Math 165 - Calculus I Christian Roettger 382 Carver Hall Mathematics Department Iowa State University www.iastate.edu/~roettger November 13, 2011
4.1 Introduction to Area Sigma Notation 4.2 The Definite Integral 4.3 First Fundamental Theorem of Calculus 4.4 Second Fundamental Theorem of Calculus 4.5 Mean Value Theorem for integrals, symmetry
Intro to Area I We all know how to calculate the area of a rectangle, a parallelogram (height x base), a triangle (height x base /2), a polygon (sum of triangles). To do that, we use common sense facts about area. The area of a rectangle is height x base. Congruent regions have equal areas. Lines have area zero. The area of two regions which overlap only in a line is the sum of the area of these regions. If one region is contained in another, then the area of the second is larger than that of the first.
Intro to Area II Area by Inscribed Polygons The following area has been evaluated by the ancient Greek mathematician Archimedes 2000 years ago. Consider the region R bounded by the parabola y = x 2, the x-axis and the vertical line x = 2.
Intro to Area III We calculate the area of the red rectangles. This will be smaller than A(R), A(r 1 ) + + A(r 4 ) = 0.4(0.4 2 + 0.8 2 + 1.2 2 + 1.6 2 ) = 1.92 A(R). and if we divide [0, 2] into n equal parts of length h, we will get in a similar way A(r 1 ) + + A(r n 1 ) = h(0h 2 + 1h 2 + 4h 2 + + (n 1) 2 h 2 ) We might guess the limit A(R) lim A(r 1) + + A(r n 1 ) n
Intro to Area IV but how do we compute it?? What happens if we take a polygon which contains R?
Intro to Area V The yellow rectangles together contain R, and their area is A(s 1 ) +... + A(s 4 ) = 0.4(0.4 2 + 0.8 2 + 1.2 2 + 1.6 2 + 2 2 ) = 2.52 A(R). If we did the same with n rectangles, dividing [0, 2] into n equal parts, we would get A(r 1 ) +... + A(r n ) = h(1h 2 + 4h 2 + 9h 2 + + n 2 h 2 ) A(R) Visually, the difference between red and yellow rectangles is just that last yellow rectangle! Its area 8/n tends to zero as n.
Sigma Notation I A convenient shorthand for handling long sums involving.... Definition For given numbers a 1, a 2, a 3,..., a n, we write n a i := a 1 +... + a n. i=1 The variable i is called index variable, 1 and n are called limits of or bounds for i. These examples show that i can start at any integer, that a i can be the value of a function at i, and that the Sigma notation can make your life if not easier, then at least shorter.
Sigma Notation II 6 i = 3 + 4 + 5 + 6, i=3 3 sin(πk/2) = sin(0) + sin(π/2) + sin(π) + sin(3π/2), k=0 100 r 2 = 1 2 + 2 2 +... + 100 2. r=1 Any programmers out there?
Sigma Notation III Many programming languages would calculate the last sum somewhat like this. S:=0; for r from 1 to 100 do S:=S+r^2; enddo; Example (constant sequence) n i=1 c = c +... + c }{{} n times = n c.
Sigma Notation IV Like D x on functions, Σ can be viewed as a linear operator on finite sequences. Theorem (A - Linearity of Σ) Given finite sequences a 1,..., a n and b 1,..., b n and a constant c, n ca i = c i=1 n (a i + b i ) = i=1 n (a i b i ) = i=1 n i=1 a i n a i + i=1 n a i i=1 n i=1 n i=1 b i b i
Sigma Notation V Example Suppose 10 i=1 a i = 3 and 10 i=1 b i = 5. Then n (2a i b i ) = 2 i=1 10 i=1 a i 10 i=1 b i = 2 3 5 = 1. Example (telescoping sums) Suppose a finite sequence a 1,..., a n+1 is given. Then n (a i+1 a i ) = (a 2 a 1 ) +... + (a n+1 a n ) = a n+1 a 1. i=1
Sums of powers I Example (special sums) n i = i=1 n j 2 = j=1 n(n + 1) 2 n(n + 1)(2n + 1) 6
Sums of powers II We can apply these formulas and linearity to work out some more complicated sums. 10 i=1 (i + 1)(i 3) = = = 10 i=1 10 (i 2 2i 3) i 2 2 i=1 10 11 21 6 10 i=1 i 10 i=1 3 10 11 30 = 245.
Archimedes area I Recall the area between the x-axis, the line x = 2, and y = x 2. Divide [0, 2] into n equal parts of length h. We got n 1 A(r 1 ) + + A(r n 1 ) = h (ih) 2 i=1 3 (n 1)n(2n 1) = h A(R). 6 Using h = 2/n, we can now let n tend to infinity! lim A(r 8(n 1)n(2n 1) 1) + + A(r n ) = lim n n 6n 3 = 8 3.
Handshake numbers and little Carl-Friedrich Gauss I Suppose 101 people are in a room and everybody shakes hands with everybody else. Let h be the number of handshakes taking place. Then h = 100 + 99 + 98 +... + 2 + 1 = 5050.
Definite Integral I Definition (Riemann Sum) Consider a function f (x) defined on a closed interval [a, b]. Suppose we have a partition P of [a, b], given by a = x 0 < x 1 <... < x n 1 < x n = b, let x i := x i x i 1. In each subinterval [x i 1, x i ], pick an arbitrary sample point x i. Then R P (f ) := n f ( x i ) x i i=1 is called a Riemann sum for f corresponding to P. Riemann sums are approximations to the area between the graph of f (x) and the x-axis. Choosing suitable sample points gives
Definite Integral II approximations from below (red rectangles) or above (yellow rectangles). For most examples, we use the regular partition which just divides [a, b] into n equal intervals of constant size h = (b a)/n. Note that the i-th interval in this partition is [a + (i 1)h, a + ih]. Definition The norm of a partition P is the maximal size of one of its intervals. It is written P. For the regular partition E, we have E = h = (b a)/n.
Definite Integral III Definition (Definite Integral) Let f be a function defined on [a, b]. If the limit exists, then b a lim R P(f ) P 0 f (x) dx := lim P 0 R P(f ) is called the definite integral of f from a to b. The variable x in the integral is a dummy variable (Las Vegas - what happens in here, stays in here). It could be p, y, z or whatever. 5 5 5 x 2 dx = y 2 dy = z 2 dz. 2 2 2
Definite Integral IV The endpoints a, b are often called limits of the integral (they are not really limits!). If f (x) is nonnegative, the definite integral is the area between the graph of f and the x-axis. If f (x) is negative, the definite integral is the negative of the area. If f is both negative and positive, the contributions may cancel each other out. b a f (x) dx = Aup A down
Definite Integral V Example Find the total area and the definite integral between the lines y = x + 1, x = 4, x = 3 and the x-axis. Solution We partition [ 4, 1] into n subintervals of length h. The i-th subinterval will be [ 4 + (i 1)h, 4 + ih] with
Definite Integral VI h = ( 1 ( 4))/n = 3/n. Then choose x i = 4 + ih. The corresponding Riemann sum is R P (x + 1) = n ( 4 + ih + 1)h = 3nh + h 2 1 n(n + 1) 2 i=1 Substitute h = 3/n and calculate the limit as n to get 1 4 x + 1 dx = 9 + 9 2 = 9 2. Do the same thing with [ 1, 3] to get h = 4/n and 3 1 x + 1 dx = lim n n ( 1 + ih + 1)h i=1 = lim 1 n h2 n(n + 1) = 8. 2
Definite Integral VII So the total area is 8 + 4.5 = 12.5 whereas 3 4 x + 1 dx = 9 2 + 8 = 3.5. The next two sections are about the Fundamental Theorem of Calculus. In many cases it allows to calculate these areas much easier and faster, without doing any limits. It is very similar to doing derivatives using rules of differentiation instead of doing limits. But unlike differentiation, this does not work in all cases. Eg for f (x) = sin(x)/x, the Fundamental Theorem of Calculus does not help and approximations are the best we can do. So Riemann sums (and some more sophisticated methods but similarly involving lots of Σ-notation, see section 4.6) are still useful to know.
Definite Integral VIII MATLAB demo of Riemann sum with 10 rectangles for sin(x)/x. Shown on top is the approximation 1.422..., below is a slider where in MATLAB you could change the number of rectangles. Theorem (A - Integrability) If a function f is bounded on an interval [a, b] and if it is continuous there except at a finite number of points, then f is integrable on [a, b]. Example
Definite Integral IX This function is not integrable on [0, 1]. { 1 for x > 0 f (x) = x 2 13 for x = 0 Definition (Limits upside down) For a < b, define a b b f (x) dx := f (x) dx. a This definition makes the following theorem work for numbers a, b, c in any order.
Definite Integral X Theorem (B - Interval Additivity Property) If f is integrable on an interval and a, b, c are points in that interval, then b c b f (x) dx = f (x) dx + f (x) dx. a a c
FFTC I When we work out integrals 1 4 t + 1 dt, 3 4 t + 1 dt, 10 4 t + 1 dt,... we see that we should do this evaluation with a variable upper boundary. Let F (x) := x 4 t + 1 dt and we find after doing our usual Riemann Sum and the limit n F (x) = 1 2 ((x + 1)2 9)
FFTC II The First Fundamental Theorem of Calculus does one better - it allows to calculate F (x) defined similarly for many integrands f (x), not just x + 1. Theorem (A - First Fundamental Theorem of Calculus) Let f (x) be continuous on the closed interval [a, b]. Then d dx x a f (t) dt = f (x). In other words... In the situation of Theorem A, consider the function defined by F (x) := x a f (t) dt
FFTC III Then F (x) is an antiderivative of f (x)!... and this is the reason behind writing f (x) dx for antiderivatives. We verify the theorem with the example f (x) = x + 1. It gives F (x) = 1 2 (x + 1)2 + C, and this agrees with the result we found above up to the constant C. What is the correct value of C? Here is a trick! F (a) = a a f (x) dx = 0 whatever f (x) may be.
FFTC IV In the example above, a = 4, and from F ( 4) = 0 we deduce 0 = 1 2 ( 4 + 1)2 + C so C = 9/2, which agrees with the result found using Riemann sums. This is much easier, faster... plain wonderful Theorem (B - Comparison Property) If f, g are integrable functions on [a, b] such that f (x) g(x) for all x in [a, b], then b a f (x) dx b a g(x) dx
FFTC V The Property Comparison Theorem (C - Boundedness Property) If f is integrable on [a, b] and m f (x) M for all x in [a, b], then m(b a) b a f (x) dx M(b a)
FFTC VI For the first inequality, take the constant function g(x) = m on [a, b] and apply Theorem B. m(b a) = b a m dx b a f (x) dx. With the constant function h(x) = M, Theorem B gives the second inequality in Theorem C. The textbook has a nice picture about the geometrical meaning of this fact. Theorem (D - Linearity of the Definite Integral) Suppose f (x), g(x) are integrable functions on [a, b] and that k is a constant. Then kf (x) and f (x) + g(x) are integrable and b a kf (x) dx = k b a f (x) dx, b b b
FFTC VII - use linearity of the summation operator and of limits.
Recall the definition So we have to calculate Proof of FFTC I F (x) := x a f (t) dt. d 1 1 F (x) = lim (F (x + h) F (x)) = lim dx h 0 h h 0 h x+h x f (t) dt. Let m, M be the minimum resp. maximum values of f (t) on the interval [x, x + h]. Note that m, M depend on h. From Theorem C, we get with b a = (x + h) x = h m 1 h x+h x f (t) dt M.
Proof of FFTC II Since f (t) is continuous, both m and M must tend to f (x) as h tends to zero. By the Squeeze Theorem for limits, 1 lim h 0 h which concludes the proof. x+h x f (t) dt = f (x)
Example Find Examples for FFTC I d dx 4 x tan 2 u cos u du. Solution Flip the limits of the integral, changing the sign. Then apply the FFTC. Example Find d x 2 3t 1 dt. dx 1
Examples for FFTC II Solution Attention - the upper limit here is x 2, not x! Define then we are looking for F (u) := u 1 3t 1 dt, d dx F (x 2 ) = d du F (u) d dx u where u = x 2. We get (3u 1) 2x = 2x(3x 2 1).
SFTC I The Second Fundamental Theorem of Calculus is a rewording of the First geared towards evaluating definite integrals. Theorem (Second Fundamental Theorem of Calculus) Let f (x) be integrable on [a, b], and let F (t) be any antiderivative of f (x). Then b a f (x) dx = F (b) F (a). Consider both sides of the equation as functions of b. Differentiate both sides to get f (b). So both sides are antiderivatives of f (x), and they differ only by a constant, b a f (x) dx = F (b) F (a) + C.
SFTC II Let b = a and you get 0 = C. Example Find b a x r dx. Solution An antiderivative of x r is 1 r+1 x r+1, so b a x r dx = 1 r + 1 (br+1 a r+1 ). Definition (Bracket) There is a shorthand for F (b) F (a) which is useful when F (t) is complicated - define [F (t)] b a := F (b) F (a).
Example Find π 0 sin(x) dx. SFTC III Solution An antiderivative of sin(x) is cos(x). So π 0 sin(x) dx = [ cos(t)] π 0 = cos(π) ( cos(0)) = 1 + 1 = 2. Check for sign errors - since sin(x) 0, the integral cannot be negative! Example Determine 2 2 3x 2 x + 4 dx.
SFTC IV Solution First find an antiderivative. 3x 2 x + 4 dx = x 3 1 2 x 2 + 4x + C. Then evaluate the antiderivative between 2 and 2. 2 [ 3x 2 x + 4 dx = x 3 1 ] 2 2 2 x 2 + 4x 2 = (2 3 + 4 2) ( ( 2) 3 + 4( 2) ) = 32. Note that the even power of x in the antiderivative cancels out. It also does not matter what C we choose, so we chose C = 0. Revision
SFTC V To evaluate a definite integral b a f (x) dx first find an antiderivative F (x) of f (x). then use the SFTC, ie take [F (x)] b a. If the integration is easy, you can do both steps in one go. For hard integrals, it sometimes helps to use
The Substitution Method I This is again the Chain Rule in reverse gear. Suppose you can rewrite the integrand f (x) as f (x) = g(u(x))u (x) and G(u) is an antiderivative of g(u) then f (x) dx = g(u(x))u (x) dx = g(u) du + C = G(u(x)) + C. In case g(u) = u r we already know this as the Generalized Power Rule! A good way to memorize the Substitution Method is u (x) dx = du
The Substitution Method II which dovetails nicely with the Leibniz notation u (x) = du dx. The problem is to find g(u) and u(x). Look out for layered parentheses to find g(u(x))! Look out for factors which could be u (x)! Example (1)
The Substitution Method III Find sin( x) x dx Solution Use u(x) = x, expand fraction by factor 2/2. Theorem (B - Substitution Method for Definite Integrals) For the appropriate limits, we memorize x from a to b u from u(a) to u(b) so for definite integrals the substitution method says b a g(u(x))u (x) dx = u(b) u(a) g(u) du
Example Find 9 The Substitution Method IV 4 sin( x) x Solution Use u = x, evaluate the antiderivative 2 sin(u) found in Example 1 between u = 2 and u = 3. Change the limits according to x 4 9 u(x) 2 3 so 9 4 sin( x) x dx = 2 3 2 dx. sin u du = 2[ cos u] 3 2 = 2(cos 3 cos 2).
The Substitution Method V You could also evaluate 2 cos( x) between x = 4 and 9 for the same result. Example (2) Find π/2 0 x sin 3 (x 2 ) cos(x 2 ) dx Solution Use u(x) = sin(x 2 ), introduce factor 2/2. Example (3)
The Substitution Method VI Find 1 0 x + 1 (x 2 + 2x + 6) 2 dx Solution Use u(x) = x 2 + 2x + 6, introduce factor 2/2. You should practice with the worked examples until one-line hints like the ones above are all you need to reproduce the whole computation. Example (My Own) Find π 0 sin(sin x) cos x dx.
The Substitution Method VII Solution Use u(x) = sin x, g(u) = sin u. Change the limits according to x 0 π u(x) 0 0 so the new limits are sin 0 = 0 and sin π = 0, the integral is zero!
MVT for integrals I Theorem (A - Mean Value Thm for Integrals) If f is continuous on [a, b], then there is a number c in (a, b) such that Define for t in [a, b] b a f (x) dx = f (c)(b a). F (t) := t a f (x) dx and apply the Mean Value Theorem to F (t). This says that there exists c in (a, b) such that F (b) F (a) = F (c)(b a)
MVT for integrals II and using the SFTC, then F (c) = f (c) we get the stated equality. Definition (Average) The number f (c) in Theorem A is called the average value of f (x) in [a, b]. Example If f (t) is the temperature at time t during a certain day, then there is at least one moment when the temperature equals the average temperature.
Symmetry, Periodicity I Theorem (B - Symmetry) If f is an even function, then a a If f is an odd function, then a Example f (x) dx = 2 a a 0 f (x) dx = 0. f (x) dx.
Symmetry, Periodicity II Evaluate 2 x sin 4 (x) + x 3 x 4 dx. 2 Solution The first two summands are odd functions, so the integral over these is zero. The integral equals 2 2 x sin 4 (x) + x 3 x 4 dx = 0 + 0 2 = 2 5 Example Find the area of a semicircle of radius 3. 2 0 x 4 dx [ x 5 ] 2 0 = 64 5.
Symmetry, Periodicity III Solution First note that a semicircle of radius 3 is the area between the graph of 9 x 2 and the x-axis. So we may write A = 3 3 9 x 2 dx. First, use the symmetry ( 9 x 2 is even), so we get twice the area of a quarter circle. The we need substitution. Substitutions like u(x) = 9 x 2 do not help here, but we can apply substitution in reverse! Use x = 3 sin u (1)
Symmetry, Periodicity IV This gives dx = 3 cos u du and new limits u = 0, u = π/2. A = 2 = 2 = 18 3 0 π/2 9 x 2 dx 0 π/2 0 9 9 sin 2 u3 cos u du cos 2 u du With the trig formula cos 2 u = (1 + cos 2u)/2 we can integrate this as A = 9 π/2 0 1 + cos 2u du = 9 [u + sin(2u)/2] π/2 0 = 9 2 π.
Symmetry, Periodicity V I know that you know this area is 9π/2! but we have rediscovered it using an integral. This is a nice way of checking that something you learned in high school is actually true - or at least consistent with the rest of Calculus. Theorem (C - Periodicity) If f is periodic with period p, then b+p a+p f (x) dx = b a f (x) dx. Example
Symmetry, Periodicity VI Evaluate 2π 0 sin x dx. Solution Split this integral by integrating from 0 to π, then from π to 2π. The periodicity theorem gives then that both integrals are equal, because sin(x + π) = sin(x).
Conclusion I 1. Suppose an object moves along a line with position x(t) at time t. Then the velocity at time t is v(t) = x (t). Conversely, if we know the velocity v(t) and the position at a given time t 0, then x(t) = t t 0 v(s) ds + x(t 0 ). So initial position x(0) and the velocity function v(t) determine position at any time t. 2. The technology projects are excellent! Even if we cannot do this as a homework, glance through it and see where Riemann sums really become exciting - namely as soon as you cannot find antiderivatives.
Conclusion II 3. In Calculus II, integrals will be used to work out the area of many different regions in the plane resp. volumes in space.