C Calculus Solutions
April 13, 2009 Workshop #11
Review for Exam 2
Instructions: I. II. III. IV. V. VI. You must submit one copy (the color copy) for the entire group. You may take the other copies (the white copies) with you. Everyone must fully understand how to solve each problem for which an answer is provided. Grading will emphasize the quality of your group work and, to a lesser extent, the number of problems your group solves. Complete the problems in any order and try to solve at least 9 problems. Please ask if you have questions about anything.
Let u=x Then, du = dx
and and
Let u=x Then, du = dx
and and
dv = x + 5dx 3 v = 2 ( x + 5) 2 3
(x
x + 5 dx = ( x ) 2 ( x + 5) 2 - 3
3
)
(
) ( (x + 5) )(dx) = 23x (x + 5)
2 3
3 2
(2)
Use integration by parts once to compute
(xe )dx = (x )(
3x
1 3
e3 x -
) (
1 3
e 3 x (dx ) =
)
x 3x 1 3x e - e +C 3 9
dv = e 3 x dx v = 1 e3 x 3
(1)
Use integration by parts once to compute
Integration by Parts:
udv = uv - vdu
straP yb noitargetnI
-------------
-------------
(xe )dx .
3x
(x
x + 5 dx .
3 2
)
-
5 4 ( x + 5) 2 + C 15
Let u = x2 Then, du = 2 xdx
and and
dv = sin( x )dx v = - cos(x )
u=x Let Then, du = dx
and and
dv = cos( x )dx v = sin( x )
= - x 2 cos( x ) + 2 x sin( x) - 2 (sin( x) )dx = - x 2 cos( x ) + 2 x sin( x ) + 2 cos( x) + C
Let u = e2x Then, du = 2e 2 x dx
and and
dv = cos(3 x)dx v = 1 sin( 3x ) 3
2 3
Let u = e2x Then, du = 2e 2 x dx
and and
dv = sin( 3 x)dx v = - 1 cos(3 x) 3
(e
2x
cos(3 x) )dx =
1 3
e 2 x sin( 3x ) + 2 e 2 x cos(3x ) 3e 2 x sin( 3 x) + 2e 2 x cos(3 x) 9 = 4 1 - (- 9 ) 13
Therefore, we have: (e 2 x cos(3x))dx = 13 e 2 x sin( 3x) + 29 e 2 x cos(3x) -
= 1 e 2 x sin( 3x ) + 2 e 2 x cos(3 x) - 3 9
4 9
(e
2x
= 1 e 2 x sin( 3x ) - 2 e 2 x (- 1 cos(3 x) ) - 3 3 3
(e
2x
cos(3 x) )dx = 1 e 2 x sin( 3x ) - 3
(( )
(e
2x
cos(3 x) )dx
(- 1 cos(3x) )(2e 2 x dx )) 3
sin( 3x ))dx
4 9
(e
2x
cos(3 x) )dx (and hint implies)
(e
2x
cos(3 x) )dx = (e 2 x )( 1 sin( 3x )) - 3
(1 sin( 3x) )(2e 2 x dx ) = 1 e 2 x sin( 3x) - 2 (e 2 x sin( 3x))dx 3 3 3
Hint:
(e
2x
cos(3 x) )dx = f ( x ) + k
(e
2x
cos(3x ))dx
(4)
Use integration by parts twice to compute
(e
(x
2
sin( x ))dx = - x 2 cos( x ) + 2 ( x )(sin( x) ) -
(
(x
2
sin( x ))dx = - x 2 cos( x ) + 2 ( x cos( x) )dx
(x
2
sin( x ) dx = x 2 (- cos( x) ) -
)
( )
(- cos( x) )(2 xdx )
(3)
Use integration by parts twice to compute
(x
2
sin( x ) dx .
)
(sin( x) )(dx ))
2x
cos(3 x) )dx .
(e
2x
cos(3 x) dx =
)
f ( x) 1- k
(5) Find the volume that is generated rotating by region R2 about the x-axis (ie, the axis of rotation is y = 0).
0
(6)
Find the volume that is generated by rotating region R3 about the line x = 4.
V = 2 (4 - x ) x - x 3 dx = 2
0
1
(
)
(7) Find the volume that is generated by rotating the region between y = 10 x , x = 0, and y = 10 about the y-axis. What common volume is formed by this rotation?
We are rotating the triangular region with vertices (0, 0), (1, 10), and (0, 10).
V = 2 ( x )(10 - 10 x )dx = 2 10 x - 10 x 2 dx = 2
0
1
The volume is a cone of height 10 and radius 1. (8) Find the volume that is generated by rotating the region between y = 5 x - x 2 and y = 0 about the line x = -1 .
V = 2 ( x + 1) 5 x - x dx = 2 - x 3 + 4 x 2 + 5 x dx = 2
2 0
5
(
)
V = 12 -
1
( x)
2
dx = (1 - x )dx =
0
1
1
(
0
5
0
weiveR 6 retpahC
2
-------------
-------------
22 44 = 15 15
)
5 10 1 = = 12 10 3 3 3
(
)
875 875 = 12 6
(9) If a 16 pound weight stretches the spring 8 inches from equilibrium, then find the total work in stretching the spring to 18 inches from equilibrium (from a starting point of 8 inches from equilibrium).
16 = 8 k 12
18 12
k = 24
8
(10) A 15 meter cable is hanging over the side of a building. If the cable has linear density 400 grams per meter, then find the total work in lifting the cable to the top of the building.
15
2 (11) A conical tank, whose height is 4 meters and radius is 1 meters, is filled with water. Using the fact that water has density 1000 kg / m 3 , find the total work in pumping out all...