Math 1710

Topics for third exam

Chapter 3: Applications of Derivatives

§ 7:
Linear approximation and differentials
Idea: The tangent line to a graph of a function makes a good approximation to the function, near the point of tangency.

Tangent line to y = f(x) at (x0,f(x0) : L(x) = f(x0)+f¢(x0)(x-x0)

f(x) » L(x) for x near x0

Ex.: [Ö27] » 5+[1/2·5](27-25), using f(x) = Öx

(1+x)k » 1+kx, using x0=0

Df = f(x0+Dx)-f(x0), then f(x0+Dx) » L(x0+Dx) translates to

Df » f¢(x0) ·Dx

differential notation: df = f¢(x0) dx

So Df » df, when dx = dx is small

In fact, Df-df = (diffrnce quot -f¢(x0))Dx = (small)·(small) = really small, goes like (Dx)2

§ 8:
Newton's method
A really fast way to approximate roots of a function.

Idea: tangent line to the graph of a function ``points towards'' a root of the function

Roots of (tangent) lines are easy to find!

L(x) = f(x0)+f¢(x0)(x-x0) ; root is x1 = x0-[(f(x0))/(f¢(x0))]

Now use x1 as starting point for new tangent line; keep repeating!

xn+1 = xn-[(f(xn))/(f¢(xn))]

Basic fact: if xn approximates a root to k decimal places, then xn+1 tends to approximate it to 2k decimal places!

BUT:

Newton's method might find the ``wrong'' root: Int Value Thm might find one, but N.M. finds a different one!

Newton's method might crash: if f¢(xn) = 0, then we can't find xn+1 (horizontal lines don't have roots!)

Newton's method might wander off to infinity, if f has a horizontal asymptote; an initial guess too far out the line will generate numbers even farther out.

Newton's method can't find what doesn't exist! If f has no roots, Newton's method will try to ``find'' the function's closest approach to the x-axis; but everytime it gets close, a nearly horizontal tangent line sends it zooming off again!

Chapter 4: Integration

§ 1:
Antiderivatives
Integral calculus is all about finding areas of things, e.g. the area between the graph of a function f and the x-axis. This will, in the end, involve finding a function F whose derivative is f.

F is an antiderivative (or (indefinite) integral) of f if F¢(x) =f(x).

Notation: F(x) = òf(x) dx ; it means F¢(x)=f(x)

``the integral of f of x dee x''

Basic list:

òxn dx = [(xn+1)/(n+1)] + C (provided n ¹ -1)

òsin(kx) dx = [(-cos(kx))/k] + C

òcos(kx) dx = [sin(kx)/k] + C

òsec2 x dx = tanx + C

òcsc2 x dx = -cotx + C

òsecxtanx dx = secx + C

òcscxcotx dx = -cscx + C

Most differentiation rules can be turned into integration rules (although some are harder than others; some will even wait until Calc II !)

Basic integration rules: sum and constant multiple rules are easy to reverse

k=constant

òk·f(x) dx = kòf(x) dx

ò(f(x)±g(x) dx = òf(x) dx ± òg(x) dx

§ 3:
Integration by substiution
The idea: reverse the chain rule!

if g(x) = u, then [d/dx]f(g(x))=[d/dx]f(u) = f¢(u) [du/dx]

so òf¢(u) [du/dx] dx = òf¢(u) du = f(u)+c

òf(g(x)) g¢(x) dx ; set u = g(x)

then du = g¢(x) dx , so òf(g(x)) g¢(x) dx = int f(u) du , where u = g(x)

Example: òx(x+2-3)4 dx ; set u = x2-3, so du=2x dx . Then

òx(x+2-3)4 dx = [1/2]ò(x+2-3)42x dx =[1/2]òu4 du |u = x2-3 =

[1/2][(u5)/5]+c |u = x2-3 = [((x2-3)5)/10]+c

The three most important points:

1. Make sure that you calculate (and then set aside) your du before doing step 2!

2. Make sure everything gets changed from x's to u's

3. Don't push x's through the integral sign! They're not constants!

§ 4:
Estimating things with sums
Idea: alot of things can estimated by adding up alot of tiny pieces.

Sigma notation: åi = 1n ai = a1+¼an ; just add the numbers up

formal properties:

åi = 1n kai = kåi = 1n ai

åi = 1n (ai±bi) = åi = 1n ai ±åi = 1n bi

length of a curve: approximate curve by a collection of straight line segments

Figure

length of curve » å(length of line segments)

distance travelled = (average velocity)(time of travel)

over short periods of time, avg. vel. » instantaneous vel.

so distance travelled » å(inst. vel.)(short time intervals)

E.g., s(t)=position, v(t)=velocity, use velocity 4 times per second

dist. travelled = s(10)-s(5) » åi = 120 v(5+[i/4]) ([1/4])

average value of a function

average of n numbers: add the numbers, divide by n

for a function, add up lots of values of f, divide by number of values

avg. value of f » [1/n]åi = 1n f(ci)

§ 5:
Definite integrals
The most important thing to approximate by sums: area under a curve.

Idea: approximate region b/w curve and x-axis by things whose areas we can easily calculate:

rectangles!

Figure

Area between graph and x-axis » å (areas of the rectangles) =åi = 1n f(ci)Dxi

We define the area to be the limit of these sums as the number of rectangles goes to ¥ (i.e., the width of the rectangles goes to 0), and call this the definite integral of f from a to b:

òab f(x) dx  =  limn®¥ åi = 1n f(ci)Dxi

When do such limits exist?

Theorem If f is continuous on the interval [a,b], then òab f(x) dx exists.

(i.e., the area under the graph is approximated by rectangles.)

§ 6:
Properties of definite integrals
Fisrt note: the sum used to define a definite integral does need to have f(x) ³ 0; the limit still makes sense. When f is bigger than 0, we interpret the integral as area under the graph.

Basic properties of definite integrals:

òaa f(x) dx =0

òba f(x) dx = -òab f(x) dx

òab kf(x) dx = kòab f(x) dx

òab f(x)±g(x) dx =òab f(x) dx ± òab g(x) dx

òab f(x) dx + òbc f(x) dx = òac f(x) dx

If m £ f(x) £ M for all x in [a,b], then

m(b-a) £ òab f(x) dx £ M(b-a)

More generally, if f(x) £ g(x) for all x in [a,b], then

òab f(x) dx £ òab g(x) dx

Average value of f : formalize our old idea!

avg(f) = [1/(b-a)]òab f(x) dx

Mean Value Theorem for integrals: If f is continuous in [a,b], then

there is a c in [a,b] so that f(c) = [1/(b-a)]òab f(x) dx

§ 7:
The fundamental theorem of calculus
Formally, òab f(x) dx depends on a and b. Make this explicit:

òax f(t) dt = F(x) is a function of x.

F(x) = the area under the graph of f, from a to x.

Fund. Thm. of Calc (# 1): If f is continuous, then F¢(x) = f(x)

(F is an antiderivative of f !)

Since any two antiderivatives differ by a constant, and F(b) = òab f(t) dt, we get

Fund. Thm. of Calc (# 2): If f is continuous, and F is an antiderivative of f, then

òab f(x) dx = F(b)-F(a) = F(x) |ab

Ex: ò0p sinx dx = (-cosp)-(-cos0) =2

Building antiderivatives:

F(x)=òax Ö{sint} dt is an antiderivative of f(x) = Ö{sinx}

G(x) = òx2x3 [Ö(1+t2)] dt = F(x3)-F(x2), where

F¢(x) = [Ö(1+x2)], so G¢(x) = F¢(x3)(3x2)-F¢(x2)(2x)...

§ 8:
substitution and definite integrals
We can use u-substitution directly with a definite integral, provided we remember that

òab f(x) dx really means òx = ax = b f(x) dx

and we remember to change all of the x's to u's!

Ex: ò12 x(1+x2)6 dx; set u = 1+x2, du = 2x dx . when x = 1, u = 2; when x = 2, u = 5 ; so

ò12 x(1+x2)6 dx = [1/2]ò25 u6 du = ...