1
Let me add an extra explanation: �
R∈P
ν(R) =�
R
(sup
R
χS−inf
R χS)ν(R) =UP(χS)−LP(χS)<ε
2 3
� 1
−1|x3|dx
= 4 3
� 1
0
x3dx+ +1
3
= 5 3 + π
2
� π 0
�� v+π
v |cosu|du
� dv =
� π2
0
�� π2
v
cosudu−
� v+π
π 2
cosudu
� dv+
� π
π 2
�
−
� 3π2
v
cosudu+
� v+π
3π 2
cosudu
� dv Method 2:
=
� π2
0
(1−sinv−sin(v+π) + 1)dv+
� π
π 2
(1 + sinv+ sin(π+v) + 1)dv
sinv+ sin(π+v) = 0
=
� π2
0
2dv+
� π
π 2
2dv
= 2π
Here I used the CoVs=u−πand that cos(s+π) =−coss
In class, we proved Fubini's theorem for the usual Darboux/Riemann integral. In particular f has to be integrable (hence bounded) to apply the result stated in class.
So we can't directly apply the theorem here.
Actually, Fubini's theorem is far more general. Here the issue is actually that the integral of f is not improperly convergent (i.e. absolutely convergent).
Oups!!!! I've just realized that I forgot this exercise...
You just have to apply the "differentiation under an integral" theorem twice.
Check all the assumptions each time!
Sorry for that!
Oups, I've just realized that I forgot this exercise...
You have to use the generalized theorem to differentiate under the integral from Exercise 9!
Sorry for that!
ADDENDUM (March 25): the full solution is next page!
Be careful, here it is improper at (0,0), but the integrand is positive.
y→0lim+ylny= 0
Here I went very fast, that's an improper integral at 0, for the lower bound, I took the limit when y goes to 0 (I didn't evaluate at 0):
In Exercise 20, we are computing line integrals of SCALAR fields (not vector fields), so the answer doesn't depend on the orientation (that's why I didn't precise any orientation in the question)!
Oups, you'll notice that I am solving for the counter- clockwise orientation, whereas I asked in the question to use the clockwise orientation...
Sorry for that, I was wearing my trigonometric watch!
That's easy to fix: we will multiply by -1 at the end!
3
−4 3 Here I use "-C" to say that
I work with the opposite orientation!
−C
Then
�
C
y
2dx − 2xdy = −
�
−C
