# M408L Exam I

1

If the graph of f is

4.

5.

which one of the following contains only graphs of anti-derivatives of f ? 1. 6. cor-

rect 2. Explanation: If F1 and F2 are anti-derivatives of f then F1 (x) − F2 (x) = constant independently of x; this means that for any two anti-derivatives of f the graph of one is just a vertical translation of the graph of the other. But no horizontal translation of the graph of an anti-derivative of f will be

Version 076 – EXAM 1 – sachse – (56620) the graph of an anti-derivative of f , nor can a horizontal and vertical translation be the graph of an anti-derivative. This rules out two sets of graphs. Now in each of the the remaining four ﬁgures the dotted and dashed graphs consist of vertical translations of the graph whose linestyle is a continuous line. To decide which of these ﬁgures consists of anti-derivatives of f , therefore, we have to look more carefully at the actual graphs. But calculus ensures that (i) an anti-derivative of f will have a local extremum at the x-intercepts of f . This eliminates two more ﬁgures since they contains graphs whose local extrema occur at points other than the x-intercepts of f . (ii) An anti-derivative of f is increasing on interval where the graph of f lies above the x-axis, and decreasing where the graph of f lies below the x-axis. Consequently, of the two remaining ﬁgures only as a deﬁnite integral. 1

2

1. limit =

6 6

2x dx 2 sin x dx

1 1

2. limit = 3. limit =

6 6

2x sin x dx 2x sin x dx correct

1 1

4. limit = 5. limit =

6 6

2 sin x dx 2x dx

1

6. limit =

Explanation: By deﬁnition, the deﬁnite integral

b

I =

a

f (x) dx

of a continuous function f on an interval [a, b] is the limit n

I = lim

n→∞

f (x∗) ∆x i

i=1

consists entirely of graphs of anti-derivatives of f . keywords: antiderivative, graphical, graph, geometric interpretation /* If you use any of these, ﬁx the comment symbols. 002 10.0 points

of the Riemann sum

n

f (x∗ ) ∆x i

i=1

formed when [a, b] is divided into n subintervals [xi−1 , xi ] of equal length ∆x and x∗ is i some point in [xi−1 , xi ]. In the given example, f (x) = 2x sin x, Thus 6

For each n the interval [1, 6] is divided into n subintervals [xi−1 , xi ] of equal length ∆x, and a point x∗ is chosen in [xi−1 , xi ]. i Express the limit n n→∞

[a, b] = [1, 6] .

lim

(2x∗ sin x∗ ) ∆x i i

i=1

limit =

1

2x sin x dx .

Version 076 – EXAM 1 – sachse – (56620) 003 10.0 points 1. f (1) = 2. f (1) = (4f (x) − 3g(x)) dx 3. f (1) = 4

3

Evaluate the deﬁnite integral

2

7 16 25 64 13 32 27 64 3 correct 8

I =

4

when

4

f (x) dx = 3 ,

2 2

g(x) dx = 1 .

4. f (1) = 5. f (1) =

1. I = −7 2. I = −6 3. I = −10 4. I = −8 5. I = −9 correct Explanation: Since a b b

Explanation: By the Fundamental Theorem of Calculus, d dx

x

f (t) dt

0

= f (x).

So by the Quotient Rule, f (x) = In this case, f (1) = 3 8 . d 4x 2+7 dx x = 28 − 4x2 . (x2 + 7)2

F (x) dx = −

F (x) dx

a

for all a, b and F , we see that

4

I = −

2

(4f (x) − 3g(x)) dx .

Using the Linearity of Integrals we can then write

4 4

keywords: indeﬁnite integral, Fundamental Theorem Calculus, FTC, function value, Quotient Rule, rational function, 005 10.0 points

I = − 4 Consequently,

2

f (x) dx − 3

g(x) dx .

2

Evaluate the deﬁnite integral

π/2

I = I = −(12 − 3) = −9 004

x

(3 cos x + 2 sin x) dx .

0

. 1. I = 8

10.0 points 2. I = 4 3. I = 5 correct 4. I = 6

If f is a continuous function such that f (t) dt =

0

x2

4x , +7

ﬁnd the value of f (1).

Version 076 – EXAM 1 – sachse – (56620) Since 5. I = 7 Explanation: By the Fundamental Theorem of Calculus, I = F (x) π/2 0

4

|x − 3| =

3 − x, x − 3,...

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