Xét \(\int\limits_{ - 1}^1 {{x^2}\sqrt {{{\left( {2 + {x^3}} \right)}^5}} dx} \), nếu đặt \(u = 2 + {x^3}\) thì \(\int\limits_{ - 1}^1 {{x^2}\sqrt {{{\left( {2 + {x^3}} \right)}^5}} dx} \) bằng

A. \({u_{10}} = - 31\).

B. \({u_{10}} = - 23\).

C. \({u_{10}} = - 20\).

D. \({u_{10}} = 15\).

A. \[36\pi \].

B. \[12\pi \].

C. \[15\pi \].

D. \[45\pi \].

A. \(\frac{{46}}{{125}}\).

B. \(\frac{{121}}{{625}}\).

C. \(\frac{{36}}{{125}}\).

D. \(\frac{{181}}{{625}}\).

A. 5.

B. 10.

C. 15.

D. 30.

A. \[S = \pi \int\limits_{ - 1}^3 {{{\left( {{x^3} - 2{x^2} - 3x} \right)}^2}dx} \].

B. \(S = \int\limits_{ - 1}^3 {\left( {{x^3} - 2{x^2} - 3x} \right)dx} \).

C.\(S = \int\limits_{ - 1}^0 {\left( {{x^3} - 2{x^2} - 3x} \right)dx} + \int\limits_0^3 {\left( {2{x^2} + 3x - {x^3}} \right)dx} \).

D. \(S = \int\limits_{ - 1}^0 {\left( {2{x^2} + 3x - {x^3}} \right)dx} + \int\limits_0^3 {\left( {{x^3} - 2{x^2} - 3x} \right)dx} \).

A. \(60^\circ \).

B. \(90^\circ \).

C. \(30^\circ \).

D. \(45^\circ \).

A. 1.

B. 2.

C.3.

D.4.