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# MIT 18 01 - Deﬁnite Integrals

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lec18lec19lec20lec22lec23lec24lec25MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 18 18.01 Fall 2006 Lecture 18: Deﬁnite Integrals Integrals are used to calculate cumulative totals, averages, areas. Area under a curve: (See Figure 1.) 1. Divide region into rectangles 2. Add up area of rectangles 3. Take limit as rectangles become thin a ba b(i)(ii)Figure 1: (i) Area under a curve; (ii) sum of areas under rectangles Example 1. f(x) = x2 , a = 0, b arbitrary 1. Divide into n intervals Length b/n = base of rectangle 2. Heights: � �2b b1st: x = , height = • n n � �22b 2b2nd: x = , height = • n n Sum of areas of rectangles: � �� �2 � �� �2 � �� �2 � �� �2b b b 2b b 3b b nb b3 + + + + = (12 + 22 + 32 + + n 2) n n n n n n · · · n n n3 · · · 1Lecture 18 18.01 Fall 2006 a=0 b2Figure 2: Area under f (x) = x above [0, b]. We will now estimate the sum using some 3-dimensional geometry. Consider the staircase pyramid as pictured in Figure 3. n = 4nFigure 3: Staircase pyramid: left(top view) and right (side view) 1st level: n × n bottom, represents volume n2 . 2nd level: (n − 1) × (n − 1), represents volumne (n − 1)2), etc. Hence, the total volume of the staircase pyramid is n2 + (n − 1)2 + + 1. · · · Next, the volume of the pyramid is greater than the volume of the inner prism: 1 1 112 + 22 + + n 2 > (base)(height) = n 2 n = n 3 · · · 3 3 · 3 and less than the volume of the outer prism: 1 112 + 22 + + n 2 < (n + 1)2(n + 1) = (n + 1)3 · · · 3 3 2� � � � Lecture 18 18.01 Fall 2006 In all, 1 1 n3 12 + 22 + + n2 1 (n + 1)3 = 3 < · · · <3 n3 n3 3 n3 Therefore, b3 1lim (12 + 22 + 32 + + n 2) = b3 , n→∞ n3 · · · 3 b3 and the area under x2 from 0 to b is .3 Example 2. f(x) = x; area under x above [0, b]. Reasoning similar to Example 1, but easier, gives a sum of areas: b2 1 n2 (1 + 2 + 3 + · · · + n) → 2 b2 (as n → ∞) This is the area of the triangle in Figure 4. bbFigure 4: Area under f (x) = x above [0, b]. Pattern: d b3 = b2 db 3 d b2 = b db 2 The area A(b) under f(x) should satisfy A�(b) = f(b). 3� � Lecture 18 18.01 Fall 2006 General Picture a bciy=f(x)Figure 5: One rectangle from a Riemann Sum Divide into n equal pieces of length = Δx = b − a • n • Pick any ci in the interval; use f(ci) as the height of the rectangle • Sum of areas: f (c1)Δx + f (c2)Δx + · · · + f(cn)Δx nIn summation notation: f(ci)Δx called a Riemann sum.← i=1 Deﬁnition: n� b lim f(ci)Δx = f(x)dx called a deﬁnite integral a n→∞ i=1 ← This deﬁnite integral represents the area under the curve y = f(x) above [a, b]. Example 3. (Integrals applied to quantity besides area.) Student borrows from parents. P = principal in dollars, t = time in years, r = interest rate (e.g., 6 % is r = 0.06/year). After time t, you owe P (1 + rt) = P + P rt The integral can be used to represent the total amount borrowed as follows. Consider a function f(t), the “borrowing function” in dollars per year. For instance, if you borrow \$ 1000 /month, then f(t) = 12, 000/year. Allow f to vary over time. Say Δt = 1/12 year = 1 month. ti = i/12 i = 1, · · · , 12. 4� Lecture 18 18.01 Fall 2006 f(ti) is the borrowing rate during the ith month so the amount borrowed is f(ti)Δt. The total is 12f(ti)Δt. i=1 In the limit as Δt 0, we have → � 1 f(t)dt 0 which represents the total borrowed in one year in dollars per year. The integral can also be used to represent the total amount owed. The amount owed depends on the interest rate. You owe f(ti)(1 + r(1 − ti))Δt for the amount borrowed at time ti. The total owed for borrowing at the end of the year is � 1 f(t)(1 + r(1 − t))dt 0 5��� ��� ���� ��� ���� Lecture 19 18.01 Fall 2006 Lecture 19: First Fundamental Theorem of Calculus Fundamental Theorem of Calculus (FTC 1) If f(x) is continuous and F �(x) = f(x), then � b f(x)dx = F (b) − F (a) a Notation: F (x) b = F (x) x=b = F (b) − F (a) a x=a � b b3 3 b3 a3x x2; x 2dx =Example 1. F (x) = F �(x) = x = 3 − 33 , 3a a Example 2. Area under one hump of sin x (See Figure 1.) � π 0 sin x dx = − cos x π = − cos π − (− cos 0) = −(−1) − (−1) = 2 0 1�Figure 1: Graph of f (x) = sin x for 0 ≤ x ≤ π. � 1 16 = 1 1 6 − 0 = 6 5dx = xExample 3. x 60 0 1��� Lecture 19 18.01 Fall 2006 Intuitive Interpretation of FTC: dx x(t) is a position; v(t) = x�(t) = is the speed or rate of change of x. dt � b v(t)dt = x(b) − x(a) (FTC 1) a R.H.S. is how far x(t) went from time t = a to time t = b (diﬀerence between two odometer readings). L.H.S. represents speedometer readings. n i=1 x(b) − x(a) = v(t) cancel each other, whereas an � ( )Δ approximates the sum of distances traveled over times Δt t tvithThe approximation above is accurate if ( ) is close to ( ) on the interval. The interpretation it tv viof ( ) as an odometer reading is no longer valid if changes sign. Imagine a round trip so that tx v 0. Then the positive and negative velocities odometer would measure the total distance not the net distance traveled. Example 4. � 2π 0 sin x dx = − cos x 2π = − cos 2π − (− cos 0) = 0. 0 The integral represents the sum of areas under the curve, above the x-axis minus the areas below the x-axis. (See Figure 2.) +-12�Figure 2: Graph of f (x) = sin x for 0 ≤ x ≤ 2π. 2Lecture 19 18.01 Fall 2006 Integrals have an important additive property (See Figure 3.) � b � c � …

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