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https://gyazo.com/60302805703764be4a572e6c2128daf0

Suppose that $g''(x)$ is continuous everywhere and that $$\int_{0}^{2\pi} g(x)\sin(x)\mathrm dx + \int_{0}^{2\pi} g''(x)\sin(x)\mathrm dx = 2$$

Given that $g(2\pi) = 1$, prove that $g(0) = 3$

Yeah I don't even know how to start.

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    Just ask her in her office hours lol – 2017-02-16
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    Its solved lol. This is much quicker imo – 2017-02-16
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    That's true to. I really wanna find out who you are man ! xD – 2017-02-16

1 Answers 1

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HINT:

Integrate by parts the integral $\int_0^{2\pi}g''(x)\sin(x)\,dx$ with $u=\sin(x)$ and $v=g'(x)$ to obtain

$$\int_0^{2\pi}g''(x)\sin(x)\,dx=-\int_0^{2\pi}g'(x)\cos(x)\,dx$$

Now, integrate by parts again.

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    Okay, we take $u=\cos x$ and $dv=g'(x)dx$. Then simplify. We eventually get $g(0)=3$. Thanks to Dr.MV's hint. (+1) – 2017-02-15
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    Wait what. $$\int_{0}^{2\pi} g''(x)cos(x)dx = sin(x)g'(x) - cos(x)g(x) - \int_{0}^{2\pi} sin(x)g(x)dx$$ What do I do now – 2017-02-15
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    You forgot to evaluate the terms $g'(x)\sin(x)$ and $g(x)\cos(x)$ at the integration limits. – 2017-02-15
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    but idk with g' pi or g'(0) is – 2017-02-15
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    We don't need to know the values of $g'(0)$ and $g'(\pi)$ since $\sin(0)=\sin(\pi)=0$. – 2017-02-15
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    oh i see so like $$-(g(2\pi)cos(360) - (1g(0)) = -1 + g(0) = 2, g(0) = 3$$ – 2017-02-15