Big idea in : approximation

′ df ∆f f (x + h) f (x) : f (x)= = − , dx ≈ ∆x h rate of change is approximately the ratio of changes in the function value and in the variable in a very short time

Linear approximation: f (x) f (x )+ f ′(x )(x x ) ≈ 0 0 − 0 The graph of nonlinear function the graph of tangent line ≈ Diﬀerentiation: Calculate the rate of small change

Integration: Accumulate all small changes How to calculate the ?

Rectangle: A = ab (a = length, b = width) 1 Triangle: A = ab 2

Polygon: divide it to triangles, calculate each then take the sum What if it is a circle?

Calculation of π: Archimedes (287–212 BC) 3.1408 223/71 <π< 22/7 3.1429 Liu Hui (about 220–280) π 3.1416,≈ ≈ Zu Chongzhi (429–500) π ≈3.1415926 ≈ Big idea

The area of non-polygon can be calculated from that of polygons! (and take limit)

Area under the curve when f (x) 0: ≥ Geometric idea

Use rectangle with certain height to approximate irregular but almost rectangular shape

Procedure: 1. divide the [a, b] into n equal subintervals with length of subinterval ∆x = (b a)/n − 2. With x = a, x = x +∆x, , xi = xi +∆x, , xn = b, and choose yi so 0 1 0 ··· +1 ··· that xi−1 yi xi . 3. Approximate the area above [xi , xi+1] by f (yi ) ∆x 4. Take the≤ sum≤ (called sum): · n f (yi )∆x = f (y )∆x + f (y )∆x + + f (yn)∆x 1 2 ··· Xi=1

How to choose yi ? Left-endpoint-sum (Ln), right-endpoint-sum (Rn), midpoint rule (Mn). Riemann sum

Example: Find area below the curve y = x2 with 0 x 1 by using Riemann sum ≤ ≤ approximation L4, R4 and M4. L =[f (0) + f (1/4) + f (1/2) + f (3/4)] 0.25 = 7/32 = 0.21875 4 · R4 =[f (1/4) + f (1/2) + f (3/4) + f (1)] 0.25 = 15/31 = 0.46875 M =[f (1/8) + f (3/8) + f (5/8) + f (7/8)]· 0.25 = 21/64 = 0.328175 4 · What if n is large? Area as limit: Deﬁnite

The area under y = f (x) is the limit of Riemann sum: n A = lim Rn = lim f (yi )∆x n→∞ n→∞ Xi=1 is the notation for sum (called sigma) P b n f (x)dx = lim f (yi )∆x (limit of Riemann sum) Z n→∞ a Xi=1 where ∆x = (b a)/n, x = a, x = x +∆x, , x = x +∆x, , xn = b, and − 0 1 0 ··· k+1 k ··· xi− yi xi . 1 ≤ ≤ When n , ∆x 0. If the limit exists, then the function f is integrable on [a, b], b →∞ → and f (x)dx is the deﬁnite integral of f from a to b. Za (Theorem: If f is continuous, or f has only a ﬁnite of jump discontinuities, then f is integrable.)

a: lower limit; b: upper limit; f (x): integrand; : integral sign (S=sum) Z deﬁnite integral is a number Meaning of the deﬁnite integral

When f (x) is positive, the deﬁnite integral is the area below f (x) (and above y = 0)

When f (x) can be both positive and negative, the deﬁnite integral is the “signed” area 2 Example: Find (2x 3)dx. Z0 − Properties of deﬁnite integral:

b b b f (x)+ g(x) dx = f (x)dx + g(x)dx, Za   Za Za b b b f (x) g(x) dx = f (x)dx g(x)dx, Za  −  Za − Za b b αf (x)dx = α f (x)dx, Za Za b c b f (x)dx = f (x)dx + f (x)dx, Za Za Zc a b c f (x)dx = f (x)dx, and f (x)dx = 0, Zb − Za Zc b b If f (x) g(x) for x [a, b], then f (x)dx g(x)dx, ≤ ∈ Za ≤ Za If m f (x) M for x [a, b], then ≤ ≤ b ∈ m(b a) f (x)dx M(b a). − ≤ Za ≤ − How to calculate the deﬁnite integral?

Evaluation Theorem: b If f (x) is continuous on [a, b], then f (x)dx = F (b) F (a), where F (x) is an Za − of f (x). Recall: F (x) is an antiderivative of f (x) if F ′(x)= f (x).

Example: 3 4 (a) (1 + 2x 4x3)dx; (b) √xdx. Z1 − Z1

F (x)= f (x)dx is also called indeﬁnite integral of f (x). Z b f (x)dx (deﬁnite integral is a number), Za and f (x)dx (indeﬁnite integral is a (family of) function) Z Basic Formula

Elementary integral formulas: (see page 277) 1 xndx = xn+1 + C sin xdx = cos x + C Z n + 1 Z − 1 dx = ln x + C cos xdx = sin x + C Z x | | Z 1 ax dx = ax + C sec2 xdx = tan x + C Z ln a Z 1 − 1 − dx = sin 1 x + C dx = tan 1 x + C Z √1 x2 Z 1+ x2 Basic Properties− of : (see page 269-271) b b b b b [f (x) g(x)]dx = f (x)dx g(x)dx c f (x)dx = c f (x)dx Za ± Za ± Za Za · Za b c b a b f (x)dx = f (x)dx + f (x)dx f (x)dx = f (x)dx Za Za Zc Zb − Za Meaning and variations of Evaluation Theorem

b f (x)dx = F (b) F (a), where F (x) is an anti-derivative of f (x) Za −

b ′ F (x)dx = F (b) F (a), where F (x) is a diﬀerentiable function Za − (meaning: the integral of rate of change is the total change)

t2 a(t)dt = v(t2) v(t1) Zt1 − (meaning: the integral of acceleration is the total change in velocity)

t2 v(t)dt = D(t2) D(t1) Zt1 − (meaning: the integral of velocity is the total change in position)

But the total distance traveled is t2 v(t) dt Zt1 | | (“mileage” and “change of position” are diﬀerent) Fundamental Theorem of Calculus

integration and diﬀerentiation are inverse process

b ′ (a) F (x)dx = F (b) F (a), here F (x) is diﬀerentiable Za − Function F (x) (diﬀerentiate) Function F ′(x) (integrate) b ⇒ ⇒ ′ F (x)dx = F (b) F (a) Za − x (b) If F (x)= f (t)dt, then F ′(x)= f (x). Za x Function f (x) (integrate) Function F (x)= f (t)dt (diﬀerentiate) ⇒ Za ⇒ d x f (t)dt = f (x) dx Za 

(c) If F (x)= f (x)dx, then F ′(x) = f (x). Z Sample problems

Problems from Math 111 Final (Fall 2002) 4 1. Consider the graph of the function f (x), and ﬁnd f (x)dx. Z0

4

3

2

1

0 1 2x 3 4 5

2. A ping-pong ball is ﬂoating down a river. You record its velocity every 5 minutes for 20 minutes as shown in the table below. Estimate the total distance the ball travelled during the ﬁrst 20 minutes by using 4 subintervals with left-endpoint values.

t (min) 0 5 10 15 20 v (m/min) 1 1.2 1.7 2.0 1.8 More problems

Distance, velocity and acceleration revisited: Suppose that a(t) = 2t 3, v(0) = 1. (1) Find the velocity function, and ﬁnd the total distance traveled during− 0 t 10. (2) Draw the graph of a(t) and v(t), and show the area on the graph representing≤ ≤ the change of velocity, position, and also distance traveled. t v(t) v(0) = a(s)ds − Z0 t D(t) D(0) = v(s)ds − Z0

Fundamental Theorem of Calculus: x 1. Find the derivative of g(x) = ln tdt Z1 10 2. Find the derivative of F (x)= tan tdt Zx x2 3. Find the derivative of h(x) = 1+ t3dt Z0 p 4. Find the average value of the function f (x) = 1/x in [1, 4], and ﬁnd c [1, 4] such ∈ that f (c) = average value. Average Value

Average value of a function on [a, b]: (think about average velocity) 1 b fave = f (x)dx b a Z − a a + a + a + + an Average value of n : 1 2 3 ··· n Mean-value Theorem for the integral: If f (x) is continuous on [a, b], then there exists b 1 b c in [a, b] such that f (c) = fave = f (x)dx, or f (c)(b a)= a f (x)dx. b a Za − − R