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Unformatted text preview: Math 1a The Definite Integral Fall, 2009 We define the definite integral of y = f ( x ) from x = a to x = b as Z b a f ( x ) dx = lim n →∞ n X i =1 f ( x * i )Δ x where x i 1 ≤ x * i ≤ x i . Note that x * i could be x i (in which case we have the limit of R n ) or x i 1 (in which case we have the limit of L n ). One key point is that it doesn’t matter which x * i we pick! Another is that, if f ( x ) ≥ 0, then the definite integral is an area. Compute the following definite integrals, using whatever x * i you like: 1 Z b x 2 dx 2 Z 1 x 3 dx Hint: n X i =1 i = n ( n + 1) 2 n X i =1 i 2 = n ( n + 1)(2 n + 1) 6 n X i =1 i 3 = n ( n + 1) 2 2 It can be simpler to compute a definite integral by recognizing that it represents an area. Use this to find the following definite integrals: 3 Z 10 10  x  dx 4 Z r r √ r 2 x 2 dx 5 Z 10 x dx 6 Z 10 10 x dx It is also useful to know the properties of the definite integral, as it can help to reduce integrals to known quantities. Here are eight properties of the definite integral: 1 Z b a c dx = c ( b a ) 2 Z b a ( f ( x ) + g ( x )) dx = Z b a f ( x ) dx + Z b a g ( x ) dx 3 Z b a cf ( x ) dx = c Z b a f ( x ) dx 4 Z b a ( f ( x ) g ( x )) dx = Z b a f ( x ) dx Z b a g ( x ) dx 5 Z c a f ( x ) dx + Z b c f ( x ) dx = Z b a f ( x ) dx 6 If f ( x ) ≥ 0 on [ a,b ], then Z a a f ( x ) dx ≥ 7 If f ( x ) ≥ g ( x ) on [ a,b ], then Z b a f ( x ) dx ≥ Z b a g ( x ) dx 8 If m ≤ f ( x ) ≤ M on [ a,b ], then m ( b a ) ≤ Z b a f ( x ) dx ≤ M ( b a ) Use these properties and the results from the first page to compute the following definite integrals: 7 Z 2 ( 4 x 2 ) dx 8 Z 2 2 x + √ 4 x 2 dx 9 Z 5 ( 3 x 3 7 x ) dx Now suppose that Z 5 f ( x ) dx = 3 , Z 10 5 f ( x ) dx = 2 , Z 10 g ( x ) dx = 6 , and Z 5 g ( x ) dx = 1 . Use the properties above to find the following integrals: 10 Z 5 10 f ( x ) dx 11 Z 10 f ( x ) dx 12 Z 10 (3 f ( x ) 2 g ( x )) dx 13 Z 5 ( g ( x ) 3 f ( x )) dx The Definite Integral – Answers / Solutions 1 We find that Z b x 2 dx = 1 3 b 3 . We’ll use the right endpoint and compute R n , so x * i = x i = a + i Δ x . In this case, Δ x = b n and a = 0, so x i = i b n . Our function is f ( x ) = x 2 , so the right endpoint sum is R n = n X i =1 f ( x * i )Δ x = n X i =1 i b n 2 b n = b 3 n 3 n X i =1 i 2 . Using the hint, we know that n ∑ i =1 i 2 = n ( n +1)(2 n +1) 6 , so R n = b 3 n 3 · n ( n +1)(2 n +1) 6 = b 3 6 · n ( n +1)(2 n +1) n 3 . Thus Z b x 2 dx = lim n →∞ R n = lim n →∞ b 3 6 · n ( n + 1)(2 n + 1) n 3 = b 3 6 · 2 1 = 1 3 b 3 , as claimed. 2 We find that Z 1 x 3 dx = 1 4 . This is very similar to Problem 1, with b = 1 and f ( x ) = x 3 . We get R n = n X i =1 i 1 n 3 1 n = 1 n 3 n X i =1 i 3 ....
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This note was uploaded on 03/27/2012 for the course MATH 1a taught by Professor Benedictgross during the Fall '09 term at Harvard.
 Fall '09
 BenedictGross
 Integrals

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