Cho hàm số \[f(x)\]có \[f'(x) = \sin (2x).co{s^2}(4x)\]và \[f(0) = 0\]. Tính \[\int\limits_0^{\frac{\pi }{2}} {f(x)dx} \] bằng:

A.\(x \le \frac{4}{3}\).

B. \(x \ge \frac{{11}}{3}\).

C.\(x \le \frac{{11}}{3}\).

D.\(x \ge \frac{4}{3}\).

A. \(\left( { - \infty \,;\,15} \right)\).

B. \(\left( {15\,;\, + \infty } \right)\).

C. \(\left( { - \infty \,;\,3} \right)\).

D. \(\left( {3\,;\, + \infty } \right)\).

A. \(S = \int\limits_{ - 1}^1 {\left( {{x^2} + x} \right)} {\rm{d}}x\).

B. \(S = \int\limits_1^{ - 1} {\left( {{x^2} + x} \right)} {\rm{d}}x\).

C. \(S = \int\limits_0^3 {\left( {{x^2} - 3x} \right)} {\rm{d}}x\).

D. \(S = \int\limits_0^3 {\left( {3x - {x^2}} \right)} {\rm{d}}x\).

A. \[{(x - 2)^2} + {(y - 2)^2} + {z^2} = 4\].

B. \[{(x + 2)^2} + {(y + 2)^2} + {z^2} = 5\].

C. \[{(x - 2)^2} + {(y - 2)^2} + {z^2} = \sqrt 5 \].

D. \({(x - 2)^2} + {(y - 2)^2} + {z^2} = 5\).

A. \(\left[ { - \frac{1}{2}; + \infty } \right)\).

B. \(\left( { - \frac{1}{2}; + \infty } \right)\).

C. \(\left( { - \infty ; - \frac{1}{2}} \right)\).

D. \(\left( { - \infty ; - \frac{1}{2}} \right]\).

A. \[y\, = \, - \,{x^3}\, + \,3x\, + \,2\].

B. \[y\, = \, - \,{x^3}\, + \,3{x^2}\, - \,2\].

C. \[y\, = {x^3}\, - \,3x\, + \,2\].

A. \(2\sqrt 3 \).

B. \(5\sqrt 2 \).

C. \(20\).

D. \(2\sqrt 5 \).