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(A) Define a function $g(x) = (1-x) \int_0^x f(t) dt$. Since $f$ is derivable on $[0, 1]$,it is continuous on $[0, 1]$,and thus the integral $\int_0^x

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there exists $c \in (0, 1)$ such that $\int_0^c f(x) dx = (1-c) f(c)$

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(A) Define a function $g(x) = (1-x) \int_0^x f(t) dt$. Since $f$ is derivable on $[0, 1]$,it is continuous on $[0, 1]$,and thus the integral $\int_0^x f(t) dt$ is differentiable. We observe that $g(0) = (1-0) \int_0^0 f(t) dt = 1 	imes 0 = 0$. Also,$g(1) = (1-1) \int_0^1 f(t) dt = 0 	imes \int_0^1 f(t) dt = 0$. Since $g(x)$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ with $g(0) = g(1) = 0$,by Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $g'(c) = 0$. Calculating the derivative: $g'(x) = -1 \int_0^x f(t) dt + (1-x) f(x)$. Setting $g'(c) = 0$,we get $-(1-c) f(c) + \int_0^c f(t) dt = 0$,which implies $\int_0^c f(t) dt = (1-c) f(c)$.

Let $f$ be a function that is derivable on the interval $[0, 1]$. Then,which of the following statements is true?

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