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