what concept am I supposed to apply if second derivative is given.

Question

Given: $f^{\prime\prime}(x)$ is continuous, $f(\pi) = 0$, and $$\int_0^\pi (f(x)+f^{\prime\prime}(x))\sin(x) \, dx = 2.$$

Find: $f(0)$.

I know integration by parts etc, but I do not know which particular concept(s) I'm supposed to apply for this one. Or is there a specific theorem I am missing?

Answer 1

Hint:

Apply integration by parts several times and you'll get the answer.

Answer 2

We need to use integration by parts.

Here is general formula :$$\int U'(x)V(x)dx = V(x)U(x)- \int U(x)V '(x)dx$$

Let's use it in the question

$$\int_0^\pi (f(x)+f^{\prime\prime}(x))\sin(x) \, dx = \int_0^\pi f(x)sin(x) \, dx + \int_0^\pi f^{\prime\prime}(x)\sin(x) \, dx=\int_0^\pi f(x)sin(x) \, dx +(f^{\prime}(\pi)\sin(\pi)-f^{\prime}(0)\sin(0))- \int_0^\pi f^{\prime}(x)\cos(x) \, dx=\int_0^\pi f(x)sin(x) \, dx - \int_0^\pi f^{\prime}(x)\cos(x) \, dx=\int_0^\pi f(x)sin(x) \, dx - (f(\pi)\cos(\pi)-f(0)\cos(0))-\int_0^\pi f(x)\sin(x) \, dx=f(0)=2$$