Bài 112297

Question

a)Cho $f$ liên tục trên $[0;\pi ]$. Chứng minh rằng : $\int\limits_{0}^{\pi } xf(\sin x)dx = \pi \int\limits_{0}^{\frac{\pi}{2} }f (\sin x) dx.$
b) Tính $M = \int\limits_{0}^{\pi }\frac{x \sin x }{1 + \cos ^2x}dx.$
c) Tính $\mathop {\lim }\limits_{}\sum_n $ trong đó:
$ \sum_n = \frac{2010}{n^2}\left ( \frac{\sin \frac{\pi}{n} }{1 + \cos ^2 \frac{\pi }{n} } + \frac{2 \sin \frac{\pi}{n} }{1 + \cos ^2 \frac{2\pi}{n} } + ...+ \frac{n \sin \frac{n\pi}{n} }{1+\cos ^2 \frac{n \pi }{n} } \right ) $

score 0 · Answer 1

a) Đặt $t = \pi -x \Rightarrow dt = -dx$
$\Rightarrow \int\limits_{\frac{\pi}{2} }^{\pi }xf(\sin x)dx = - \int\limits_{\frac{\pi}{2} }^{0}(\pi - t)f(\sin t)dt = \int\limits_{0 }^{\frac{\pi}{2}}(\pi -x)f(\sin x)dx $
$\Rightarrow \int\limits_{0}^{\pi}x.f(\sin x)dx = \int\limits_{0}^{\frac{\pi}{2} }xf(\sin x)dx + \int\limits_{\frac{\pi}{2} }^{\pi}xf(\sin x)dx$
                                        $= \int\limits_{0}^{\frac{\pi}{2} }xf (\sin x)dx + \int\limits_{0}^{\frac{\pi}{2} }(\pi -x)f(\sin x) dx$
$\Rightarrow \int\limits_{0}^{\pi } xf(\sin x)dx = \pi \int\limits_{0}^{\frac{\pi}{2} }f(\sin x)dx.$
b) Ta có : $ \int\limits_{0}^{\pi }\frac{x \sin x}{1 + \cos ^2 x}dx = \pi \int\limits_{0}^{\frac{\pi}{2} }\frac{\sin x}{1 + \cos ^2x}dx$
Đặt $ t = \cos x \Rightarrow dt = - \sin xdx$
$\Rightarrow \int\limits_{0}^{\pi }\frac{x \sin x}{1 + \cos ^2 x}dx = \pi \int\limits_{1}^{0}\frac{-dt}{1+t^2} = \pi \int\limits_{0}^{1}\frac{dt}{1+t^2}$
Tiếp tục , đặt $ t = \tan \alpha \left ( 0\leq \alpha < \frac{\pi}{2} \right ) \Rightarrow dt = \frac{d\alpha}{\cos ^2 \alpha}$
$\Rightarrow \int\limits_{0}^{\pi }\frac{x \sin xdx}{1 + \cos ^2 x} = \pi \int\limits_{0}^{\frac{\pi }{4} } \frac{\frac{d\alpha}{\cos ^2 \alpha} }{1+\tan ^2 \alpha}= \frac{\pi^2}{4}.$
c) * Chia đoạn $[0;\pi ]$ thành $n$ đoạn nhỏ bằng nhau bởi các điểm:
                    $x_0 = 0 <x_1 = \frac{\pi}{n}<...<x_1=\frac{\pi i}{n}<...< x_n = \pi $
    * Chọn $\epsilon_i = x_{i+1} = \frac{\pi (i+1)}{n}, i = \overline{0,n-1}.$
    * Lập tổng tích phân:
                $ S_n = \sum\limits_{i = 0}^{n - 1}f(\xi _i)(x_{i+1}-x_i) ( với f(x) = \frac{x \sin x}{1+ \cos ^2x} x \in R$
                     $=\frac{\pi}{n} \sum\limits_{i = 0}^{n - 1}\frac{\xi _i. \sin \xi _i}{1 + \cos ^2 \xi _i}=\frac{\pi^2}{n^2} \sum\limits_{i = 0}^{n}\frac{i \sin \frac{\pi i}{n} }{1 + \cos ^2 \frac{\pi i}{n} } $
Mà $ \mathop {\lim }\limits_{n \to +\infty } S_n= \int\limits_{0}^{\pi }f(x)dx = \frac{\pi^2}{4}$
Nên $\mathop {\lim }\limits_{n \to +\infty } \sum\nolimits_n= \mathop {\lim }\limits_{n \to +\infty }\frac{2010}{n^2}\sum\limits_{i = 0}^n \frac{i \sin \frac{\pi i}{n} }{1 + \cos ^2 \frac{i\pi}{n}} = \frac{2010}{\pi ^2} . \frac{\pi^2}{4} = \frac{1005}{2}$

Bài 112297

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Bài 112297

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