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