.
Try again, this is true of the function, f, the derivative of F .
Not correct. Choice (d) is false.
Try again, look at the way the derivative changes sign at these points.

Question 3

Consider the graph of y = f(x) below.
PIC

Suppose F′(x ) = f(x) and F(5) = 4 . Determine the nature of the critical point of F at x = 9 and its coordinates.
Which one of the following statements is correct?

a)
The point (9,10) is a local minimum of F.
b)
The point (9,10) is a local maximum of F.
c)
The point (9,- 2) is a local minimum of F.
d)
The point (9,- 2) is a local maximum of F .

 

Not correct. Choice (a) is false.
Try again, you have not determined the nature of the critical point correctly.
Your answer is correct.
∫ 9 5 f(x)dx = F (9)- F(5) = F (9)- 4. From the diagram we know that
∫ 9 5 f(x)dx = 6. Hence F (9) = 6+ 4 = 10 and the critical point is (9,10).
The derivative changes from positive to negative at x = 9 so the critical point is a maximum.
Not correct. Choice (c) is false.
Try again, you have not calculated the critical point or the nature of the point correctly.
Not correct. Choice (d) is false.
Try again, you have not determined the coordinates of the critical point correctly.
Recall ∫ b f(x)dx = F (b) - F (a). a

Question 4

Consider the graph of y = f(x) below.
PIC

Suppose F′(x ) = f(x) and F(2) = 1 . Determine the nature of the critical point of F at x = 5 and its coordinates.
Which one of the following statements is correct?

a)
The point (5,4) is a local minimum of F .
b)
The point (5,4) is a local maximum of F .
c)
The point (5,- 2) is a local minimum of F.
d)
The point (5,- 2) is a local maximum of F .

 

Not correct. Choice (a) is false.
Try again, you have not calculated the critical point correctly.
Recall ∫ b a f(x)dx = F (b) - F (a).
Not correct. Choice (b) is false.
Try again, you have not calculated the critical point or the nature of the point correctly.
Your answer is correct.
∫ 5 2 f(x)dx = F(5)- F(2) = F (5)- 1. From the diagram we know that
∫ 5 f(x)dx = - 3. 2 The area is beneath the x -axis so the integral is negative.
Hence F(5) = - 3+ 1 = 2 and the critical point is (5,- 2).
The derivative changes from negative to positive at x = 5 so the critical point is a minimum.
Not correct. Choice (d) is false.
Try again, you have the correct point but you have not determined the nature of the critical point.
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