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Integral function

Let f, g:\mathbb{R} \rightarrow \mathbb{R} be two continuous functions such that

    \[\int_{\int_{x}^{1} f(t) \, \mathrm{d}t}^{\int_{0}^{x}f(t) \, \mathrm{d}t} g(t) \, \mathrm{d}t>0 \quad \text{for all} \;\; x \in \mathbb{R} \setminus \{0, 1 \}\]

If g(x) + g(2-x)=2 and g(x) \neq 0 for all x \in \mathbb{R}, then prove that:

  1. \displaystyle \int_{0}^{x} f(t) \, \mathrm{d}t > \int_{x}^{1} f(t) \, \mathrm{d}t
  2. \displaystyle \int_{0}^{1} f(t)\, \mathrm{d}t = 0
  3. f(1)=f(0)=0
  4. the equation \displaystyle f(x) \int_{x}^{1} f(t) \, \mathrm{d}t = f(x) f'(x) has at least a root in (0, 1) if f is differentiable.
  5. the equation \displaystyle xf(x) = \int_{x}^{1} f(t) \, \mathrm{d}t has at least a root in (0, 1).
  6. the area of g bounded by the axis x'x and the lines x=0 , x=2 is 2 square meters.

Solution

  1. Since g(x) \neq 0 for all x \in \mathbb{R} it follows that the sign of g is constant. For x=1 we get that g(1)=1>0 hence g(x)>0 for all x \in \mathbb{R}. Therefore,

        \[\int_{\int_{x}^{1} f(t) \, \mathrm{d}t}^{\int_{0}^{x}f(t) \, \mathrm{d}t} g(t) \, \mathrm{d}t>0 \overset{g(x)>0}{\Leftarrow \! =\! =\! =\! =\! \Rightarrow } \int_{0}^{x} f(t) \, \mathrm{d}t > \int_{x}^{1} f(t) \, \mathrm{d}t\]

  2. Taking limits we have

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