Area and Integration

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1 Area and Integration Professor Richard Blecksmith Dept. of Mathematical Sciences Northern Illinois University richard/math229. Two Fundamental Problems of Calculus First Problem: Study how one variable changes with respect to another How does area change with respect to radius How does temperature change with respect to altitude How does y change with respect to x Second Problem: Find the area under a curve We use derivatives to solve Problem We use integrals to solve Problem 2 Given a curve y = f(x) 2. The Area Problem For now we will assume that f(x) is 0 How do we find the area under the curve and above the x-axis, between two vertical sides x = a and x = b? y = f(x) x = a Area x = b

2 2 3. A case we can do If the curve is linear, then we can use the formula for the area of a triangle A = 2 base height Example. y = f(x) = 2x between a = 0 and b = 5 base = 5 height = 0 A = = Graph of y = 2x (5, 0) y = 2x Area = 25 height=0 base = 5

3 3 5. A case we can t do (3,9) (2,4) y = f(x) = x 2 Area (,) a = 0 b = 3 6. Rectangular Approximation y = x 2 3 rectangles

4 4 7. Rectangular Approximation y = x 2 6 rectangles 8. Rectangular Approximation y = x 2 2 rectangles 9. Sub-dividing [a, b] Set-up: divide the interval [a,b] into n equally spaced segments, whose division points are x 0,x,...,x n.

5 5 x a=x 0 x x 2 x 3 x 4 x n x n =b The width of each segment is x = b a n The division points x k are x 0 = a, x = a+ x, x 2 = a+2 x, x 3 = a+3 x and, in general, x k = a+k x When a = 0, x 0. Special case: a = 0 0=x 0 x x 2 x 3 x 4 x n x n =b these forumlas become x = b n and x k = k x Note that x n = n x = n b n = b, as it should.. Riemann Method In the Riemann method, we use n rectangles to approximate the area under the curve y = f(x) Each rectangle has as its base the subinterval [x k,x k ] of width x The height of each rectangle is f(x k ) Since the area of a rectangle is base height, the k-th rectangle has area

6 6 x f(x k ) Our approximation, called a Riemann Sum is the sum of the areas of these n rectangles 2. Riemann Method Picture x 0=x 0 x x 2 x 3 x n x n=b 3. Using 3 subdivisions For the function f(x) = x 2, and interval [a,b] = [0,3] suppose we use n = 3 rectangles We divide [0,3] into 3 equally spaced subintervals, each of width x = 3/3 = with division points x k = k x = k 0=x 0 x = x 2 =2 x 3 = 3 Notice that the height of the kth rectangle with base [x k,x k ] is f(x k ) = x 2 k

7 7 k 2 3 x k 2 3 f(x k ) Rectangles Each rectangle has width x = and height f(x k ) = x 2 k The area of the 3 red rectangles is A(3) = x f(x )+ x f(x 2 )+ x f(x 3 ) = x x 2 + x x x x 2 3 = = +4+9 = 4 5. Rectangular Approximation 9 y = x rectangles

8 Rectangles Each rectangle has width x =, subdivision point x 2 k = k x = k, and 2 height f(x k ) = x 2 k The area of the 6 bluish green rectangles is A(6) = x f(x )+ x f(x 2 )+ x f(x 3 )+ x f(x 4 )+ x f(x 5 )+ x f(x 6 ) = x x 2 + x x x x x x x x x x 2 6 ( ) 2 ( 3 ) 2 ( 5 ) 2 = x + x () 2 + x + x (2) 2 + x + x (3) = = = = Using 6 Rectangles y = x 2 6 rectangles 8. Notation Time Out It is getting very tedious writing these sums

9 9 The sum with 2 rectangles will be even worse! To ease the writing, mathematicians invented the summation symbol It works like this: underneath we write the starting index above we write the ending index to the right of we write the form of the things we are adding 3 a k is the sum a +a 2 +a 3 k= 9. So What? It doesn t seem so bad to write these out! 50 a k is the sum k= a +a 2 +a 3 +a 4 +a 5 +a 6 +a 7 +a 8 +a 9 +a 0 +a +a 2 +a 3 +a 4 +a 5 + a 6 +a 7 +a 8 +a 9 +a 20 +a 2 +a 22 +a 23 +a 24 +a 25 +a 26 +a 27 +a 28 +a 29 +a 30 +a 3 +a 32 +a 33 +a 34 +a 35 +a 36 +a 37 +a 38 +a 39 +a 40 +a 4 +a 42 +a 43 + a 44 +a 45 +a 46 +a 47 +a 48 +a 49 +a 50 Nobody wants to write all 50 terms. Though, in fairness we could have written this last sum as a +a 2 +a 3 + +a 49 +a Summation Examples The sum of the first 0 integers can be written S = 0 k Evaluate this sum

10 0 S = 0 k = = Summation Examples What is S = 0 2k This is the sum of the first 0 even integers: S = = 0 55? Can you think of an easier way, using our previous calculation S = 0 k = Factor out the 2 from each term S = 0 2k = 2 0 k = 2 55 = Rectangular Approx of Area Using summation notation, we can write the sum of the areas of the 6 rectangles used to approximate the area under the curve A(6) = x f(x )+ x f(x 2 )+ x f(x 3 )+ x f(x 4 )+ x f(x 5 )+ x f(x 6 ) as A(6) = 6 xf(x k ) To go from 6 to 2 rectangles is easy: A(2) = 2 xf(x k ) = x 2 f(x k ) by factoring out x.

11 23. Example y = x 2 Revisited y = f(x) = x 2 n = number of rectangles [a,b] = [0,3] x = 3 n x k = x k A(n) = x f(x k ) 24. Computing the exact area The following table gives the values of A(n) for various values of n: n A(n) It looks like the exact area under the curve y = x 2 between 0 and 3 is Calculating Areas The exact area A under the curve y = f(x) from a to b is defined to be the limit of the Riemann Sums A(n) as the number of rectangles n goes to infinity. That is, A = lim n A(n) = lim n x f(x k ) k=

12 2 For many simple functions, such as f(x) = x n, the limit that gives the area under the curve has been worked out. 26. Example y = x 2 For example, suppose f(x) = x 2 and we wish to compute the area from a = 0 to b. We begin by computing the area under the n rectangles, using right end points. Here x = b n and x k = k x A(n) = x f(x k ) k= = x f(x k ) [factor out x] k= A(n) = x = x = x f(x k ) = x k= (k x) 2 k= k 2 x 2 k= 27. Example Part 2 x 2 k [f(x) = x 2 ] k= [because x k = k x] [algebra] = ( x) 3 k 2 [factor out ( x) 2 ] = b3 n 3 k= k 2 [because x = b] n k=

13 3 28. Example Part 3 At this point we need a formula for n Page 309, formula 6 : k= k= k2. k 2 = n(n+)(2n+) 6 Verify formula 6 for n =, 2, 3, 4, 5. Thus, A(n) = b3 n 3 k= k 2 = b3 n(n+)(2n+) n 3 6 ( ) ( ) = b3 n+ 2n+. 6 n n [note n cancels.] 29. Example Conclusion The exact area A under the curve is the limit of the rectangular appoximations A(n) as n : A = lim n A(n) b 3 = lim n 6 b 3 = lim n 6 = b3 6 ( n+ )( 2n+ ) n n ( + )( 2+ ) n n (+0)(2+0) = b3 3 When b = 3, A = 33 3 = 9, a fact that we laboriously worked out earlier.

14 4 30. Variables versus Constants In mathematics, variables represent quantities which are allowed to vary, while constants are quantities that never change. We usually use letter at the beginning of the alphabet, such as a, b, c to represent constants, and variables close to x (the unknown) to represent variables. In computing the area from a to b, if we want to allow the right endpoint b to vary, we should probably rename it to a variable letter, such as x. This creates a problem, however, since we usually use the letter x when we write the function as f(x). To avoid confusion, we rewrite the function as f(t). That is, simultaneously, search and replace b x and x t. 3. Changing Letters For example, suppose f(t) = t 2, the squaring function, and we wish to calculated the area from a = 0 to b = x. Note that we are using the letter x for the right endpoint. To avoid confusion, we are using the letter t to describe the function. Using x instead of b for the right endpoint, we have x 3 ( A = lim A(n) = lim + )( 2+ ) = x3 x3 (+0)(2+0) = n n 6 n n Area under y = x n Suppose f(t) = t n for any non-negative integer. We want to find the area A under the curve from a = 0 to b = x. Case: n = 0

15 5 Our function is just f(t) = t 0 =. The region is one long rectangle, whose dimensions are x by. The area of this rectangle is A = x. 33. Area under y = x Case: n = Our function is just f(t) = t = t. (5, 0) The region is an isosceles triangle whose dimensions are x by x. The area of this rectangle is A = 2 x2. y = x A = 2 x2 base = x height = x 34. Area under y = x 2 Case: n = 2 Our function is just f(t) = t 2. We have already worked this out. The area of this region is A = 3 x3.

16 6 a = 0 A = 3 x3 y = f(x) = x 2 b = x 35. Area under y = x n, n = 0,...,6 For every possible value of the n, the area under the curve y = f(x) = x n from 0 to x has been calculated. These areas are given in the following chart: f(x) A x 0 = x x x 2 x 3 x 4 x 5 x 6 Differentiate the second column and compare with the first column. 2 x2 3 x3 4 x4 5 x5 6 x6 7 x7 36. Big Surprise! The second column is the anti-derivative of the first column. Equivalently, the derivative of the area A(x) is just the function f(x). This connection between area and derivatives is so, well fundamental, that it is given the following name: The Fundamental Theorem of Calculus

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