Notes

Winter Semester
Calculus 2 ( 1014 )

5.4 Indefinite Integrals and Net Change Theorem
- When we work in definite integrals we use F(b) - F(a) but with indefinite integrals it works differently
Instead of going through F(b) - F(a) we can just begin changing it into its antiderivative.
- Note though after we find the antiderivative with must add a constant +C at the end
- There are multiple indefinite integral identities in the chart.

Net Change Theorem
- The theorem states that when a quantity changes the final value is equal to the initial value + integral of the rate of change which is commonly a definite integral (Meaning it will equal a number since definite integrals will always give a value)

5.5 U-Substitution
- Substitution rule for integration means we can put another variable in the place of a value
u = g(x) meaning du = g'(x)
Its easiest to think of dx and du as differentials
You can decide which value to be u by using LIPET
Logarithmic
Inverse
Polynomials
Exponential
Trigonometric

Substitution with indefinite integrals

Steps to solve!
- First start by deciding on what value will be equal to u
- Next we find du (the derivative of u) and after solve for dx
- We rewrite and input the variable u and value of du or dx
- Move the coefficient outside of the integral and begin solving by finding the antiderivatives
- Input the u value back and add +C to your answer

Substitution with definite integrals
Note! Whenever you use u substitution you need to change the upper and lower bounds in respect to u
since without it would be in respect to x still
- We solve it normally calculating F(b) - F(a)
We can follow the steps above used to indefinite integrals nearly the exact same.

Symmetry
- If there is symmetry you can expect that one will be equal to the other


6.1 Areas between curves
- We begin using Riemann's sum again
but in this case they're split up into integrals
- We integrate them normally such as solving definite integrals in respect to x

Integrating in respect to y
- There isn't anything that differs too greatly other than instead of xr - xl the values change to y and you solve it as a normal definite integral

6.2 Volumes
- Again we are using Riemann's sums and integrals but we are adding a twist with including radius

Volumes of Solids in revolution
- Now we will be solving both inner and outside for area
In an example of a disk
Here we use A = pi(outer radius)^2 - pi(inner radius)^2
these then will be rewritten as a definite integral to then solve

Finding volume with cross-sectional area
- Cross section is a shape or what will be left if you cut a solid shape in half.
- This isn't any different than solving like with the others but broken into more parts

A cross section of a solid is a plane figure obtained by the intersection of that solid with a plane. The cross section of an object therefore represents an infinitesimal "slice" of a solid, and may be different depending on the orientation of the slicing plane.



6.4 Work
- We know work from physics its a force
W = F * d
F = ma or F = mg most commonly
it can also be written as m ( d^2s/dt^2) for F = ma

Now lets express this in integrals!
Work = upper b lower a f(x)dx
- We treat it as a definite integral and solve
Now we know that f = kx when applying hooke's law
where k is always the spring constant

6.5 Average value of a function
- Commonly when we solve for the average of something we add and then divide by the amount of numbers we added
- but for functions we write it as x = (b - a)/ n (n being any number)
- for integrals we will rewrite it as f avg = 1/b-a (upper b lower a) f(x)dx
a being the value of x and b the value of y
- Then we can simply solve for the average!

7.1 Integration by parts
- IBP short form is a way for us to solve integrals when we struggle and don't entirely know the antiderivative off the top of our heads
- the formula for integration by parts looks as such
integral sign u dv is equal to and can be written as
= u(v) integral sign v du

Indefinite integrals
- We need to identify u, du, dv, v
u can be picked using lipet and dv is whatever is left over
- We can then apply all the rules we know of integration and at the end remember to add +C


Definite integrals
- This works nearly similar to normal definite integrals but just like indefinite integrals
we need to identify the values and input them
- After we can solve as usual with F(b) - F(a)

Reduction formulas!
- Whenever we may have a trigonometric function to the power we can reduce because one may equal another such as
cos^2 x = 1 - sin^2 x
- This way we can cancel variables out to make it easier to solve
- Then we can solve normally


7.2 Trigonometric Integrals
- Trigonometric integrals differ from normal integrals due to using trig identities
- There are many trig identities that correlate with another meaning we can sometimes simplify to cancel out variables

Integrals of Powers of Sine and Cosine
- These work similarly to where we use U- Substitution to solve the integral and cancel
- We also can convert trig identities or factor or separate sin and cos to powers to change it into either
making it easier to substitute or identify a u value

Integrals of Powers of Secant and Tangent
- Secant and Tangent work similarly to sine and cosine power integrals not differing too greatly
- We again convert, factor/separate to make it easier for us to solve the integral

Using Product Identities
- These are examples of product identities that can help when evaluating trig integrals
- Sin(A)Cos(B) = 1/2 [sin(A-B) + sin(A+B)]
- Sin(A)Sin(B) = 1/2 [cos(A-B ) cos(A+B )]
- Cos(A)Cos(B) = 1/2 [cos(A-B ) cos(A+B )]

7.3 Trigonometric substitution
- This works somewhat similar to u substitution
- The only difference is you use trig identities in this case
- This type of substitution is called Inverse substitution


7.4 Integration of Rational functions by partial fractions

Method of Partial fractions
- first off lets imagine a rational function f(x) = P(x)/Q(x)
where P and Q are polynomials

In case 1 where Q(x) is the product of distinct linear factors
- We need to factor the denominator Q(x) to then be able to express it as A/(ax - b)^i or Ax + B / (ax^2+bx+c)
- We can factor the denominator to make it simpler to solve for A, B, C
- To determine the values of them we multiply both side of the equation
- Then by expanding on the right side we can rewrite into standard form showing us that the values of the left are equal to the right side
- In this case x^2 = (everything in the brackets corresponding to its variable) and the same for x and the constant

In case 2 where Q(x) is a product of linear factors some of which are repeated
- It works similarly to case 1 but instead the denominators are different where it goes
A/a + B/a^2 and C/the entire function aka C



7.8