Calculus Review Topics

Area Between Curves
This topic is difficult because I often forget to take the anti-derivative of the functions before solving for the area.

Example:  Find area of the region bounded by: f(x) = 2x - 2 and g(x) = -x² - 2x + 3

Find area of the region bounded by: f(x) = 2x³ - 1 and g(x) = 2x - 1

Find area of the region bounded by f(x) = x and g(x) = √x

Volume of a Region Revolved Around an Axis Using Disks
This topic can be difficult when you forget which is the width and which is the height. Usually this is just a foolish mistake, however.

Example: Using the region bounded by f(x) = √x and x = 4, find the volume revolving about the x-axis.

Using the region bounded by f(x) = 2x and x=5, find the volume revolving about the x-axis.

Volume of a Region Revolved About an Axis Using Washers

For me, this topic is usually hard as a result of changing the width and height.

Example: Find the volume of area bounded by x = 0, y = x + 4 and y = (1/2)x² revolved around x = -1.

Find the volume of the solid obtained by rotating the portion of the region bounded by y = x^(1/3), x = 0 and y = x / 4.

Volume of a Region evolved About an Axis Using Cylindrical Shells
As a result of spending a lot of time on washers and disks, one hard part is remembering to switch which values we use based on what the region is being revolved around.

Example Find the volume of region bounded by y = 2x³, y = 2x, x =1, and x=2; revolved about the y-axis.

Find the volume of region bounded by y = 5x², x = 1 and y = 1; revolved about the y-axis.

Applying the Fundamental Theorem of Calculus Parts 1 and 2

Sometimes problems can arise when you forget to multiply by the derivative if you are not dealing solely with a variable when using the FTC part 1. For the FTC part 2, sometimes you might forget to use the anti-derivative to evaluate integrals.

Example: Evaluate f'(x) when f(x) = ∫_-1^4x (4t - 10t²)dt.

Evaluate f'(x) when f(x) = ∫_x^{3x^2} cos(θ)dθ

Evaluate the following definite integral: f(x) = ∫₀¹x³dx

Volume through Integration (Washers and Disks)

Rotating Around a Horizontal and Vertical Line

When a function is rotated around a vertical line, the washers or disks are stacked in a tower-like structure. They are stacked horizontally. When a function is rotated around a horizontal line, the washers or disks are stacked next to each other. They are lined up vertically. The variable that goes with rotation around a vertical line is y and the variable that goes with rotation around a horizontal line is x.

Washers vs. Disks

The main difference between creating disks or washers when rotating a function is that washers have space in the middle. Disks have no space in the middle of the shape.

Examples

Find the volume of the region bounded by y = x², y = 0, and x = 2 rotated about x = 3.

Find the volume of the region bounded by y = 2x - 2, y = 0, and x = 3 rotated about the x-axis.

Anti-derivatives and Areas Under Curves Review

U-Substitution to Integrate Functions

In order to integrate using U-substitution, there are several steps that must be followed.

1) Set u equal to a part of the problem.

2) Find the derivative of . This is du.

3) Use substitution to remove all variables besides u from the problem. If there are still other variables, solve for the variable using the definition of u and plug that in.

4) If there are limits, use the definition of u to find the limits with respect to u.

5) Find the anti-derivative.

6) (NO LIMITS) Substitute for u so that the answer is in terms of the original variable.

6) (LIMITS) Evaluate u at the upper and lower limits. Then subtract them.

For me, the hardest part of U-substitution is remembering to change the limits and finding the anti-derivative.

Example:  Integrate the function f(x) = x / √(2x-5) dx on [3, 7]

u = 2x - 5           Find u

du = 2dx            Find du

x dx / √u            Substitute u

1/2 | 2dx x / √u  Plug in a two and pull out a 1/2

1/2 | du x / √u    Substitute du

u = 2x - 5
2x = 5 + u
x = (5 + u) / 2    Solve for x and substitute it in

1/2 | du ((5 + u) / 2) / √u

u = 2(3) - 5

u = 1                  Find the new lower limit

u = 2(7) -5

u= 9                   Find the new upper limit
[1, 9]

1/4 | (5 + u)(u^-1/2)    Rewrite the equation so it is easier to find the antiderivative
1/4 | 5u^-1/2 + u^1/2

1/4 | 10u^1/2 +(2/3)u^3/2          Find the anti-derivative

1/4 (30 + 18) - 1/4 (10 + 2/3) Plug in the limits and solve for the answer.

1/4(48) - 1/4 (32/3)

12 - (8/3)

28 / 3      

Fundamental Theorem of Calculus Part II to Find the Area Under a Curve
The FTC part II says that if f(x) on [a, b] then f(x)dx on [a, b] = F(b) - F(a)
In order to do this, there are some steps that must be followed.
1) Find the anti-derivative of the function.
2) Plug the upper limit into the anti-derivative.
3) Plug the lower limit into the anti-derivative.
4) Subtract the two for the answer.

For me, the hardest thing to remember for this learning target is to always take the anti-derivative and not the derivative by mistake.

Example: Find the area under the curve of f(x) = 4 - 2x² on [0, 12]

4x - (2x / 3)³ + C                                Find the anti-derivative

(4(12) - (2/3)12³) - (4(0) - (2/3)(0)³) Plug in the upper and lower limits

(48 - 1152) - 0                                    Simplify

-1104                                                  This is your answer

Area Under a Curve Using Left and Right Endpoints
In order to do this, there are several steps that must be followed.
1) Determine the width of your rectangles. The more rectangles you use, the more accurate the area will be.
2) Set up a summation the properly shows how you are solving the problem. When you write the summation, the n is the number of rectangles you are using and it goes at the top. The i shows what point you are starting with and it goes at the bottom. To the right of the summation is Δx f(lower limit + iΔx). Δx = interval / number of rectangles
3) Solve the summation using either the left or right endpoints.

For me, the hardest part of these problems is to mess up which points to use for left or right.

Example: Approximate the area under the curve using 6 rectangles for the function
f(x) = 4 - 2x² on [0, 12]

Using Right Endpoints

n = 6 i = 1 Δx = 12 / 6 = 2

  6
  Σ 2(0 + 4 - 2(2i)²)
i = 1

  6
  Σ 2(4 - 2(2i)²)
i = 1

2(-4 - 28 - 68 - 124 - 196 - 284)

2(-704)

-1408 is the answer using right endpoints

Using Left Endpoints

  5
  Σ 2(0 + 4 - 2(2i)²)
i = 0

  5
  Σ 2(4 - 2(2i)²)
i = 0

2(4 - 4 - 28 - 68 - 124 - 196)

2(-416)

-832 is the answer using left endpoints

Optimization

Process of Optimization

1) You must find the derivative because you are looking for the maximum or minimum. Sometimes this requires using two equations if you have more than one variable. You may have to use information in the problem to create two equations and then determine which one is used to find the derivative and which one is used to find a variable. For example, if you are finding a point that is closest, you are finding the minimum and will use the distance formula to find the derivative.

2) Once you find the derivative, you must set it equal to 0.

3) Next, you solve to find the minimum or maximum. Sometimes you will have more than one possible answer and must determine which is correct based on information in the problem.

Here is an Example

Find the point on the line y = 2x + 3 that is closest to the origin (0,0).

1) The first step is to find the derivative. Due to the fact that we are finding the point closest to the origin, the distance formula is one of our equations. 

d = √(x₂ - x₁)² + (y₂ - y₁)²

We are going to find the derivative of this equation because we are looking for the minimum of a point that is closest to the origin. We will plug in 2x + 3 for y and we will also plug in (0,0).

d = √(x₂ - 0)² + (2x + 3 - 0)²

Now we will square both sides in order to remove the square root.

d² = (x)² + (2x + 3)²

Now we will find the derivative using the power rule and the chain rule.

2d = 2x + 2(2x + 3)(2)


2d = 2x + 4(2x + 3)

2d = 2x + 8x + 12

2d = 10x + 12

Now we will set the derivative equal to 0 and will solve for x.

0 = 10x + 12

-12 = 10x

x = -12/10 = -6/5

Now we will plug in this x-value in order to find the y-value of the point.

y = 2(-6/5) + 3

y = 3/5

This means that the closest point on the line y = 2x + 3 to the origin is (-6/5, 3/5).

If f'(x) = 6x² - 10x - 1 give two options of what f(x) could be

f(x) = 2x³ - 5x² - x

f(x) = 2x³ - 5x² - x + 1000

Seven Swans a Swimming Christmas Post

Where the Derivative Doesn't Exist
There are four functions/graphs where the derivative doesn't exist.

One of these is the absolute value function. The derivative doesn't exist at the corner.
f(x) = |x|

Another graph where the derivative doesn't exist is a cusp.

Piece-wise functions where there is a jump in the graph also don't have derivatives at the jump.

Rational functions lack a derivative at the asymptote.


Implicit Differentiation
Implicit differentiation is where you take the derivative of a function and you have multiple variables. When you find the derivative of each piece, you must multiply it by d(variable)/d(variable you are finding with regards to). Here is an example:

y² + x² = 3 Find dy/dx
2y(dy/dx) + 2x(dx/dx) = 0
2y(dy/dx) + 2x = 0
2y(dy/dx) = -2x
dy/dx = -x/y


The most important thing to remember while solving related rates problems is multiplying each piece by the correct d(variable)/d(variable). Otherwise, you will most likely not get the correct answer.

The Chain Rule and the Derivative

1) What is f' = 0?

f'(x) = 0 represents the maximum or minimum of f(x).

2) Identifying Where a Function Increases or Decreases

In order to identify where a function increases or decreases, you must first find the critical values of the function. This will give you the critical points which are possible maximums or minimums. Then you must plug in points before and after the critical points in the derivative. This will tell you if the points are negative or positive. You can use a number line to help visualize this. If the points go from negative to positive, then the function is increasing. If the points go from positive to negative, then the function is decreasing.

3) The Chain Rule

The chain rule is the process of finding the derivative of a composite function. Here are the steps to using the chain rule:
1) Take the derivative of the outside function.
2) Rewrite the interior function.
3) Multiply by the derivative of the inside.

Here is an example where we find the derivative and then the equation of the tangent line:

Find the tangent line of the function f(x) = (2x² + x)³ at the point x = -2

f'(x) = (2x² + x)³
        =  3(2x² + x)² • (4x + 1)
f'(x) = 3(2x² + x)² • (4x + 1)

f(-2) = (2(2)² + -2)³
        = (2(4) + -2)³
          = (8 + -2)³
        = (6)³
          = 216

slope = 3(2(-2)² + -2)² • (4(-2) + 1)
         = 3(6)² • (-7)
         = 3(36) • (-7)
         = 756
Our tangent line is:
y - 216 = 756(x + 2)
y - 216 = 756x + 1512
y = 756x + 1728

4) h(x) = f(g(x))
g(-4) = 5, g'(-4) = 2, and f'(5) = 20. Find h'(-4).

h'(x) = f'(g(x)) • g'(x)

h'(-4) = f'(g(-4)) • g'(-4)

h'(-4) = f'(5) • 2

h'(-4) = 20 • 2

h'(-4) = 40

Continuity and the Derivative

Continuous Functions
A function f(x) is continuous at x=a if:
1) The limit exists at x=a
2) f(a) exists (there is no hole/asymptote)
3) The limit at x=a is equivalent to f(a)

When a function isn't continuous:
          {x²+3, 0 < x
f(x) = {5, 0 = x
          {x+3, 0 > x

In order to check to see if a function is continuous, you must go through all three steps mentioned above. 

Step 1

We will see if the limit exists at x=a. We must look at the limit as x approaches the left and right of 0. The limit of the function as x approaches 0 from the left is 3. The limit of the function as x approaches 0 from the right is 3. This means that the limit exists.

Step 2

Next, we must make sure f(0) exists. By looking at the equation, we can see that f(x) = 5 when x = 0. This means that f(0) = 5 and f(a) exists. There isn't a hole or an asymptote.

Step 3

Finally, we must check to see if the limit at x=a is equivalent to f(a). Our limit as x approaches 0 is 3. However f(0) = 5. 5 doesn't equal 3. Therefore, the function isn't continuous.

Intermediate Value Theorem

If a function is continuous on a closed interval [a,b], and f(a) < N < f(b) then there is a < c < b such that f(x) = N.

We can use this theorem to tell if there is a solution on the interval [-2, 3] for the function f(x) = 2x² +3x - 5

 For this equation, f(-2) = -3 and f(3) = 22. Due to the fact that we know this function is continuous on the interval and the values are on opposite sides of the x-axis, we know there must be a solution in between them. We write this as: since f is continuous on [-2, 3], and f(-2) = -3 < 0 < 22 = f(3), then there exists c ∈[-2, 3] such that f(c) = 0.

Let's use this theorem on another function.

f(x) = 3x³ + 2 on the interval [-5, -1]

For this equation, f(-5) = -373 and f(-1) = -1. Due to the fact that both numbers are negative, we can't prove there is a solution on this interval. However, this doesn't mean there isn't one.

Derivatives

The derivative is basically the slope of a certain point in a function. There are two ways to find the derivative. One is by using the difference quotient and the other is by finding the derivative in terms of slope. We will find a linear derivative to demonstrate these concepts.

Difference Quotient

When we find the derivative by using the difference quotient, we use:

 lim   f(x+h)-f(x)

h→0        h

Find the derivative of f(x) = 3x + 5

3(x + h) + 5 - (3x + 5)

                h

3x + 3h + 5 -3x -5

             h

3x + 3h + 5 -3x -5

             h

3h

h

The derivative of this function is 3.

The Derivative in Terms of Slope

When we find the derivative in terms of slope, we use:

lim   f(x) - f(a)  at f(x) = a

x→a    x - a

Find the derivative of f(x) = 3x + 5 at f(x) = 2
3x + 5 - (3(2) + 5)
          x - 2

3x + 5 - 11
    x - 2

3x - 6
 x - 2

3(x - 2)
1(x - 2)

3(x - 2)
1(x - 2)

3
1

The derivative of this function is 3.

For me, the hardest thing about finding the derivative is foiling correctly while using the difference quotient.

Instantaneous Velocity and Average Velocity
Instantaneous velocity is the slope (or derivative) of an individual point. Average velocity is the derivative that can be used to find any point's slope for a particular function.