Solutions to midterm 3.
I have decided to quote word-for-word the exam solutions of one member of
the class, this is what I really wanted to see. Notice that he/she oftens writes in
sentences and writes every step.
1. (a) Calculate the area underneath the curve y = 7 x between x = 0 and
x = =2.
Area between y = 7 x and y = cos(x) is:
Solution:
Z
0
(7
(b) Evaluate the integral
Solution:
Z
9
4
p2 dt =
t
Therefore
Z
4
9
Z
9
4
Z
Z
Z
7 dx
x dx
cos(x) dx
0
0
0
= 7xj0 12 x2 j0 sin(x)j0
= 7 0 [ 12 2 0] [sin sin 0]
= 7 12 2
9
4
p2 dt.
t
2t 12 dt = 2 [2t 21 ]94 = 4(9 21 4 21 ) = 4(3 2) = 4:
p2 dt = 4.
t
2. (a) Evaluate the integral
Solution:
Z
cos(x)) dx =
x
Z s(s + 1)2
ps ds.
Z 2
Z s(s + 1)2
ps ds = s(s +p2ss + 1) ds
Z s3 + 2s2 + s
=
Z
ps
ds
+ 2s 23 + s 12 ds
= 27 s 72 + 45 s 52 + 23 s 32 + c:
Z s(s + 1)2
ps ds = 72 s 72 + 45 s 25 + 23 s 32 + c.
Therefore
Z
(b) Evaluate the integral 3x7 + 8 sec(x) tan(x) dx.
=
1
5
s2
Solution:
Z
Z
Z
sin x dx:
3x7 + 8 sec(x) tan(x) dx = 3x7 dx + 8 cos1 x cos
x
Set u = cos x, du = sin x dx so
Z sin x
Z sin x
8 cos2 x dx = 8 cos2 x dx
Z
= 8 u12 du
Z
Therefore,
= 8 u 2 du
= 8[ u 1 + c0]
= cos8 x + c00]:
Z
3x7 + 8 sec(x) tan(x) dx = 38 x8 + cos8 x + c:
3. Answer 'true' (T) or 'false' (F) by circling the appropriate letter.
F
\If
Z
a
b
f (x) dx = 0 then f (x) = 0 for all x in [a; b].
If the area of f above the x-axis equals the area below then
the integral is zero. For example, the integral of cos(x) from 0 to is
zero, but cos x isn't zero for all x 2 [0; ].
Z1
\ jxj dx = 1:"
Solution:
T
1
Just draw a picture and add up the area of the pair of triangles.R
\ f 0(x) dx = f (x) for every function f (x)."
Take f (x) = 1 say, then f 0(x) = 0. But R f 0(x) dx = c for
any constant c, not just c = 1.
Z rp
\The area of a circle of radius r is 4
r2 x2 dx.
Solution:
F
Solution:
T
0
The given integral compute the area in the part of the circle
lying in the rst quadrant. This is exactly one quarter of the area, so
four times this does compute the area.
Z 3
\ (x + 1) tan12(3x2 + 6x) sec2(3x2 + 6x) dx = 1=6."
Solution:
F
Solution:
3
Both limits on the integral are 3 so the integral must be zero.
2
dy
=
4. Consider the dierential equation dx
rx
y
.
(a) Calculate the general solution [Hint: this should involve a constant]
Solution:
dy
dx
=
rx
y
=) py dy = px dx, so
Z
y 2 dy
1
=
Z
x 2 dx
1
2 y 32 + c0 = 2 x 32 + c00 :
3
3
Therefore y 32 = x 32 + c.
(b) Find the particular solution of the dierential equation that satises the
condition y = 4 at x = 1.
When y = 4 at x = 1, 4 23 = 1 23 + c, so 8 = 1 + c, c = 7.
Therefore
3
3
y 2 = x 2 + 7:
Solution:
5. Consider the integral
Z
6
0
3x + 2 dx.
(a) Calculate the Riemann sum for this integral using right endpoints and
n = 6.
x = 6=n = 6=6 = 1 and x1 = 1; x2 = 2; x3 = 3; x4 =
4; x5 = 5; x6 = 6. Hence
R1 = f (x1 )x = (3 1 + 2) 1 = 5;
R2 = f (x2 )x = (3 2 + 2) 1 = 8;
R3 = f (x3 )x = (3 3 + 2) 1 = 11;
R4 = f (x4 )x = (3 4 + 2) 1 = 14;
R5 = f (x5 )x = (3 5 + 2) 1 = 17;
R6 = f (x6 )x = (3 6 + 2) 1 = 20:
Therefore Area= 5 + 8 + 11 + 14 + 17 + 20 = 75.
(b) Do the same for the general case using right endpoints where the interval
is divided into n pieces of length 6=n.
x = 6=n, xi = 6i=n for i = 1; : : : ; n. Then
6i 6 18i 6 108i 12
+2 =
+ :
Rn = f (xi )x = 3 + 2 =
Solution:
Solution:
n
i
12
Therefore, Area = Pni=1 108
n2 + n .
3
n
n
n
n2
n
Z
6
(c) i Compute the integral 3x + 2 dx by computing the limit as n ! 1
0
of your answer to (b); you may use the formula Pni=1 i = 21 n(n + 1);
The integral is
n
X
108
i 12
108
1
12
lim
+ n = nlim
!1 n2 2 n(n + 1) + n n
n!1
n2
i=1
54 2
= nlim
!1 n2 (n +n) + 12
1 + 12
: = lim 54 1 +
Solution:
n!1
= 54 + 12 = 66:
4
n