This proof can be modifed as in Problem 5 to take care of the cases f a 0 or f

# This proof can be modifed as in problem 5 to take

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Unformatted text preview: This proof can be modifed as in Problem 5 to take care of the cases f ( a ) > 0 or f ( b ) > 0. 12. Since f is continuous on [ a, b ], f takes on a minimum value m at some point c 1 ∈ [ a, b ], and a maximum value M at some point c 2 ∈ [ a, b ]. We may assume that m < M (and c 1 negationslash = c 2 ) for if m = M , then f ( x ) ≡ α is constant and 1 b- a integraldisplay b a f = 1 b- a integraldisplay b a αdx = 1 b- a α ( b- a ) = α = f ( c ) for all c ∈ [ a, b ]. With m < M we have, By Problem 30.11, m ( b- a ) ≤ integraldisplay b a f ≤ M ( b- a ) so m ≤ 1 b- a integraldisplay b a f ≤ M Since f ( c 1 ) = m ≤ 1 b- a integraltext b a f ≤ M = f ( c 2 ), there is a number c between c 1 and c 2 such that f ( c ) = 1 b- a integraltext b a f by the Intermediate-Value Theorem. 13. Suppose f ( x ) negationslash = g ( x ) for all x ∈ [ a, b ]. Then f ( x )- g ( x ) negationslash = 0 for all x . Since f and g are continuous, f- g is continuous and so either f ( x )- g ( x ) > for all x , or f ( x )- g ( x ) < 0 for all x . In either case, integraldisplay b a [ f ( x )- g ( x )] dx = integraldisplay b a f ( x ) dx- integraldisplay b a g ( x ) dx negationslash = 0, which implies integraldisplay b a f ( x ) dx negationslash = integraldisplay b a g ( x ) dx , a contradiction. Section 31 4. For each x ∈ [ a, b ], integraldisplay x a f + integraldisplay b x f = integraldisplay b a f , which implies that integraldisplay b x f = integraldisplay b a f- integraldisplay x a f 2 Since integraldisplay b a f is differentiable (the derivative is 0 since integraldisplay b a f is a constant) and integraldisplay x a f is differentiable, integraldisplay b x f is differentiable. d dx parenleftbiggintegraldisplay b x f parenrightbigg = d dx parenleftbiggintegraldisplay b a f parenrightbigg- d dx parenleftbiggintegraldisplay x a f parenrightbigg = 0- f ( x ) . =- f ( x ) Alternatively, integraldisplay b x f ( t ) dt =- integraldisplay x b f ( t ) dt so d dx parenleftbiggintegraldisplay b x f parenrightbigg =- d dx parenleftbiggintegraldisplay...
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• Fall '08
• Staff
• Continuous function, dt, dx