The Final Countdown

Anti derivatives are tough because I sometimes confuse them with regular derivatives.
f(x)=x^2-sin(x)
F(x)= x^3/3 + cos(x) + c
f(x)= tan^2(x) + 3dx
F(x)= 2x + tan(x) + c

U-sub is tough because it is hard to determine what u will be
x(1 - 3x^2)^4dx
u= 1-3x^2
du = -6xdx
answer: -1/30(1-3x^2)^5 + c

(1+sec(x))^2(sec(x)tan(x))dx
u= sec(x)
du=sec(x)tan(x)
answer: sec^3(x)/3 + sec^2(x) + sec(x)+ c

sqrt(x2-3x)dx
u= 2-3x
du = -3dx
answer: -2/135(2-3x)^3/2(9x+4) + c

Summation is tough because it's a lot of algebra and I'm not good at that, nor do I remember how to set up summation.
I'm lost on how to set up the problem given because of the given number of rectangles... Do I use that for the n value above the E symbol? And what is the change in x?

Limit definition is tough because we went over a much harder way to do it and I failed that quiz so now I feel cheated doing it.
f(x)= 4x-5 on [1,5]
Antiderivative: 2x^2 - 5x
2(5)^2 - 5(5) - (2(1)^2 - 5(1))
25 - (-3)
28

Using the disk method is tough because it has a lot of moving parts to the set up. 
f(x) =sqrt(x) and x = 4, find the volume revolving about the x-axis
sqrt(x)^2 - (4)^2 from 0 to 4

Volume using Integration

1. To rotating around a vertical and horizontal line changes which function is the bigger or smaller radius. X goes with horizontal line. Y goes with vertical. 

2. A washer has two functions, a disk has one function. 

3a. Find the volume of the region bounded by: y=x^2-3x+4, x=2, x=4 and rotated about the x axis

3b. Find the volume of the region bounded by: y=sqrt(x), y=x^2 and roared about x=1 

Thank God this is for more points - CSlatt

• I can find the constant of an antiderivative given a coordinate pair that is included in that function.
• EX: Find the antiderivative of f(x) = 6x^2-4x+5when F(1) = 2. Then find F(2).
• So, to start find the antiderivative, which was the other learning target but I'll explain. For 6x^2, add one to the exponent and keep a 2 in front. For 4x, you add one to the exponent and keep a 2 in front. Make 5 5x. Then plug in 1 for x of the antiderivative because of the F(1) and set that equal to 2. Solve and get -3. Do the same as you did for F(1) but for F(2) and solve without setting equal to anything, and get 18.
  
  
  What trips me up is solving for the constant on an F(x)=?
• I can approximate the area under a curve via left and right endpoints for a given amount of rectangles.
• EX: Approximate the area under the curve using 6 rectangles for the function f(X) = 4 - 2x^2   on  [0, 12] Use both right and left endpoints.
• So, start by setting up a summation, using the number of rectangles for n. Use the equation for f(i/n) and plug in for f. Break the summation apart with subtraction and continue to break it down, pulling out 12/n and further solving for i^2. Eventually pull out what you can and cancel the n's out.
  
  
  What throws me off is not having an n at the top of the summation and having to solve for that
• I can use U-Substitution to integrate functions.
• EX: Integrate the function f(x) = x/sqroot(2x-5) on [3, 7].
• Assuming Christian forgot a dx on the end of the problem, make u everything under the square rooot, 2x-5. Du in turn, equals the derivative of that, 2, in terms of dx. Next plug in 7 and 3 into u to make them in terms of u, to now get 9 and 1. To balance out du=2dx, place a 1/2 in front of the integral, and set x= to 5/2. Now take the anitderivative of what you have and get 5/2u over u^3/2 over 3/2. Next plug x back in for u and do fundamental part 2 for 9 and 1.
  
  
  I am thrown off when having to choose the u and then plugging it back in

Optimization in terms of CSlatt

1. To find optimization, you will first find the derivative of the equation. Next, set that derivative equal to 0 and solve for the variable.
2. Find the point on the line y=2x+3 that is closet to the origin. To optimize, you will first need to use the distance formula. d=sqr[(x)^2-(2x+3)^2]. Next, find the derivative of the new distance formula using a combination of the power rule and chain rule. (d^2)'=[2x+2(2x+3)*2] = (2x+8x+6)= 10x+12. Next set this equation equal to 0 and solve for x. 10x+12=0, 10x=-12, x=-6/5 and plug back into the original y= equation. y=2(-6/5)+3 = 3/5. The final answer comes to [(-6/5),(3/5)]
3. f'(x)=6x^2-10x-1
a) 2x^3-5x^2-x
b)2x^3-5x^2-x+69

Tis' The Season Blog Post

1. Derivatives do not exists in corners, cusps, jumps, and asymptotes.

2. Take the derivative of the equation, but for the derivative of each constant, write "d of the constant" (For example: dy) over dx. Then solve for that constant. When deriving x, dx/dx will be equal to one, thus you do not have to write it while solving.
3. Writing the "dy/dx" or other constants while taking the derivatives of those constants.
4. I shared it.

2 Chainz Rule #TRU

1. f'(0) represents the critical point where the slope of the point is 0
2. Make a number line with values greater than and less than the x value you found. Plug and chug and find whether it is positive or negative. If it goes negative to positive, it is increasing. Positive to negative is decreasing.
3. Take the derivative of the outside function multiplied by the inside function multiplied by the derivative of inside function.
f(x)=(4-2x)^3
f"=3(4-2x)^2*-2 = -6(4-2x)^2
Tangent line at x=3
m=-24 (Plug 3 into derivative)
y= -8 (Plug 3 into original equation)
y+8=-24(x-3)
4. h'(-4)=f'(g(-4))g'(-4)

h'(-4)=f'(-5)g'(-4)

h'(-4)=20*2=40

Blog Post 4

1. f(x) 3- {sqr(x) where x is greater than or equal to 4
               {x^2+3 where x is less than 4
Plug 4 into each equation, approaching the left and right
Equation #1 comes to 19
Equation #2 comes to 1
They are not equal, therefore f(x) is not continuous

2. f(x)=2x^2+5x-3 [0,4]
Plug 0 and 4 into the equation
f(0) comes out to equal -3
f(4) comes out to equal 49
Since one is negative and the other is positive, there is indeed a solution

f(x)= -3x^2+14x-8 [-3,0]
Plug -3 and 0 into the equation
f(-3) comes out to equal -77
f(0) comes out to equal -8
Since they are both negative, a solution cannot be proven.

3. Two types - Different quotient and slope of a tangent line
Linear derivative - f(x)=2x-3 as x approaches 1
2x-3- f(1) / x-1
2x-3 - 1 / x-1
2x-2 / x-1
(x-1)(x+1) / x-1
x+1 = 2
A derivative is the slope of a line at a certain point
The hardest part is the difference quotient and doing the algebraic part of distributing the xs and hs.

4. Instantaneous velocity is the slope of one certain point. Average velocity is the average slope of a point.