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Compute the value of a function satisfying a given integral equation

by RoRi

Let f(x) be a function satisfying

    \[ \int_0^{\pi} (f(x)+f''(x)) \sin x \, dx = 5. \]

Furthermore, let

    \[ f(\pi) = 2. \]

Compute f(0).

First, using linearity of the integral we write,

    \[ \int_0^{\pi} (f(x) + f''(x)) \sin x \, dx = \int_0^{\pi} f(x) \sin x \, dx + \int_0^{\pi} f''(x) \sin x \, dx = 5. \]

Next, we will use integration by parts twice to obtain an expression for the first integral in terms of f''(x) (and it will turn out the integrals with f''(x) will cancel which will allow us to solve this problem). For the first integration by parts let,

    \begin{align*} u &= f(x) & du &= f'(x) \\ dv &= \sin x \, dx & v &= -\cos x \end{align*}

Therefore,

    \begin{align*} \int_0^{\pi} f(x) \sin x \, dx &= -f(x) \cos x \Big \rvert_0^{\pi} + \int_0^{\pi} f'(x) \cos x \, dx \\ &= (f(\pi) + f(0)) + \int_0^{\pi} f'(x) \cos x \, dx \\ & = 2 + f(0) + \int_0^{\pi} f'(x) \cos x \, dx. \end{align*}

Now we want to integrate by parts again to evaluate the integral \int_0^{\pi} f'(x) \cos x \, dx. To that end let

    \begin{align*} u &= f'(x) & du &= f''(x) \\ dv &= \cos x & v &= \sin x. \end{align*}

Then we have,

    \begin{align*} 2 + f(0) + \int_0^{\pi} f'(x) \cos x \, dx &= 2 + f(0) + f'(x) \sin x \Big \rvert_0^{\pi} - \int_0^{\pi} f''(x) \sin x \, dx \\ &= 2 + f(0) - \int_0^{\pi} f''(x) \sin x \, dx. \end{align*}

Finally, plugging this back into our original expression we have,

    \begin{align*} 5 &= \int_0^{\pi} f(x) \sin x \, dx + \int_0^{\pi} f''(x) \sin x \, dx \\ &= 2 + f(0) - \int_0^{\pi} f''(x) \sin x \, dx + \int_0^{\pi} f''(x) \sin x \, dx \\ &= 2 + f(0). \end{align*}

Thus, we have

    \[ f(0) = 3.\]

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