Cho hàm số \(y = f( x ) = { \begi{array}{l}{2}{{x + 1}}\;{\rm{khi}}\;0 \le x < 1\\2x - 1

a) Đ, b) S, c) Đ, d) Đ

a) \(\int\limits_0^1 {f\left( x \right)dx} = \int\limits_0^1 {\frac{2}{{x + 1}}dx} \).

b) \(\int\limits_2^3 {f\left( x \right)dx} = \int\limits_2^3 {\left( {2x - 1} \right)dx} \).

c) \(\int\limits_1^3 {f\left( x \right)dx} = \int\limits_1^3 {\left( {2x - 1} \right)dx} = \left. {\left( {{x^2} - x} \right)} \right|_1^3 = 9 - 3 - \left( {1 - 1} \right) = 6\).

d) \(\int\limits_0^3 {f\left( x \right)dx} = \int\limits_0^1 {f\left( x \right)dx} + \int\limits_1^3 {f\left( x \right)dx} \)\( = \int\limits_0^1 {\frac{2}{{x + 1}}dx} + 6\)\( = \left. {2\ln \left| {x + 1} \right|} \right|_0^1 + 6 = 2\ln 2 + 6 = \ln 4 + 6\).