Home » Blog » Find values for a function satisfying a given integral equation

Find values for a function satisfying a given integral equation

Let f be an everywhere continuous function satisfying

    \[ \int_0^x f(t) \, dt = -\frac{1}{2} + x^2 + x \sin (2x) + \frac{1}{2} \cos (2x) \]

for all x. Find the values of

    \[ f \left( \frac{\pi}{4} \right) \qquad \text{and} \qquad f' \left( \frac{\pi}{4} \right). \]


Let

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

By the first fundamental theorem of calculus we know the derivative A'(x) exists (since f is continuous by assumption) and we have

    \[ A'(x) = f(x). \]

Therefore, taking the derivative of A(x) we have,

    \[ f(x) = A'(x) = 2x + \sin (2x) + 2x \cos (2x) - \sin (2x) = 2x + 2x \cos (2x) \]

Then, evaluating at \frac{\pi}{4},

    \begin{align*}  f \left( \frac{\pi}{4} \right) &= A' \left( \frac{\pi}{4} \right) \\  &= \frac{\pi}{2} + \frac{\pi}{2} \cos \left( \frac{\pi}{2} \right) \\  &= \frac{\pi}{2}. \end{align*}

Then, taking the derivative of f(x) we have,

    \[ f(x) = 2x + 2x \cos (2x) \quad \implies \quad f'(x) = 2 + 2 \cos(2x) - 4x \sin (2x). \]

Therefore,

    \[ f' \left( \frac{\pi}{4} \right) = 2 + 2 \cos \left( \frac{\pi}{2} \right) - \pi \sin \left( \frac{\pi}{2} \right) = 2 - \pi. \]

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):