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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?

Michael Hardy
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TPR
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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
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    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
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    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. –  Apr 29 '12 at 23:41

2 Answers2

2

Hint:

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

chemeng
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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

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

Mathlover
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