Tính thể tích tròn xoay do \[\frac{{{{\rm{x}}^{\rm{2}}}}}{{{{\rm{a}}^{\rm{2}}}}}{\rm{ + }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{{\rm{b}}^{\rm{2}}}}}{\rm{ = 1}}\]quay quanh Oy

A. 0

B. \[\ln \frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}\]

C. \[\ln \frac{{\sqrt 2 + 1}}{3}\]

D. \[\ln \frac{{\sqrt 2 + 1}}{{3(\sqrt 2 - 1)}}\]

A. \[\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f(x)dx}} = 2\mathop \smallint \limits_0^{\rm{a}} {\rm{f(x)dx}}\]

B. \[\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f(x)dx}} = - \mathop \smallint \limits_0^{\rm{a}} {\rm{f(x)dx}}\]

C. \[\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f(x)dx}} = \mathop \smallint \limits_0^{\rm{a}} {\rm{f(x)dx}}\]

D. \[\mathop \smallint \limits_{ - {\rm{a}}}^{\rm{a}} {\rm{f(x)dx}} = 2\mathop \smallint \limits_{ - a/2}^{{\rm{a/2}}} {\rm{f(x)dx}}\]

A. \[\left( {\forall {\rm{x}} \in \left[ {{\rm{a,b}}} \right]} \right){\rm{f(x)}} < {\rm{g(x)}} \Rightarrow \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f(x)dx}} > \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{g(x)dx}}\]

B. \[\left( {\forall {\rm{x}} \in \left[ {{\rm{a,b}}} \right]} \right){\rm{f(x)}} \le {\rm{g(x)}} \Rightarrow \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f(x)dx}} \le \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{g(x)dx}}\]

C. \[\left( {\forall {\rm{x}} \in \left[ {{\rm{a,b}}} \right]} \right){\rm{f(x)}} \le {\rm{g(x)}} \Rightarrow \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f(x)g(x)dx}} \le \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{g(x)dx}}\]

D. \[{\rm{f(x)}} \le {\rm{g(x)}} \Rightarrow \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{g(x)dx}} \le \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{g(x)dx}}\]

A. f'(0) = 0

B. f'(0) = -1

C. f'(0) = 1

D. Không tồn tại

A. \[\frac{1}{5}\]

B. 0

C. \(\infty \)

D. \[\frac{1}{{10}}\]

A. -1

B. +∞

C. 0

D. e-1/2

A. 0

B. -1/80

C. 10

D. 20