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Workshop scribe

Workshop scribe

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  • 1. Hello, just Van here doing the  scribe for GreyM. So, I will  declare that, tomorrow,  GreyM is scribing, for sure.  Hopefully.
  • 2. So, today's class began  with Chris's scribe, and  the Carnival of  Mathematics, and how  they wanted to include  his scribe.  Congratulations to him.
  • 3. So, as promised by Mr.  K, we had a workshop,  with review questions to  help prepare us for the  test on Wednesday. Don't  forget to BOB!
  • 4. And also, since I don't have equation  editor on Microsoft Word on my  computer, a lot of formulas/equations  might look a little... pixelated. I  appologize in advance.
  • 5. Just this moment after 45 minutes of work  I have just found the notebook's equation  editor... sorta. By accident. So, now the  pictures aren't going to be pixelated! ... but they do look a little odd. Better than  blur! In your face Microsoft Word!
  • 6. In order to approximate the integral below with the greatest possible error of  0.0001, how large must  n be if you use a: 2   (a) trapezoid sum? 1 (b) midpoint sum? ∫ 2 dx 1 x How'd we get 71? Show you in the  next slide(s). (With explanations)
  • 7. So, a few days back, we use this formula (thank god we don't have to  remember it... yet) and apply it to our question. b 3 M(b ­ a) ______ 2 ∫ f(x)dx ­ Trap n ≤ 2 12n a 2   1 ∫ 2 dx 1 x
  • 8.   b 3 2 1 M(b ­ a) ______ 2 ∫ 2 dx ∫ f(x)dx ­ Trap n ≤ 12n 2 1 x a So, first we find the 2nd derivative of the  function, to determine M 2 
  • 9. Solve for M 2 (1) ­4 M2(1) = 6(1) = 6 So, now we have M 2  and try to solve for  n  now. And we substitute our values in...
  • 10. b 3 M(b ­ a) ______ 2 ∫ f(x)dx ­ Trap n ≤ 2 12n a 3 0.0001 ≤ 6(2­1) Since... 2 12n 0.0012n 2 ≤6 or n 2 ≤5000 n ≤ 71 Then...
  • 11. 1 __ Since... 0.0001 ≤ 2n 2 2 10000 ≥ 2n ______ √5000 ≥ n Then... 71 ≥n
  • 12.   b 3 2 M(b ­ a) ______ 2 1 ∫ f(x)dx ­ Trap ≤ ∫ 2 dx n 12n 2 1 x a b 3 M(b ­ a) ______ 2 ∫ f(x)dx ­ Trap Mid n ≤ 2 12n 24 a (b) midpoint sum? 71 Only thing that  = 35.5 changed is the divisor.  So, instead of  n = 71,  2 (round up) divide by 2 and round  up (since n must be a  positive integer) n = 36
  • 13. Okay, so, those first 12 slides took  me 2 hours. But, I'm busy chatting  with people on msn, discovery of  the Equation Editor, and not  having a writing utility of  somesort. This is actually kinda  fun. So, now, to find some  derivatives.
  • 14. Find these derivatives: d dx sin ­1 ( x) 1 1 ­2 1 2 ­x Applying chain rule, we get  √1 ­ ( ) x our final answer to be 1 1 2 ­x 2 √1 ­ ( ) x
  • 15. Find these derivatives: d 1 arctan x dx 4 ( 4) 1 x 4 [ d arctan dx ( 4) ] Again, applying chain rule,  we get our answer to be: 16 (1 1 + 1 ( x 2 4) )
  • 16. d Find these derivatives: arc  cot(x) dx Let First we make a  arc  cot(x) =y substitution Then take the  cot  of both  cot(arccot(x)) = cot(y) sides to solve for x Then we differentiate  x = cot(y) both sides 1 = ­ csc (y)2 y ' (Continues next page...)
  • 17. 1 = ­ 2 csc (y) y ' 1 Rearrange equation to solve for y` y = ­' 2 csc   (y) Then input what csc (y) is 2 1 √1 + x 2 ' y = ­ 1 2 1 ( √1 + x ) 2 y x y = ' ­ 1­x 2
  • 18. Okay, and at this point in time, I am  now 3 hours into the work. Getting  used to this repetitive grouping/ copy/  paste/ capture/ enlarging/ moving/ lots  of other things. So, now to the aunty derivatives! And don't forget to put +C
  • 19. Find these antiderivatives: dx ∫ 4+x 2 First, we factor  1 = ∫4 1 1+ 4 x(2 ) dx out 1/4 then 2 make x /4 to be  2 (x/2)  so we can  1 1 ( ) = 4 ∫ 1 + ( x ) 2 dx antidifferentiate  2 the quot;thingquot; into       ­1 1 ­1 ( x ) tan  and place  = 4 tan 2 + c (x/2) back in
  • 20. dx ∫ √4 ­ x 2 dx First factor out  4 √ =∫ x2 2√ ­ 1 √4 take the square root    of x 2/2 and square it,  1 1 to make it  = 2 ∫ 1 ­ x 2 dx differentiateable √ (√2 ) anti­differentiate the  1 arc sin x (√2 ) + C quot;thingquot; = 2
  • 21. Find these antiderivatives:   dx ∫ x  2 + 4x + 5 quot;Complete the  1 squarequot; or add 0  =∫ dx to make a  x2 +4 x + 5­ 1 + 1 difference of  1 squares = ∫ (x + 2) 2 + 1 dx Antidifferentiate  into arctan and  = arctan(x+2) + C substitute x+2  back in
  • 22. Evaluate the given integral. Hint: let x = sin θ 1 ∫ √1 ­ x 2 dx 0 x LET x = sin θ dx = cos θ of  θ when x = 0 , θ = 0 x = 1 , θ =  π 2 π 2