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

b) \(\int\limits_1^2 {\frac{1}{{g\left( x \right)}}dx} = \int\limits_1^2 {\frac{1}{{x + 1}}dx} = \left. {\ln \left( {x + 1} \right)} \right|_1^2 = \ln 3 - \ln 2 = \ln \frac{3}{2}\).

c) \(\int\limits_1^2 {f\left( x \right)dx} + \int\limits_2^1 {g\left( x \right)dx} = \int\limits_1^2 {\left( {2{x^2} + 3x + 2} \right)dx} + \int\limits_2^1 {\left( {x + 1} \right)dx} \)\[ = \left. {\left( {\frac{{2{x^3}}}{3} + \frac{3}{2}{x^2} + 2x} \right)} \right|_1^2 + \left. {\left( {\frac{{{x^2}}}{2} + x} \right)} \right|_2^1\]

\( = \frac{{67}}{6} - \frac{5}{2} = \frac{{26}}{3}\).

d) \(\int\limits_0^2 {\frac{{f\left( x \right)}}{{g\left( x \right)}}dx} = \int\limits_0^2 {\frac{{2{x^2} + 3x + 2}}{{x + 1}}dx} \)\( = \int\limits_0^2 {\left( {2x + 1 + \frac{1}{{x + 1}}} \right)dx} \)\( = \left. {\left( {{x^2} + 2x + \ln \left( {x + 1} \right)} \right)} \right|_0^2\)\( = 8 + \ln 3\).

Suy ra a = 8; b = 1. Khi đó a + b = 9 > 8.

Đáp án: a) Sai; b) Đúng; c) Đúng; d) Đúng.

Biết I = 2 ∫ 1 3^x dx = a/ ln 3 . Giá trị của a bằng