Tính: a) A = 2 ∫ − 1 ( x − 4 x 2 ) d x + 4 2 ∫ − 1 ( x 2 − 1 ) d x ; b) B = 0 ∫ − 1 ( x 3 − 6 x ) d x + 1 ∫ 0 ( t 3 − 6 t ) d t .

6/13

Tính:

a) \[A = \int\limits_{ - 1}^2 {\left( {x - 4{x^2}} \right)dx + 4\int\limits_{ - 1}^2 {\left( {{x^2} - 1} \right)dx} } \];

b) \[B = \int\limits_{ - 1}^0 {\left( {{x^3} - 6x} \right)dx} + \int\limits_0^1 {\left( {{t^3} - 6t} \right)dt} \].

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

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

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

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

\[ = \left. {\left( {\frac{{{x^2}}}{2} - 4x} \right)} \right|_{ - 1}^2 = - \frac{{21}}{2}\].

Vậy \[A = - \frac{{21}}{2}\].

b) \[B = \int\limits_{ - 1}^0 {\left( {{x^3} - 6x} \right)dx} + \int\limits_0^1 {\left( {{t^3} - 6t} \right)dt} \]

\[ = \int\limits_{ - 1}^0 {\left( {{x^3} - 6x} \right)dx} + \int\limits_0^1 {\left( {{x^3} - 6x} \right)dx} \]

\[ = \int\limits_{ - 1}^1 {\left( {{x^3} - 6x} \right)dx} = \left. {\left( {\frac{{{x^4}}}{4} - 3{x^2}} \right)} \right|_{ - 1}^1 = 0\].