#ATTENDANCE QUIZ for Lecture 14 of Math251(Dr. Z.)
#EMAIL  RIGHT AFTER YOU WATCHED THE VIDEO
#BUT NO LATER THAN  Oct. 26, 2020, 8:00PM (Rutgers time)
#THIS .txt FILE (EDITED WITH YOUR ANSWERS)
#TO:
#DrZcalc3@gmail.com
#Subject: aq14
#with an ATTACHMENT CALLED:
#aq14FirstLast.txt
#(e.g. aq14DoronZeilberger.txt)




#LIST ALL THE ATTENDANCE QUESTIONS FOLLOWED BY THEIR ANSWERS

Question 1. 
a[I]= I-th digit of RUID and if it is 0 make it 1

Find the number:
Int(int(int(a[1]*x+a[2]*y+a[3]*z^2, x=y*z...2*y*yz), z=0...y^2, y=0...a[6])

Answer to Question 1. 

With my RUID, the problem is the following:
Int(int(int(x+9*y+6*z^2, x=y*z...2*y*yz), z=0...y^2, y=0...1)

Solving for the first integral: = (3/2)*y^2*z^2 +9*y^2*z + 6*z^3*y

Solving for the second integral: = (1/2)*y^8+(9/2)*y^6+(3/2)*y^9

Solving for the third integral: = 1/18 +9/14 +3/20 ~ 1069/1260 ~ 0.85


Question 2. 
Problem from a previous final at the end of handout 15.3

Answer to Question 2. 
Int(int(int( 1, z=0...1-y), y=x^2...1),x=-1..1)
First integral:= 1-y
Second integral:= (1/2)-x^2+(1/4)*x^4
Third integral:= (1/2)*x-(x^3)(1/3)+(x^5)*(1/20)
Question 3. 
Finish it up by plugging in x=1 and x=-1 and subtracting and hopefully get 8/15

Answer to Question 3. 
[(1/2)-(1/3)+(1/20)]- [-(1/2)+(1/3)-(1/20)] = 5/18