Workshop scribe

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