Cho hình phẳng ( H ) giới hạn bởi đồ thị hàm số \(y = f( x ) = {x + 1}}{x}\), trục hoành

a) Đ, b) S, c) Đ, d) S

a) Ta có \(S = \int\limits_2^6 {\left| {f\left( x \right)} \right|dx} = \int\limits_2^6 {\left| {\frac{{x + 1}}{x}} \right|dx = } \int\limits_2^6 {\frac{{x + 1}}{x}dx} = \int\limits_2^6 {\left( {1 + \frac{1}{x}} \right)dx} \)

\( = \left. {\left( {x + \ln x} \right)} \right|_2^6 = 6 + \ln 6 - \left( {2 + \ln 2} \right) = 4 + \ln 3\).

b) \(S = \int\limits_2^6 {\left| {f\left( x \right) - 1} \right|dx} = \int\limits_2^6 {\left| {\frac{{x + 1}}{x} - 1} \right|dx = } \int\limits_2^6 {\frac{1}{x}dx} \)\( = \left. {\ln x} \right|_2^6 = \ln 6 - \ln 2 = \ln 3\).

c) Ta có \(V = \pi {\int\limits_2^6 {\left( {\frac{{x + 1}}{x}} \right)} ^2}dx\)\( = \pi {\int\limits_2^6 {\left( {1 + \frac{1}{x}} \right)} ^2}dx\)\( = \pi \int\limits_2^6 {\left( {1 + \frac{2}{x} + \frac{1}{{{x^2}}}} \right)} dx\)

\( = \left. {\pi \left( {x + 2\ln x - \frac{1}{x}} \right)} \right|_2^6\)\( = \pi \left( {6 + 2\ln 6 - \frac{1}{6} - 2 - 2\ln 2 + \frac{1}{2}} \right) = \pi \left( {4 + 2\ln 3 + \frac{1}{3}} \right)\)\( = \frac{{\left( {13 + 6\ln 3} \right)\pi }}{3}\).

d) \(V = \pi \int\limits_2^6 {\left( {{f^2}\left( x \right) - 1} \right)dx} \)\( = \pi \int\limits_2^6 {\left[ {{{\left( {\frac{{x + 1}}{x}} \right)}^2} - 1} \right]dx} \)\( = \pi \int\limits_2^6 {{{\left( {\frac{{x + 1}}{x}} \right)}^2}dx} - \pi \int\limits_2^6 {1dx} \)

\( = \frac{{\left( {13 + 6\ln 3} \right)\pi }}{3} - \left. {\pi x} \right|_2^6\)\( = \frac{{\left( {13 + 6\ln 3} \right)\pi }}{3} - 4\pi = \frac{{\left( {1 + 6\ln 3} \right)\pi }}{3}\).