Review for Midterm 2, Concepts

Let f be continuous on [ a , b ].

Definitions

Let f be a function with domain D.

f has an absolute maximum on D at x = c if f (x) ≤ f (c) for all x in D .

f has an absolute minimum on D at x = c if f (x) ≥ f (c) for all x in D . f has a relative maximum at x = c if there exist an interval ( r , s ) containing c such that f (x) ≤ f (c) for all x in both D and ( r , s ). f has a relative minimum at x = c if there exist an interval ( r , s ) containing c such that f (x) ≥ f (c) for all x in both D and ( r , s ).

Extreme Value Theorem

If f is continuous on a closed interval [ a , b ], then f has an absolute maximum and an absolute minimum on [ a , b ] If f has a local maximum or minimum value at x = c and f '(c) exists, then f '(c)= 0 .

Suppose f is continuous on [ a,b ] . How do we find its maximum and minimum values?

1. Check where f '(x)= 0 or where f '(x) does not exist. 2. Check f (a) and f (b) .

Mean Value Theorem

Suppose f is continuous on the interval [ a,b ] and differentiable on the interval ( a,b ) . Then there exists at least one point c in ( a,b ) f (b) f (a) such that f '(c) = − . b − a

The average rate of change in f over [ a,b ] is equal to the instantaneous rate of change at some point in ( a,b ). Theorem

If f '(x) > 0 for every x in an interval ( r , s ) then f is strictly increasing on the interval ( r , s ).

Theorem

If f '(x) < 0 for every x in an interval ( r , s ) then f is strictly decreasing on the interval ( r , s ).

Theorem

If f '(x) = 0 for every x in an interval ( r , s ) then f is a constant function on the interval ( r , s ).

Theorem

If f '(x) = g'(x) for every x in an interval ( r , s ) then there exist a constant c such that f (x) = g(x)+ c for all x in ( r , s ).

First Derivative Test:

Suppose x = c is a critical point for f .

If f ' changes from negative to positive at c then f has a local minimum at c .

If f ' changes from positive to negative at c then f has a local maximum at c .

If f ' does not change sign at c then f has no local extremum at c .

Concavity and Curve Sketching

Example

Consider the function f defined by f (x)= x3. Note that f '(x)= 3x2 , f "(x)= 6x f "(x)> 0 for x > 0 f "(x) 0 for x 0 < < f ' is increasing on ( 0, ∞ ) f ' is decreasing on ( − ∞, 0 ) f "changes sign at x = 0

Definition

The function f is concave up on the interval ( a,b ) if f ' is increasing on ( a,b ).

The function f is concave down on the interval ( a,b ) if f ' is decreasing on ( a,b ).

Definition

( c, f (c)) is a point of inflection if the graph of f has a tangent line at ( c, f (c)) and the concavity of f changes at ( c, f (c)).

Second Derivative Test

Suppose f " is continuous near x = c .

1. If f '(c)= 0 and f "(c)< 0 then f has a local maximum at x = c . 2. If f '(c)= 0 and f "(c)> 0 then f has a local minimum at x = c . 3. If f '(c)= 0 and f "(c)= 0 then the test fails.

Indeterminate Forms and L’Hopital’s Rule

Remember lim sinx . x→0 x

L’Hopital’s Rule: Suppose f and g are differentiable, f (a)= g(a)= 0 and g'(x)≠ 0 when x ≠ a Then

f (x) f '(x) lim = lim . x→a g(x) x→a g'(x)

For the indeterminate forms 0 , ∞ , the limit of the quotient of two 0 ∞ functions is equal to limit of the quotient of the derivatives. Other 0 ∞ indeterminate form can often be reformulated to fit or . 0 ∞

Related Rates, Applied Problem Solving

An office building is located right on a straight riverbank. A power plant is on the opposite bank, 1500 feet downstream from the office building. The river is 300 feet wide. If we want to connect the power plant and the building by cable, which costs $1700 per foot to lay down underwater and $800 per foot to lay underground, what is the least expensive path for the cable?

Problem Solving Steps

1. Understand what is given. 2. Choose a variable (s). 3. Obtain a mathematical description of the problem. 4. Do the math. 5. Interpret the results.

Antiderivatives

⎡ ⎤ The answer is ⎣ ⎦ , what is the question? d a) ⎡ ⎤ = x7 dx ⎣ ⎦

d 1 b) ⎡ ⎤ = dx ⎣ ⎦ x

d c) ⎡ ⎤ = 3cos2θ + 2sec2θ tanθ dx ⎣ ⎦

d d) ⎡ ⎤ = xk dx ⎣ ⎦

Definition

If F '(x) = f (x) for all x, then F is an antiderivative for f .

Suppose F '(x) = f (x) for all x . Then any antiderivative of f can be represented by F(x)+ c . These antiderivatives are denoted by ∫ f (x)dx , called the indefinite integral of f .

xk+1 xk dx = + c (Power Rule) ∫ k +1

∫ k f (x) dx = k ∫ f (x) dx

∫ [ f (x) ± g(x)]dx = ∫ f (x)dx ± ∫ g(x)dx

2 ∫ e−x dx = ?

Integrals

Archimedes Diagram:

Archimedes used the method of exhaustion to find an approximation to the area of a circle. This is an early example of integration that led to approximate values of π.

Example

Let’s approximate the area below the graph of y = f (x) = 1− x2 between x = 0 and x =1.

Upper Sums, Lower Sums, Midpoint Sums

Riemann sum calculator: https://www.desmos.com/calculator/tgyr42ezjq

Sums

n n(n 1) 1+ 2+ 3+ 4 +...+ (n−1)+ n = ∑ i = + . (Gauss) i =1 2

n n(n 1)(2n 1) 12 + 22 + 32 + 42 + ... + k2 + ...+ n2 = ∑ k2 = + + k =1 6

n n2(n 1)2 13 + 23 + 33 + 43 + ... + k3 + ...+ n3 = ∑ k3 = + k =1 4

Riemann Sums, Definite Integral

How should we approximate with areas of rectangles?

1. We need to partition the interval [ a,b ] into small subintervals.

2. We must then use the function f to determine the height of each rectangle and decide whether to count the area positively or negatively.

Definition

A partition of [ a,b ] is a set of points

{ x0 , x1 , x2 , x3 ... xk−1 , xk ,..., xn−1 , xn } such that

a = x0 < x1 < x2 < x3 < ... < xk−1 < xk <...< xn−1 < xn = b .

Our goal is to approximate areas with ever increasing degree of accuracy; so we will want our partitions to define a large number of subintervals with small width. If P is a partition of [ a,b ] then determine how good the partition is by considering the length of the largest subinterval.

Some more notation:

Let P = { x0 , x1 , x2 , x3 ... xk−1 , xk ,..., xn−1 , xn } be a partition of [ a,b ].

So a = x0 < x1 < x2 < x3 < ... < xk−1 < xk <...< xn−1 < xn = b .

The subintervals are

[ x0 , x1 ],[ x1 , x2 ],[ x2 , x3 ],...,[ xk−1 , xk ],...,[ xn−1 , xn ] . Let

x x x , x x x , ... , x x x , ... , x x x Δ 1 = 1 − 0 Δ 2 = 2 − 1 Δ k = k − k−1 Δ n = n − n−1 The length of the largest subinterval is equal to

max { x , x ,..., x , ... , x } Δ 1 Δ 2 Δ k Δ n and is denoted by P , called the norm of the partition P.

If we want our approximation to be accurate, then we want P to be small (close to zero).

For the heights of the rectangles we will choose a point

ck from each[ xk−1 , xk ] and evaluate f ( ck ) .

We now obtain a Riemann sum:

n ∑ f ( ck ) Δk . What happens when P → 0 ? k =1

Definition

Suppose f is a continuous function on [ a,b ]. The definite integral of f over [ a,b ] is

n b lim ∑ f ( ck ) Δk and is denoted by f (x) dx . P →0 k =1 ∫a

We read “the integral of f from a to b with respect to x”. Formally,

n lim ∑ f ( ck ) Δk = L means P →0 k =1

for each ε > 0, there exists δ > 0 such that

n ∑ f ( ck ) Δk − L < ε whenever P < δ k =1

As long as the norm of the partition is small enough (norm less than delta), it doesn’t matter what point you choose from each subinterval. The Riemann sum will be within epsilon of L.

Suppose f is a continuous function on [ a,b ] . The average value of f on [ a,b ] can be computed in terms of a definite integral.

b Average value of f on [ a,b ] is equal to 1 f (x)dx b − a ∫a

b How do we compute f (x) dx ? ∫a We are going to show that if f is continuous on [ a,b ] , then

b b f (x) dx = F(x) = F(b)− F(a) where F is any ∫a a antiderivative of f .

Fundamental Theorem of Calculus

I. Suppose f is a continuous function on [ a,b ] . Let

x F (x)= f (t)dt for a ≤ x ≤ b . Then F '(x) = f (x) for each x. Also ∫a b note that F (b)= f (x)dx . ∫a

II. Suppose f is a continuous function on [ a,b ] . Let G(x) be any antiderivative of f (x) . Then

b b f (x)dx = G(x) = G(b)− G(a). ∫a a

Definition Suppose f is a differentiable function. Let dx be a variable called the differential dx. The differential dy is dy = f '(x)dx .

Method of Substitution to find ∫ f (x)dx

Definite Integrals and Areas Between Curves

Definition

A function f is even provided f (−x) = f (x) for each x.

a a If f is even then f (x)dx = 2 f (x)dx . ∫−a ∫0 A function f is odd provided f (−x) = − f (x) for each x.

a If f is odd then f (x)dx = 0 . ∫−a

Areas Between Curves Find the area of the region enclosed by the curve defined by x − y3 = 0 and the line whose equation is y = x .

(Graph and Integrate)