Hello, just Van here doing the scribe for GreyM. So, I will declare that, tomorrow, GreyM is scribing, for sure. Hopefully.
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.
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!
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.
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!
In order to approximate the integral below with the greatest possible error of 0.0001, how large must n be if you use a:
(a) trapezoid sum? (b) midpoint sum?
How'd we get 71? Show you in the next slide(s). (With explanations)
dx1
2
∫ 1x 2
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.
dx1
2
∫ 1x 2
b
af(x)dx Trap n∫ ≤
M2(b a)______12n
3
2
So, first we find the 2nd derivative of the function, to determine M 2
b
af(x)dx Trap n∫ ≤
M2(b a)______12n
3
2
dx1
2
∫ 1x 2
Solve for M 2(1)
M2(1) = 6(1)4
= 6
So, now we have M 2 and try to solve for n now. And we substitute our values in...
0.0001 ≤6(21) 3
12n 2
0.0012n 2≤ 6n2 5000≤n 71≤
or
Since...
Then...
b
af(x)dx Trap n∫ ≤
M2(b a)______12n
3
2
0.0001 ≤12n 2__
10000 ≥2n 2
Since...
Then...
√5000 ≥n______
71 ≥ n
b
af(x)dx Trap n∫ ≤
M2(b a)______12n
3
2
(b) midpoint sum?
Mid 24
Only thing that changed is the divisor. So, instead of n = 71, divide by 2 and round up (since n must be a positive integer)
712
= 35.5
= 36(round up)
n
b
af(x)dx Trap n∫ ≤
M2(b a)______12n
3
2
dx1
2
∫ 1x 2
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.
1
1 1x( )√ 2x 2
Find these derivatives:
x 21
1 1x( )√ 2 Applying chain rule, we get
our final answer to be
sinddx
1x( )1
ddx
14
arctan x4( )Find these derivatives:
41 arctand
dxx4( )[ ]
)( x4
2+111
16 ( )
Again, applying chain rule, we get our answer to be:
Find these derivatives: arc cot(x)ddx
Let
arc cot(x) = y
cot(arccot(x)) = cot(y)
x = cot(y)
First we make a substitution
Then take the cot of both sides to solve for x
Then we differentiate both sides
(Continues next page...)1 = csc (y) y '2
y ' = 1csc (y)
y1
x
√1 + x 2
Rearrange equation to solve for y`
Then input what csc (y) is
y ' = 1
( ) 21 + x√1
2
=y ' 1 x 2
22
1 = csc (y) y '2
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
Find these antiderivatives:
∫ dx4+x 2
First, we factor out 1/4 then make x /4 to be (x/2) so we can antidifferentiate the "thing" into tan and place (x/2) back in
22
1
∫( )1+ 2
14 1 x
2( ) dx=
dx4
∫ 14 ( )1
1 + x 2=
+14 )( 2
xtan c
1=
First factor out √4
take the square root of x /2 and square it, to make it differentiateable
antidifferentiate the "thing"
dx∫ 4 x√ 2
2
∫dx
2 1 x4√ √
2=
2
11 x 2 dx∫√
1( )√2
=
12
arc sin ( x ) + C√2=
Find these antiderivatives:
∫ dxx + 4x + 52
dx1
∫ x 2 + + 1 + 1x4 5
x∫ 1
+ 22
+ 1dx
( )
arctan(x+2) + C=
=
=
"Complete the square" or add 0 to make a difference of squares
Antidifferentiate into arctan and substitute x+2 back in
Evaluate the given integral. Hint: let x = sin θ1
0∫ √1 x 2
xdx
LET x = sin θdx = cos θ of θ
whenx = 1 θ = π2,x = 0 θ = 0,
π2