SOLUTIONS TO FIRST MIDTERM REVIEW SHEET
1. Express the definite
integral as the limit of a Riemann sum. Explicitly define all expresR −2
sions you use. −5 sin(x)dx
Rb
P
Recall the definition of the Riemann integral is a f (x)dx = limn→∞ ni=1 f (xi )∆x, where
Pn
3i 3
∆x = b−a
i=1 sin(−5 + n ) n .
n and xi = a + i∆x. This gives us limn→∞
2. Approximate the definite integral
right endpoints.
R4
2
x−3
x dx
as a sum. Use four equal subintervals and
P
Need to find 4i=1 xix−3
∆x using the same notation as above. This gives ∆x =
i
P
i
1
1 1
27
xi = 2 + 2 and our sum is 4i=1 xix−3
∆x = ( −1
5 + 0 + 7 + 4 ) 2 = 280 .
i
2
4
=
1
2
and
R5
dy
3. Find the derivative dx
if y = x ln(t3 )dt. Be sure to show your work and give the
statement of any theorems that you use.
Rx
Rewrite the integral as − 5 ln(t3 )dt and then using the Fundamental Theorem of Calculus,
dy
Part 1 we have dx
= −ln(x3 ).
4.A.) Given the graph of y = f (x) (from the review sheet), find
R8
3
f (x)dx.
The integral is approximately 27 .
B.) Find the average value of f on [3, 8].
The average value is given by fave. =
1
b−a
Rb
a
f (x)dx. Then fave =
7
10 .
5.A.) Find the general antiderivatives for the following functions: f (x) =
B.) g(θ) = sec2 (θ)
3
C.) h(x) = (ln(x))
.
x
2
A.) x −3x+1
= x − 3 + x1 , which gives F (x) =
x
B.) tan(θ) + C.
Date: October 21, 2012.
1
x2
2
x2 −3x+1
x
− 3x + ln(x) + C, here C is a constant.
2
SOLUTIONS TO FIRST MIDTERM REVIEW SHEET
C.) Use u-substitution, where u = ln(x) and du = x1 dx. This gives the integral
u4
4 + C.
R
u3 du =
6.A.) √
Use the Fundamental Theorem of Calculus to evaluate the following definite integrals.
R 4 1+ x
√ .
1
R 1x x2
B.) 0 xe dx.
A.) Break up as
√
1+
√ x dx
1
x
R4
=
R4
1
√1
x
R1
2
B.) 0 xex dx use the u-substitution given by u = x2 , du = 2xdx to get
1
e
2 − 2.
7. Find N (t) if
dN
dt
1
+ 1dx, which integrates to become 2x 2 + x|41 = 5.
1
2
R1
0
eu du =
eu 1
2 |0
=
= 3e−5t and N (0) = 50.
R
R u
Integrating 3e−5t dt using the u-substitution u = −5t, du = −5dt we get −3
e du to
5
−3 u
−3 −5t
−3 −5t
give 5 e + C = 5 e
+ C. Then N (t) = 5 e
+ C and plugging t = 0 we get
3
253
−5t + 253 .
50 = N (0) = −3
+
C
and
gives
C
=
50
+
=
.
Then
N (t) = −3
5
5
5
5 e
5
8. Find the total area bounded by y = x and y = x2 .
The first thing one should do is graph
two curves to determine what the enclosed area
R 1 these
2
will be. The total enclosed area is 0 x−x dx. Integrating we have 21 x2 − 31 x3 |10 = 12 − 13 = 16 .
9. Find the volume of the solid obained by rotating the region bounded by y = ex , y = e−x ,
and x = ln(2) about the X-Axis.
Now, just like the above problem, one should first sketch the above curves and see where
they intersect and what the enclosed region will be. From looking at the enclosed region it
looks like the easiest method to use is the method of washers and to integrate them along
the X-Axis. The area of a single washer at the point x is given by A(x) = πR2 − πr2 =
πe2x − πe−2x and we want to integrate these washers from x = 0 to x = ln(2), so our
R ln(2) 2x
ln(2)
volume is V = π 0
e − e−2x dx = π( 12 e2x + 21 e−2x )|0
= π2 (e2ln(2) + e−2ln(2) − 2) =
−2
π ln(22 )
+ eln(2 ) − 2) = π2 (4 + 41 − 2) = 9π
2 (e
8 .
10. A tank is in the shape of a cone with vertex down. The diameter is 16m away and the
height is 10m. The tank is filled with water to a depth of 3m and the density of water is
kg
1000 m
3 . Set up the integral necessary to find the work done pumping the water over the
edge of the cone.
Notice that the approach we want to use is the same as the example presented on pg. 448
of the text. First, as usual, draw the inverted cone and label everything accordingly. Next,
SOLUTIONS TO FIRST MIDTERM REVIEW SHEET
3
take your coordinates to be at the top of the cone so that the positive x direction moves
from the top of inverted cone downwards to the point of the cone. In order to calculate
the work you want to think of lifting thin cyclinders of water from a given height of 10 − xi
up to the top of the cone. The volume for each of these thin slices of water is given by
ri
8
= 10
and we get
Vi = πri2 ∆x. Using the property of similar triangles we have 10−x
i
4
16π
2
ri = 5 (10 − xi ). This gives us our thin layer of volume to be Vi = πri ∆x = 25 (10 − xi )2 ∆.
Now, our ultimate goal is to calculate the work, which is given by the equation W =
a F (x)dx, where F (x) is the force or weight of each cylindrical slice of water. Since force
is given by F = ma = mass × acceleration we need to calculate the mass of this slice.
2
mi = Vi × density = 16000π
25 (10 − xi ) ∆x and the force of a single slice is given by the
equation Fi = mi gi = (9.8)(16000π)
(10 − xi )2 ∆x. Then using the definition of the Riemann
25
R 10
Pn (9.8)(16000π)
(10 − xi )2 ∆x = 7 (9.8)(16000π)
(10 − x)2 dx.
sum we have W = limn→∞ i=1
25
25
Rb