Additional Problems 1 Let f a b R be continuous and suppose that

# Additional problems 1 let f a b r be continuous and

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Unformatted text preview: Additional Problems 1. Let f : [ a, b ] → R be continuous and suppose that integraldisplay b a f ( x ) dx = 0. Which of the following statements are true, which are false. Give reasons to support your answer. (a) f ( x ) = 0 for all x ∈ [ a, b ]. False: integraldisplay π cos xdx = 0, and cos x is not 0 for all x ∈ [0 , π ]. (b) f ( x ) = 0 for some x ∈ [ a, b ]. True: If f ( x ) negationslash = 0 for all x ∈ [ a, b ], then either f ( x ) > 0 or f ( x ) < 0 for all x . In the first case, integraldisplay b a f ( x ) dx > 0, in the second case integraldisplay b a f ( x ) dx < 0. 4 (c) For any partition P , U ( f, P ) ≥ 0. True: For any partition P , U ( f, P ) ≥ integraldisplay b a f ( x ) dx = 0 (d) There does not exist a partition P , such that L ( f, P ) > 0. True: For any partition P , L ( f, P ) ≤ integraldisplay b a f ( x ) dx = 0 2. Let f, g : [ a, b ] → R be continuous and suppose that integraldisplay b a f ( x ) dx = integraldisplay b a g ( x ) dx . Prove that there exists a point c ∈ [ a, b ] such that f ( c ) = g ( c ). Set h ( x ) = f ( x )- g ( x ). Then h is continuous and integraldisplay b a h ( x ) dx = integraldisplay b a ( f ( x )- g ( x ) dx = integraldisplay b a f ( x ) dx- integraldisplay b a g ( x ) dx = 0 Therefore, there must be a number c ∈ [ a, b ] such that h ( c ) = 0 as shown in 6(b) above. 3. The function f ( x ) = 4 + integraldisplay 3 x radicalbig 7 + 2 t 2 dt has an inverse. ( f- 1 ) prime (4) = ? ( f- 1 ) prime (4) = 1 f prime ( a ) where f ( a ) = 4. f (3) = 4 and f prime ( x ) = √ 7 + 2 x 2 . Therefore, ( f- 1 ) prime (4) = 1 f prime (3) = 1 5 . 4. Given that integraldisplay 2 f ( x ) dx = 8 3 , integraldisplay 2 1 f ( x ) dx = 4 3 , and integraldisplay 3 f ( x ) dx = 11 3 , find integraldisplay 1 3 f ( x ) dx . integraldisplay 3 1 f ( x ) dx = integraldisplay 3 f ( x ) dx- bracketleftbiggintegraldisplay 2 f ( x ) dx- integraldisplay 2 1 f ( x ) dx bracketrightbigg = 11 3- bracketleftbigg 8 3- 4 3 bracketrightbigg = 7 3 . Therefore, integraldisplay 1 3 f ( x ) dx =- 7 3 5. If f is a continuous function and F ( x ) = integraldisplay x bracketleftbigg (2 t + 3) integraldisplay 2 t f ( u ) du bracketrightbigg dt , then F primeprime (2) = F prime ( x ) = (2 x + 3) integraldisplay 2 x f ( u ) du ; F primeprime ( x ) = (2 x + 3)[- f ( x )] + 2 integraldisplay 2 x f ( u ) du ; F primeprime (2) =- 7 f (2) 5...
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• Fall '08
• Staff
• Continuous function, dt, dx
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