Math 0120 Examination #3 Sample Name (Print)

Transcription

Math 0120 Examination #3 Sample Name (Print)
ArringtonSample
Math 0120
Examination #3
Sample
Name (Print)
Signature
PeopleSoft #
Score
.
.
TA (Circle one)
Instructions:
1. Clearly print your name and PeopleSoft number and sign your name in the
space above.
2. There are 8 problems, each worth the specified number of points, for a total of
100 points. There is also an extra-credit problem worth 5 points.
3. Please work each problem in the space provided. Extra space is available on
the back of each exam sheet. Clearly identify the problem for which the space
is required when using the backs of sheets.
5. Write neatly and legibly. Cross out any work that you do not wish to be
6 No calculators, headphones, tables, books, notes, or computers may be
used. All derivatives and integrals are to be found by learned methods of
calculus.
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1
1. Let f ( x )  .
x
(a) (5 pts.) Approximate the area under the curve y = f(x) from a = 1 to b = 13 using a Riemann sum with 3
left rectangles. (Write the sum; you need not evaluate it.)
13  1
 4 . X1 = 1, x2 = 5, x3 = 9
3
1 1
4 4
A  RL(3) = [f(1) + f(5) + f(9)](4) = [1 +  ](4) = 4  
5 9
5 9
x 
(b) (5 pts.) Find the exact area under the curve y = f(x), 1 ≤ x ≤ 13, using the Fundamental Theorem
of Calculus.
13 1
A= 
1x
13
dx = ln | x | 1 = ln(13) – ln(1) = ln(13)
2. (15 pts.) Find the following integrals:
3
7
1
4 4 x
4 3 1
4
(a)  ( x  ) dx =  ( x  ) dx  x   C
e
e
7
e
1
(b)  (   x 1 ) dx = (x – ln|x|)  13 = (- – ln(1)) – (-3 – ln(3)) = 2 + ln(3)
3
2
x2
(c)  x 1 dx =  ( x  1 ) dx =
 ln|x| + C
x
x
2
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3. (16 pts.) Use substitution to find the following integrals:
(a)  ( x 2  1)( x 3  3 x ) 2 dx =
1 2
1 1
1
C
 u du   u  C  
3
3
3( x 3  3 x )
Let u = x3 – 3x
du =( 3x2 – 3) dx = 3(x2 – 1) dx
du
= (x2 – 1) dx
3
3
e x
3 2
1 x
8
(b) 
2
=  3eu du  3eu 12  3(e2  e)  3e(e  1)
dx
1
1
3
Let u = x  x
2
1 3
du = x
dx
3
3 du =
3
1
2
3
x
dx 
1
3 2
dx
x
4. (6 pts.) A company’s marginal cost function is MC(x) = 4x - 6e-0.1x where x is the number of units
and fixed costs are \$65. Find the cost function.
C(x) =  MC( x ) dx   ( 4 x  6e .1x ) dx
C(x) = 2x2 + 60 e-.1x + K
C(0) = 60 + K = 65
K=5
C(x) = 2x2 + 60 e-.1x + 5
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5. (12 pts.) Set up, but do not evaluate, integrals for the area
(a) Between the curves y = x and y = x2 on [-1,1].
x = x2
x2 – x = x(x – 1) = 0: x= 0, x = 1
0
1
1
0
A =  ( x 2  x ) dx   ( x  x 2 ) dx
(b) Bounded by the curves y = 4x – x2 and y = x2.
x2 = 4x - x2
2x2 – 4x = 2x(x – 2) = 0: x= 0, x = 2
2
2
0
0
A =  ( 4 x  x 2  x 2 ) dx   ( 4 x  2x 2 ) dx
ArringtonSample
6. (9 pts.) Given a demand function of d(x) = 300 – 0.4x and a supply function of s(x) = 0.2x.
(a) Find the market demand (the positive value of x at which the demand function intersects the
supply function).
300 - .4x = .2x
300 = .6x
x = 500 = A
(b) Evaluate a definite integral for the producers’ surplus at the market demand.
B = d(500) = s(500) = 100
500
P.S. =  (100  .2x ) dx  (100 x  .1x 2 )
0
500
0
 50,000  .1(250,000)  50,000  25,000  25,000
7. (24 pts.) Use integration by parts to find:
(a)  ( x 2 ln x ) dx =
dv = x2 dx
u = lnx
du =
dx
x
v=
(b)  xe 2x dx = 
x3
3
x 2x 1 2x
x
1
 e
e
dx   e 2x  e 2x  C
2
2
2
4
dv = e-2x dx
u=x
v= 
du = dx
(c) 
x2
x3
x3
x3
lnx - 
dx 
lnx +C
3
3
3
9
x
x5
u=x
du = dx
e 2 x
2
3
dx
4
= 2x x  5 - 2  x  5 dx = 2x x  5 - ( x  5) 2 + C
3
dv =
dx
x5
v=2 x5
ArringtonSample
8. (8 pts.) f(x,y) = ln(x) + 2y. Find fx, fy, fxx and fyx.
fx =
1
1
; fy  (ln 2)2 y ; fxx=  2 ; fyx  0
x
x
2
(5 pts) Extra-Credit : You may earn an extra 5 points by evaluating  ( x  4) dx without using the
0
Fundamental Theorem of Calculus.

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