Prove $\exists \xi \in [0,\pi]$ such that $ \int_0^\pi e^{-x}\cos(x)\,dx = \sin(\xi)$
Well, I was thinking on using the Generalized MVT, but it doesnt seem I obtain the right answer. How do you guys would tackle this problem?
Prove $\exists \xi \in [0,\pi]$ such that $ \int_0^\pi e^{-x}\cos(x)\,dx = \sin(\xi)$
Well, I was thinking on using the Generalized MVT, but it doesnt seem I obtain the right answer. How do you guys would tackle this problem?
$\left|\int_0^\pi\mathrm e^{-x}\cos(x)\,\mathrm dx\right|\lt\int_0^\pi\mathrm e^{-x}\,\mathrm dx=\left[-\mathrm e^{-x}\right]_0^\pi=1-\mathrm e^{-\pi}\lt1\;.$
Since you want to prove existence, one example will suffice.
Note that $\int_0^\pi {{e^{ - x}}\cos (x)dx} = \frac{{{e^{ - \pi }}(1 + {e^\pi })}}{2}$ (you can use complex exponential trick to evaluate the integral) and let $\xi = {\sin ^{ - 1}}\left( {\frac{{{e^{ - \pi }}(1 + {e^\pi })}}{2}} \right) \in \left[ {0,\pi } \right].$