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

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?

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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. – Austin Mohr Apr 29 '12 at 23:37
You could try applying integration by parts twice to $\int_0^{\pi}f''(x)\sin x\,dx$. – Gerry Myerson Apr 29 '12 at 23:39
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. – Hurkyl Apr 29 '12 at 23:41

Hint:

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

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Provided you apply it to the right thing.... – Gerry Myerson Apr 29 '12 at 23:40
The fact that integrating $f''(x)$ and differentiating $f(x)$ lead to the same thing is what strikes me. – Ross Millikan Apr 29 '12 at 23:43

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

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

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