Cho $f(x) = \frac{x^2}{8x^3+1} , (x\geq 0)$
a) Khảo sát sự biến thiên và vẽ đồ thị $(C) : y = f(x).$
b) Tính diện tích hình phẳng giới hạn bởi $Ox, (C)$ và đường thẳng $ x=1$.
c) Đặt $a_n = \frac{1^2}{2^3+n^3} + \frac{2^2}{4^3+n^3}+...+ \frac{n^2}{(2n)^3+n^3}, n \in  N$.  Tính $ \mathop {\lim }\limits_{n \to +\infty} $
a) + MXĐ : $D = [0;+\infty  )$
    + $y' = f'(x) =\frac{2x(8x^3+1)-24x^4}{(8x^3+1)^2} = \frac{2x - 8x^4}{(8x^3+1)^2}=\frac{2x(1-4x^3)}{(8x^3+1)^2}=0$
                                     $ \Leftrightarrow  \left[ \begin{array}{l}
x = 0\\
x = \frac{{\sqrt[3]{2}}}{2}
\end{array} \right.$
    + Bảng biến thiên :
    + Đồ thị:

b) $ S = \int\limits_{0}^{1}\frac{x^2dx}{8x^3+1} = \frac{1}{24}\int\limits_{0}^{1}\frac{(8x^3+1)dx}{8x^3+1}=\frac{1}{24}\left ( \ln (8x^3+1)\left| \begin{array}{l}
1\\
0
\end{array} \right. \right )=\frac{\ln 3}{12}$ (đvdt)
c) * Hàm số $f$ liên tục trên $[0;1]$
    * Phân hoạch đoạn $[0;1]$ :
            $ 0 = x_0 < x_1 = \frac{1}{n}< x_2 = \frac{2}{n}<...< x_n = 1 $
    * Chọn $ \xi _i = x_{i+1} = \frac{i+1}{n}, i = \overline{0,n-1}.$
    * Tổng Riemann của $f$ :
       $\sum\limits_{i = 0}^{n - 1}f(\xi _i)(x_{i+1}-x_i) = \sum\limits_{i = 0}^{n - 1}\frac{\left ( \frac{i+1}{n}  \right )^2 }{8 \left ( \frac{i+1}{n}  \right )^3+1 }.\frac{1}{n} = \sum\limits_{i = 1}^{n} \frac{i^2}{8i^3+n^3}$
                                             $= \frac{1}{2^3+n^3} + \frac{2^2}{4^3+n^3}+...+\frac{n^2}{(2n)^3+n^3} = a_n.$
Vậy $\mathop {\lim }\limits_{n \to +\infty} a_n = \int\limits_{0}^{1}f(x)dx = \frac{\ln 3}{12}.$
hay quá! –  tangnobi 04-11-14 09:09 PM
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