MATH 150: CALCULUS math San Diego MesaCollege GRADED … · San Diego MesaCollege QUIZ 12 f. javier marquez Assume the diﬀerentials behave as algebraic quantities, in the sense

MATH 150: CALCULUSGRADED QUIZ

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San Diego Mesa CollegeQUIZ 12 f. javier marquez

1. CAREFULLY Select ALL APPLICABLE choices:

By a judicious choice of a trigonometric function substitution for x, the quantity

1 + x2

could become:

A.

1− sin2(u)

B.

1− cos2(u)

C.

sec2(u)− 1

D.

csc2(u)− 1

E.1 + tan2(u)

F.1 + cot2(u)

G. none of these



5 + x2

could become:

A.

5− 5 sin2(u)

B.

5− 5 cos2(u)

C.

5 + 5 tan2(u)

D.

5 + 5 cot2(u)

E.1 + tan2(u)

F.1 + cot2(u)

G. none of these



pg. 1 c©2007-2011 MathHands.com v.15


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5 + 7x2

could become:

A.

5− 5 sin2(u)

B.

5− 5 cos2(u)

C.

5 + 5 tan2(u)

D.

5 + 5 cot2(u)

E.7 + 7 tan2(u)

F.7 + 7 cot2(u)

G. none of these


Consider integrating∫

1

9− 5x2dx

the following choices outline potential strategies to compute the integral, select the successful strategy/ies.

A. make the trig substitution:

x =3√5tan(u)

B. make the trig substitution:

x =3√5cot(u)

C. make the trig substitution:

x =3√5sin(u)

D. make the trig substitution:

x =3√5cos(u)

E. make the trig substitution:

x =3√5sec(u)

F. make the trig substitution:

x =3√5csc(u)



mathhands



Consider diagram below and answer the true statement/s: assume by ’net area’ we mean the positive total area, not’the integral’, which may be negative sometimes... assume in this problem all ’areas’ are net areas..

1

2

3

4

5

−1

−2

−3

−4

−5

−6

1 2 3 4 5 6 7−1−2−3−4−5−6−7−8

f =√25− x2

g = −√25− x2

dA

A. based on the diagram:

dA = [g − f ] dx

B. based on the diagram:

dA = [f − g] dx

C. based on the diagram:

dA = 2fdx

D. based on the diagram:

dA = 2 [0− g] dx

E. based on the diagram:

dA = 2gdx

F. based on the diagram:

dA = −2gdx

G. none of these



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1

2

3

4

5

6

−1

1 2 3 4 5 6 7 8 9−1

y1 = 5 + sin(x)

y2 = 5

y3 − 5 = −1

2(x− 5)

y4 = 2

y5 = 1

The entire gray area is equal to30.75 square units

A. True

B. False





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1

2

3

4

5

−1

−2

−3

−4

−5

−6

1 2 3 4 5 6 7−1−2−3−4−5−6−7−8

f =√x+ 4− 3

g = −√x+ 4− 3

h = x− 5

A. total gray area is given by:∫ 5

−4

[f − h] dx

B. total gray area is given by:∫

5

−4

[f − g] dx

C. total gray area is given by:

∫

5

−4

[g − h] dx

D. none of these





mathhands


1

2

3

4

5

−1

−2

−3

−4

−5

−6

1 2 3 4 5 6 7−1−2−3−4−5−6−7−8

f =√25− x2

g = −√25− x2 h = x− 5

dA

A. based on diagram throughout the gray areadA is given by:

dA = [g − f ] dx

B. based on diagram throughout the gray areadA is given by:

dA = [h− f ] dx

C. based on diagram throughout the gray areadA is given by:

dA = [f − g] dx

D. none of these





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1

2

3

−1

−2

ππ2

3π2

2π

f = cos(x) + 2

g = 2

dA


dA = [g − f ] dx


dA = [f − g] dx

C. the area in gray is given by∫

3π/2

π/2

[2− (cos(x) + 2)] dx

D. the area in gray is given by

∫

3π/2

π/2

[(cos(x) + 2)− 2] dx

E. none of these





mathhands


1

2

3

−1

−2

ππ2

3π2

2π

f = cos(x) + 2

g = 2

A. the area in gray is given by∫ 2π

0

[f − g] dx

B. the area in gray is given by∫ 2π

0

[g − f ] dx

C. the area in gray is given by

∫ π/2

0

[g − f ] dx+

∫ 3π/2

π/2

[g − f ] dx+

∫ 2π

3π/2

[g − f ] dx

D. none of these





mathhands


1

2

3

−1

−2

ππ2

3π2

2π

f = cos(x) + 2

g = 2

dA


dA = [g − f ] dx


dA = [f − g] dx

C. none of these




mathhands


1

2

3

4

5

6

−1

1 2 3 4 5 6 7 8 9−1

y1 = 5 + sin(x)

y2 = 5

y3 − 5 = −1

2(x− 5)

y4 = 2

y5 = 1

Horizontal rectangles, can not be used to set up integral/s representing the area represented above, thus verticalrectangles must be used.

A. True

B. False


1

2

3

4

5

6

−1

1 2 3 4 5 6 7 8 9−1

y1 = 5 + sin(x)

y2 = 5

y3 − 5 = −1

2(x− 5)

y4 = 2

y5 = 1 dA

Over entire gray area, dA is given bydA = [y3 − y5]dx

A. True

B. False



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y1

y2

dA

Consider partitioning an area into rectangles oriented as in the diagram above. The area for each such rectanglewould best be described by:

A.dA = [x1 − x2] dx

B.dA = [y1 − y2] dx

C.dA = [y2 − y1] dx

D.

dA = [x2 − x1] dx

E.None ofthese

15. CAREFULLY Select ALL APPLICABLE choices: The area bounded by y = 1

x2 , y = 0, x ∈ [1, 2] is equal to1/2

A. True

B. False

16. CAREFULLY Select ALL APPLICABLE choices: The area bounded by y = x2 − x, y = 0, x ∈ [3, 8] is equalto 805/6

A. True

B. False



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17. CAREFULLY Select ALL APPLICABLE choices: The area bounded by y = 1

x2 , y = 0, x ∈ [1, 2] is equal to3/2

A. True

B. False

18. CAREFULLY Select ALL APPLICABLE choices: The area bounded by y = x2 − x − 6, y = 0, x ∈ [0, 2] isequal to 34/4

A. True

B. False

19. CAREFULLY Select ALL APPLICABLE choices: The area bounded by y = 2 sin(x), y = sin(2x), 0 ≤ x ≤ πis equal to 4

A. True

B. False

20. CAREFULLY Select ALL APPLICABLE choices: The area bounded by y = 2− x2 and y = −x is equal to9/2

A. True

B. False


1

2

3

4

5

−1

−2

−3

−4

−5

−6

1 2 3 4 5 6 7−1−2−3−4−5−6−7−8

x1 =√

25− y21

x2 = −√

25− y22

y3 − x3 + 5 = 0



mathhands


A. the gray area is given by:

∫

5

−5

[x1 − x2] dy

B. the gray area is given by:

∫ 0

−5

[x3 − x2] dy +

∫ 5

0

[x1 − x2] dy

C. the gray area is given by:

3

4· π(5)2 + 5 · 5

2

D. the gray area is given by:

∫ 5

−5

[x1 − x3] dy

E. the gray area is given by:

∫

0

−5

[x2 − x3] dy +

∫

5

0

[x1 − x2] dy

F.None ofthese

22. CAREFULLY Select ALL APPLICABLE choices: The arc-length of the curve defined by

x =y4

4+

1

8y2

as y goes from 1 to 2 is equal to 123

32

A. True

B. False


dx

dyds



mathhands


Assume the differentials behave as algebraic quantities, in the sense that the usual real-number axioms apply tothem, and assume they are all positive in length, then select the true statement/s.

A.ds =

√

dx2 + dy2

B.

ds =

√

1 +

(

dy

dx

)2

dx

C.

ds =

√

(

dx

dy

)2

+ 1dy

D.

ds =

√

(

dx

dt+

dy

dt

)2

dt

E.ds2 = dx2 + dy2

F. None ofthese


y = x3

2

1

2

3

4

5

6

7

−1

−2

1 2 3 4 5 6−1−2


MATH 150: CALCULUSGRADED QUIZ math

hands


Select the true statement/s.

A. The total arc-length for the curve over the x-interval [0, 4] is given by

∫ y=8

y=0

√

4

9y2/3+ 1 · dy

B.∫ y=8

y=0

√

4

9y2/3+ 1 · dy =

8

27

(

103/2 − 1)

C.ds2 = dx2 + dy2

D.

ds =

√

4y−2/3

9+ 1 · dy

E.None ofthese


x(t) = cos(2t)

y(t) = sin(2t)

1

−1

−2

1 2−1−2−3


A.

ds =

√

(

dx

dt

)2

+

(

dy

dt

)2

· dt

B.ds =

√

cos(2t) + sin(2t) · dt

C. the total arc-length as t goes from 0 to π/2 is given by

π

pg. 15 c©2007-2011 MathHands.com


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x(t) = et cos(t)

y(t) = et sin(t)20

40

60

80

−20

−40

−60

−80

−100

20 40 60 80−20−40−60−80−100


A.

ds =

√

(

dx

dt

)2

+

(

dy

dt

)2

· dt

B. the total arc-length as t goes from 0 to π/2 is given by

√2eπ/2 −

√2


x =y3

3+

1

4y

as y goes from 1 to 3 is equal to 53

6

A. True

B. False




mathhands


y = 1/x

1

2

3

−1

−2

1 2 3 4 5 6−1−2

ds

Assume the differentials behave as algebraic quantities, select the true statement/s.

A. The total arc-length for the curve over the interval [1, 5] is given by

∫

ds =

∫ y=1

y=.20

√

1

y4+ 1dy

B.

ds =

√

1

y4+ 1dy

C.∫ y=5

y=1

√

1

y4+ 1dy =

∫ y=1

y=.20

√

1

y4+ 1dy

D.ds2 = dx2 + dy2

E.

ds =

√

1

x4+ 1dx

F. The total arc-length for the curve over the interval [1, 5] is given by

∫

ds =

∫ x=5

x=1

√

1

x4+ 1dx

G. None ofthese


y =2

3(x2 + 1)3/2

from 1 to 4 is equal to 50.



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A. True

B. False


y =1

2(ex + e−x)

from 0 to ln 2 is equal to .75.

A. True

B. False


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MATH 150: CALCULUS math San Diego MesaCollege GRADED … · San Diego MesaCollege QUIZ 12 f. javier marquez Assume the diﬀerentials behave as algebraic quantities, in the sense