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Find a function whose graph has given properties

Let f(x) be a function on the interval [0,1] that is nonnegative and differentiable, and such that f(1) = 0. If for each a in the open interval (0,1) the line given by the equation x = a divides the ordinate set of f(x) into regions A and B where A denotes the leftmost region. If the areas of the two regions obey the equation

    \[ A - B = 2f(a) + 3a + b, \]

where b is a constant that does not depend on a, find f(x) and b.


Incomplete.

2 comments

  1. Evangelos says:

    First, let’s define the regions A and B as functions of the variable a

        \begin{align*} A = \int_{0}^{a} f(x) dx \\ B = \int_{a}^{1} f(x) dx \\ \end{align*}

    Now, we take our given equation:

        \begin{align*} A - B = 2f(a) + 3a + b \\ \int_{0}^{a} f(x) dx - \int_{a}^{1} f(x) dx = 2f(a) + 3a + b \\ \int_{0}^{a} f(x) dx + \int_{1}^{a} f(x) dx = 2f(a) + 3a + b \\ \end{align*}

    If we take the derivative with respect to a of both sides of the equation, we get

        \begin{align*} 2f(a) = 2f'(a) + 3 \\ f'(a) = f(a) - \frac{3}{2} \\ \end{align*}

    This equation is satisfied by the function

        \begin{align*} f(a) =  Ce^{a} + \frac{3}{2} \\ \end{align*}

    Referring back to our initial condition that f(1) = 0, we can solve for the constant C

        \begin{align*} f(1) = Ce + \frac{3}{2} \\ = 0 \end{align*}

        \begin{align*} C = -\frac{3}{2e} \end{align*}

        \begin{align*} f(a) =  -\frac{3}{2e}e^{a} + \frac{3}{2} \\ = \frac{3}{2}(1 - e^{a-1}) \end{align*}

    As requested.

    • Evangelos says:

      And to find the scalar constant b, all we need to do is use our newly found function and evaluate the above integrals.

      Recall,

          \begin{align*} A = \int_{0}^{a} f(x) dx \\ B = \int_{a}^{1} f(x) dx \end{align*}

          \begin{align*} A - B = 2f(a) + 3a + b \end{align*}

      And we now now that the unknown function f(x):

          \begin{align*} f(x) = \frac{3}{2}(1 - e^{x-1}) \end{align*}

      The above equation becomes

          \begin{align*} A - B &= \int_{0}^{a} f(x) dx - \int_{a}^{1} f(x) dx\\ &= \int_{0}^{a} f(x) dx + \int_{1}^{a} f(x) dx \\ &= \frac{3}{2} \int_{0}^{a} (1 - e^{x-1}) dx + \frac{3}{2} \int_{1}^{a} (1 - e^{x-1}) dx \\ &= \frac{3}{2}[x - e^{x-1}]_{0}^{a} + \frac{3}{2}[x - e^{x-1}]_{1}^{a}\\ & =  \frac{3}{2}[(a - e^{a-1}) + \frac{1}{e}] + \frac{3}{2}[(a - e^{a-1})]\\ & = 3(a - e^{a-1}) + \frac{3}{2e} \\ &= 2f(a) + 3a + b \end{align*}

          \begin{align*} 3(a - e^{a-1}) + \frac{3}{2e} &=  3(1 - e^{a-1}) + 3a + b\\ \end{align*}

          \begin{align*} \frac{3}{2e} &= 3 + b\\ \end{align*}

          \begin{align*} b = \frac{3}{2e} - 3 \quad \blacksquare \end{align*}

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