unit3_who_sept24

# unit3_who_sept24 - MIT OpenCourseWare http://ocw.mit.edu...

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Unformatted text preview: MIT 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 b a b (i) (ii) Figure 1: (i) Area under a curve; (ii) sum of areas under rectangles Example 1. f ( x ) = x 2 , a = 0, b arbitrary 1. Divide into n intervals Length b/n = base of rectangle 2. Heights: 2 b b 1 st : x = , height = n n 2 2 b 2 b 2 nd : x = , height = n n Sum of areas of rectangles: 2 2 2 2 b b b 2 b b 3 b b nb b 3 + + + + = (1 2 + 2 2 + 3 2 + + n 2 ) n n n n n n n n n 3 1 Lecture 18 18.01 Fall 2006 a=0 b 2 Figure 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 = 4 n Figure 3: Staircase pyramid: left(top view) and right (side view) 1 st level: n n bottom, represents volume n 2 . 2 nd level: ( n 1) ( n 1), represents volumne ( n 1) 2 ), etc. Hence, the total volume of the staircase pyramid is n 2 + ( n 1) 2 + + 1 . Next, the volume of the pyramid is greater than the volume of the inner prism: 1 1 1 1 2 + 2 2 + + n 2 &gt; (base)(height) = n 2 n = n 3 3 3 3 and less than the volume of the outer prism: 1 1 1 2 + 2 2 + + n 2 &lt; ( n + 1) 2 ( n + 1) = ( n + 1) 3 3 3 2 Lecture 18 18.01 Fall 2006 In all, 1 1 n 3 1 2 + 2 2 + + n 2 1 ( n + 1) 3 = 3 &lt; &lt; 3 n 3 n 3 3 n 3 Therefore, b 3 1 lim (1 2 + 2 2 + 3 2 + + n 2 ) = b 3 , n n 3 3 b 3 and the area under x 2 from 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: b 2 1 n 2 (1 + 2 + 3 + + n ) 2 b 2 (as n ) This is the area of the triangle in Figure 4. b b Figure 4: Area under f ( x ) = x above [0 ,b ]. Pattern: d b 3 = b 2 db 3 d b 2 = 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 b c i y=f(x) Figure 5: One rectangle from a Riemann Sum Divide into n equal pieces of length = x = b a n Pick any c i in the interval; use f ( c i ) as the height of the rectangle Sum of areas: f ( c 1 ) x + f ( c 2 ) x + + f ( c n ) x n In summation notation: f ( c i ) x called a Riemann sum . i =1 Definition: n b lim f ( c i ) x = 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 ]....
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## unit3_who_sept24 - MIT OpenCourseWare http://ocw.mit.edu...

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