Tính các tích phân sau: a) 2 ∫ 0 ( 3 x − 2 ) ( 3 x + 2 ) d x ; b) 2 ∫ 1 t 2 ( 5 t 2 − 2 ) d t ; c) 1 ∫ − 1 ( x − 2 ) ( x 2 + 2 x + 4 ) d x .

a) \[\int\limits_0^2 {\left( {3x - 2} \right)\left( {3x + 2} \right)dx} = \int\limits_0^2 {\left( {9{x^2} - 4} \right)dx} \]

                                        \[ = \left. {\left( {3{x^3} - 4x} \right)} \right|_0^2\]

                                        = (3.2³ – 4.2) – (3.0³ – 4.0) = 16.

b) \[\int\limits_1^2 {{t^2}\left( {5{t^2} - 2} \right)dt} = \int\limits_1^2 {\left( {5{t^4} - 2{t^2}} \right)dt} \]

                             \[ = \left. {\left( {{t^5} - \frac{2}{3}{t^3}} \right)} \right|_1^2\]

                             \[ = \left( {{2^5} - \frac{2}{3}{{.2}^3}} \right) - \left( {{1^5} - \frac{2}{3}{{.1}^3}} \right)\]

                                          \[ = \frac{{79}}{3}\].

c) \[\int\limits_{ - 1}^1 {\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)dx} = \int\limits_{ - 1}^1 {\left( {{x^3} - 8} \right)dx} \]

                                               \[ = \left. {\left( {\frac{{{x^4}}}{4} - 8x} \right)} \right|_{ - 1}^1 = - 16\].