Chapter 3: Applications of Derivatives
Tangent line to y = f(x) at (x_{0},f(x_{0}) : L(x) = f(x_{0})+f^{¢}(x_{0})(x-x_{0})
f(x) » L(x) for x near x_{0}
Ex.: [Ö27] » 5+[1/2·5](27-25), using f(x) = Öx
(1+x)^{k} » 1+kx, using x_{0}=0
Df = f(x_{0}+Dx)-f(x_{0}), then f(x_{0}+Dx) » L(x_{0}+Dx) translates to
Df » f^{¢}(x_{0}) ·Dx
differential notation: df = f^{¢}(x_{0}) dx
So Df » df, when dx = dx is small
In fact, Df-df = (diffrnce quot -f^{¢}(x_{0}))Dx = (small)·(small) = really small, goes like (Dx)^{2}
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(x_{0})+f^{¢}(x_{0})(x-x_{0}) ; root is x_{1} = x_{0}-[(f(x_{0}))/(f^{¢}(x_{0}))]
Now use x_{1} as starting point for new tangent line; keep repeating!
x_{n+1} = x_{n}-[(f(x_{n}))/(f^{¢}(x_{n}))]
Basic fact: if x_{n} approximates a root to k decimal places, then x_{n+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^{¢}(x_{n}) = 0, then we can't find x_{n+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
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:
òx^{n} dx = [(x^{n+1})/(n+1)] + C (provided n ¹ -1)
òsin(kx) dx = [(-cos(kx))/k] + C
òcos(kx) dx = [sin(kx)/k] + C
òsec^{2} x dx = tanx + C
òcsc^{2} 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
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 = x^{2}-3, so du=2x dx . Then
òx(x+2-3)^{4} dx = [1/2]ò(x+2-3)^{4}2x dx =[1/2]òu^{4} du |_{u = x2-3} =
[1/2][(u^{5})/5]+c |_{u = x2-3} = [((x^{2}-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!
Sigma notation: å_{i = 1}^{n} a_{i} = a_{1}+¼a_{n} ; just add the numbers up
formal properties:
å_{i = 1}^{n} ka_{i} = kå_{i = 1}^{n} a_{i}
å_{i = 1}^{n} (a_{i}±b_{i}) = å_{i = 1}^{n} a_{i} ±å_{i = 1}^{n} b_{i}
Some things worth adding up:
length of a curve: approximate curve by a collection of straight line segments
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) » å_{i = 1}^{20} 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]å_{i = 1}^{n} f(c_{i})
Idea: approximate region b/w curve and x-axis by things whose areas we can easily calculate:
rectangles!
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:
ò_{a}^{b} f(x) dx = lim_{n®¥} å_{i = 1}^{n} f(c_{i})Dx_{i}
When do such limits exist?
Theorem If f is continuous on the interval [a,b], then ò_{a}^{b} f(x) dx exists.
(i.e., the area under the graph is approximated by rectangles.)
Basic properties of definite integrals:
ò_{a}^{a} f(x) dx =0
ò_{b}^{a} f(x) dx = -ò_{a}^{b} f(x) dx
ò_{a}^{b} kf(x) dx = kò_{a}^{b} f(x) dx
ò_{a}^{b} f(x)±g(x) dx =ò_{a}^{b} f(x) dx ± ò_{a}^{b} g(x) dx
ò_{a}^{b} f(x) dx + ò_{b}^{c} f(x) dx = ò_{a}^{c} f(x) dx
If m £ f(x) £ M for all x in [a,b], then
m(b-a) £ ò_{a}^{b} f(x) dx £ M(b-a)
More generally, if f(x) £ g(x) for all x in [a,b], then
ò_{a}^{b} f(x) dx £ ò_{a}^{b} g(x) dx
Average value of f : formalize our old idea!
avg(f) = [1/(b-a)]ò_{a}^{b} 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)]ò_{a}^{b} f(x) dx
ò_{a}^{x} 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) = ò_{a}^{b} f(t) dt, we get
Fund. Thm. of Calc (# 2): If f is continuous, and F is an antiderivative of f, then
ò_{a}^{b} f(x) dx = F(b)-F(a) = F(x) |_{a}^{b}
Ex: ò_{0}^{p} sinx dx = (-cosp)-(-cos0) =2
Building antiderivatives:
F(x)=ò_{a}^{x} Ö{sint} dt is an antiderivative of f(x) = Ö{sinx}
G(x) = ò_{x2}^{x3} [Ö(1+t^{2})] dt = F(x^{3})-F(x^{2}), where
F^{¢}(x) = [Ö(1+x^{2})], so G^{¢}(x) = F^{¢}(x^{3})(3x^{2})-F^{¢}(x^{2})(2x)...
ò_{a}^{b} f(x) dx really means ò_{x = a}^{x = b} f(x) dx
and we remember to change all of the x's to u's!
Ex: ò_{1}^{2} x(1+x^{2})^{6} dx; set u = 1+x^{2}, du = 2x dx . when x = 1, u = 2; when x = 2, u = 5 ; so
ò_{1}^{2} x(1+x^{2})^{6} dx = [1/2]ò_{2}^{5} u^{6} du = ...