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Prove that a function satisfies a particular integral equation

Let g(x) be an everywhere continuous function with

    \[ g(1) = 5 \qquad \text{and} \qquad \int_0^1 g(t) \, dt = 2. \]

Define another function f(x) by

    \[ f(x) = \frac{1}{2} \int_0^x (x-t)^2 g(t) \, dt. \]

Prove that

    \[ f'(x) = x \int_0^x g(t) \, dt - \int_0^x t g(t) \, dt. \]

Also, compute the values, f''(1) and f'''(1).

Proof. We start with the equation for f(x) and use properties of the integral and the fundamental theorem of calculus to rearrange,

    \begin{align*}  f(x) &= \frac{1}{2} \int_0^x (x-t)^2 g(t) \, dt \\  &= \frac{1}{2} \int_0^x (x^2 g(t) - 2xt g(t) + t^2 g(t)) \,dt \\  &= \frac{1}{2} \int_0^x x^2 g(t) \, dt - \int_0^x xt g(t) \, dt + \frac{1}{2} \int_0^x t^2 g(t) \, dt &(\text{linearity of integral})\\  &= \frac{x^2}{2} \int_0^x g(t) \, dt - x \int_0^x t g(t) \, dt + \frac{1}{2} \int_0^x t^2 g(t). \end{align*}

We can pull the x out of the integral since it does not depend on t (which is what we are taking the integral over). Then, we want to take the derivative with respect to x; however, we need to be careful and use the product rule. Looking closely at the first term we have

    \begin{align*}    \frac{d}{dx}\left( \frac{x^2}{2} \int_0^x g(t) \, dt\right) &= \left( \frac{x^2}{2} \right)' \cdot \int_0^x g(t) \, dt + \frac{x^2}{2} \left( \int_0^x g(t) \, dt \right)' \\  &= x \int_0^x g(t) \, dt + \frac{x^2}{2} g(x). \end{align*}

Where we used the fundamental theorem to obtain \left( \int_0^x g(t) \, dt \right)' = g(x). So, using this method to take the derivative for f(x) we have

    \begin{align*}  f'(x) &= x \int_0^x g(t) \, dt + \frac{x^2}{2} g(x) - \int_0^x t g(t) \, dt - x^2 g(x) + \frac{x^2}{2} g(x) \\  &= x \int_0^x g(t) \, dt - \int_0^x t g(t) \, dt. \qquad \blacksquare \end{align*}

Now, to evaluate the second and third derivatives at 1 as requested, we first get expressions for f''(x) and f'''(x).

    \begin{align*}    f''(x) &= \int_0^x g(t) \, dt + x g(x) - x g(x) = \int_0^x g(t) \, dt. \\  f'''(x) &= \left( \int_0^x g(t) \, dt \right)' = g(x). \end{align*}


    \begin{align*}  f''(1) &= \int_0^1 g(t) \, dt = 2 \\  f'''(1) &= g(1) = 5. \end{align*}

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