Let $f:[0, +\infty) \rightarrow \mathbb{R}$ be a twice differentiable function with a continuous second derivative. If
\begin{equation} \int_{0}^{\pi}\left ( f(x) \sin x + f''(x) \sin x \right )\, {\rm d}x =2 \end{equation}
and $f(\pi)=1$ hold , then evaluate $f(0)$.
Solution
Back at $(1)$ we have successively after applying IBP
\begin{align*} 2 &= \int_{0}^{\pi}f(x)\sin x \, {\rm d}x + \int_{0}^{\pi}f''(x)\sin x \, {\rm d}x \\ &= \int_{0}^{\pi} f(x) \sin x \, {\rm d}x + \cancelto{0}{\left [ f'(x) \sin x \right ]_0^\pi }- \int_{0}^{\pi}f'(x) \cos x \, {\rm d}x \\ &= \int_{0}^{\pi}f(x) \sin x \, {\rm d}x - \left [ f(x) \cos x \right ]_0^\pi - \int_{0}^{\pi} f(x) \sin x \, {\rm d}x\\ &= - f(\pi) \cos \pi + f(0) \cos 0 +\cancel{\int_{0}^{\pi}f(x) \sin x \, {\rm d}x - \int_{0}^{\pi} f(x) \sin x \, {\rm d}x}\\ &= 1 + f(0) \end{align*}
Hence $f(0)=1$.
\begin{equation} \int_{0}^{\pi}\left ( f(x) \sin x + f''(x) \sin x \right )\, {\rm d}x =2 \end{equation}
and $f(\pi)=1$ hold , then evaluate $f(0)$.
Solution
Back at $(1)$ we have successively after applying IBP
\begin{align*} 2 &= \int_{0}^{\pi}f(x)\sin x \, {\rm d}x + \int_{0}^{\pi}f''(x)\sin x \, {\rm d}x \\ &= \int_{0}^{\pi} f(x) \sin x \, {\rm d}x + \cancelto{0}{\left [ f'(x) \sin x \right ]_0^\pi }- \int_{0}^{\pi}f'(x) \cos x \, {\rm d}x \\ &= \int_{0}^{\pi}f(x) \sin x \, {\rm d}x - \left [ f(x) \cos x \right ]_0^\pi - \int_{0}^{\pi} f(x) \sin x \, {\rm d}x\\ &= - f(\pi) \cos \pi + f(0) \cos 0 +\cancel{\int_{0}^{\pi}f(x) \sin x \, {\rm d}x - \int_{0}^{\pi} f(x) \sin x \, {\rm d}x}\\ &= 1 + f(0) \end{align*}
Hence $f(0)=1$.
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