5
$\begingroup$

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?

  • 1
    I've formatted your question. Please make sure that I haven't changed your intended meaning. Also, if you are interested in learning how these things are written, you can click "edit" to see the code I typed.2012-04-29
  • 1
    You could try applying integration by parts twice to $\int_0^{\pi}f''(x)\sin x\,dx$.2012-04-29
  • 2
    This looks like an exercise in applying the knowledge creatively to produce information that could be useful. You are not expected to **know** what to do. Instead, you're expected to conjecture and experiment until you find something that is useful.2012-04-29

2 Answers 2

2

Hint:

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

  • 0
    Provided you apply it to the right thing....2012-04-29
  • 3
    The fact that integrating $f''(x)$ and differentiating $f(x)$ lead to the same thing is what strikes me.2012-04-29
0

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

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