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Integration Practice for Brief Calculus | MAT 210, Assignments of Mathematics

Arizona State University (ASU) - TempeMathematics

Material Type: Assignment; Class: Brief Calculus; Subject: Mathematics; University: Arizona State University - Tempe; Term: Unknown 1990;

Typology: Assignments

Pre 2010

Uploaded on 09/02/2009

koofers-user-pdk
koofers-user-pdk 🇺🇸

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bg1
IntegrationPractice-Mat210
ByScottSurgent(email:
[email protected].edu
ifyouseeamistake)
I. Basic antiderivatives. Using the
propertiesfromyourtext,determinethe
generalantiderivativebelow.
a)
dx2
b)
(
)
−dxx 1
2
c)
(
)
++ dxxx 724
2
d)
(
)
+dxx
x
1
e)
+dx
x
x
3
22
f)
+dxxx
762.0545.1
713.3782.4
g)
dx
x
)2(35
h)
dx
x
2
)335.3(782.6
(Hint:
rewritethebase)
i)
dxe
x
65.0
86.26
 (Samehintas
above)
Answers:
a) F(x)=2x+C
b) F(x)=
Cxx +−
3
3
1
c)
CxxxxF +++= 7)(
23
3
4

d) CxxxF ++= ln)(
2/3
3
2
e)
CxxF
x
+−=
2
1
ln2)( 
f) CxxxF
++=
762.1545.2
107.2878.1)(
g) CxF
x
+=
)2(49.50)(
h) CxF
x
+=
)122.11(815.2)(
i) CxF
x
+=
)916.1(323.41)(
II. Initial conditions. For each function
below, find the particular antiderivative
thatsatisfiesthegiveninitialcondition.
a)
=+ 3)2(;74 Fdxx
b)
=
+5)1(;
11
2
Fdx
x
x
c) The velocity of a particle is
governed by the equationv(t) =
2t
2
+t,wheretisinminutesand
v(t) is in feet/minute. Find the
distance function D(t) for this
particle given that after 5
minutes, the particle had
traveled30feet.
Answers:
a) 1872)(
2
−+=
xxxF
b) 6ln)(
1
+−=
x
xxF
c) 833.65)(
2
2
1
3
3
2
−+=
tttD
III. Definite Integrals. Evaluate the
followingdefiniteintegrals.
a)
( )
−
2
0
65 dxx
b)
(
)
−
−+
5
1
2
14 dxxx
c)
9
4
dxx
d)
+
4
1
3dxx
x
e)
5ln
4.2ln
dxe
x
pf2

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Download Integration Practice for Brief Calculus | MAT 210 and more Assignments Mathematics in PDF only on Docsity!

Integration Practice - Mat 210

By Scott Surgent (email: [email protected] if you see a mistake)

I. Basic antiderivatives. Using the properties from your text, determine the general antiderivative below.

a)  2 dx

b)  ( x^2 − 1 ) dx

c)  (^4 x^2 + 2 x + 7 ) dx

d)  ( x + x^1 ) dx

e)  

 (^) + dx x x^3

f)  4. 782 x^1.^545 + 3. 713 x^0.^762 dx

g)  35 ( 2 ) xdx

h)  6. 782 ( 3. 335 )^2 xdx (Hint:

rewrite the base)

i)  26. 86 e^0.^65 xdx (Same hint as

above)

Answers: a) F ( x ) = 2 x + C

b) F ( x ) = 31 x^3 − x + C

c) F ( x )= 34 x^3 + x^2 + 7 x + C

d) F ( x )= 32 x^3 /^2 +ln x + C

e) F ( x )= 2 ln x − x^1 2 + C

f) F ( x )= 1. 878 x^2.^545 + 2. 107 x^1.^762 + C

g) F ( x )= 50. 49 ( 2 ) x^ + C

h) F ( x )= 2. 815 ( 11. 122 ) x^ + C

i) F ( x )= 41. 323 ( 1. 916 ) x^ + C

II. Initial conditions. For each function below, find the particular antiderivative that satisfies the given initial condition.

a)  4 x + 7 dx ; F ( 2 )= 3

b)   =

x x^2 dx F

c) The velocity of a particle is governed by the equation v ( t ) = 2 t^2 + t , where t is in minutes and v ( t ) is in feet/minute. Find the distance function D ( t ) for this particle given that after 5 minutes, the particle had traveled 30 feet.

Answers: a) F ( x )= 2 x^2 + 7 x − 18 b) F ( x )= ln x −^1 x^ + 6 c) D ( t )= 32 t^3 + 21 t^2 − 65. 833

III. Definite Integrals. Evaluate the following definite integrals.

a)  ( − )

2 0 5 x^6 dx

b)  ( )

−

5 1 4 x^2 x 1 dx

c) 

9 4 xdx

d)  

4 1

x dx x

e) 

ln 5 ln 2. 4 e dx

x

f) 

  1. 3
  2. 71.^488 (^2.^716 ) dx

x

g) The rate of change in a town's population is given by the function f ( x ) = 169.147(1.07) x , where x is years since 1980 and f ( x ) is people/year. What was the overall change in the town's population between 1980 and 1990?

Answers to definite integrals: a) - b) 174 c) (^383)

d) 7.5 + 3 ln 4 e) 2. f) 2. g) 2,417 people.

IV. Area Between Two Curves. Set up the integrals and determine the area between the two given curves between the specified bounds. If no bounds are given you will need to determine them (points of intersection).

a) f ( x ) = x^2 , g ( x ) = x - 2, -2 ≤ x ≤ 3

b) f ( x ) = 2 e x, g ( x ) = (1.05)x, 2 ≤ x ≤ 4

c) Area enclosed within the curves f ( x ) = x^2 - 3 and g ( x ) = x - x^2

Answers: a) (^1156) b) 92. c) The intersection points are x = -1 and x = 1.5. The integral would be

− (^ (^ x − x )^ −(^ x − ))^ dx

  1. 5 1