Cho hàm số f(x). Biết f(0) = 4 và \(f'\left( x \right) = 2{\cos ^2}x + 1\)

Đáp án C
Ta có: $f\left( x \right) = \int {f'\left( x \right)dx} = \int {\left( {2{{\cos }^2}x + 1} \right)dx} = \int {\left( {2 + \cos 2x} \right)} dx = 2x + \frac{1}{2}\sin 2x + C$.
Theo bài: \(f\left( 0 \right) = 4 \Leftrightarrow 2.0 + \frac{1}{2}.\sin 0 + C = 4 \Leftrightarrow C = 4\). Suy ra\(f\left( x \right) = 2x + \frac{1}{2}\sin 2x + 4\).
\(\begin{array}{l}\int\limits_0^{\frac{\pi }{4}} {f\left( x \right)dx} = \int\limits_0^{\frac{\pi }{4}} {\left( {2x + \frac{1}{2}\sin 2x + 4} \right)dx} \\ = \left. {\left( {{x^2} - \frac{{\cos 2x}}{4} + 4x} \right)} \right|_0^{\frac{\pi }{4}}\\ = \left( {\frac{{{\pi ^2}}}{{16}} + \pi } \right) - \left( { - \frac{1}{4}} \right) = \frac{{{\pi ^2} + 16\pi + 4}}{{16}}\end{array}\)