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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?

  • 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

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Hint:

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

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    The fact that integrating $f''(x)$ and differentiating $f(x)$ lead to the same thing is what strikes me.2012-04-29
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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

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