Cho hàm số f(x) có f (0) =4 và f'(x) = 2 cos ^2 x _ 1

8/22

Cho hàm số \(f(x)\) có \(f(0) = 4\) và\(f'(x) = 2{\cos ^2}x + 1,\forall x \in \mathbb{R}\) Khi đó \(\int\limits_0^{\frac{\pi }{4}} {f(x)dx} \) bằng.

\(\frac{{{\pi ^2} + 16\pi + 16}}{{16}}\).

\(\frac{{{\pi ^2} + 4}}{{16}}\).

\(\frac{{{\pi ^2} + 14\pi }}{{16}}\).

\(\frac{{{\pi ^2} + 16\pi + 4}}{{16}}\).

Giải thích

Chọn D

Ta có: \[\int\limits_0^1 {\left( {3x + 1} \right)\left( {x + 3} \right){\rm{d}}x} = \int\limits_0^1 {\left( {3{x^2} + 10x + 3} \right){\rm{d}}x} = \left. {\left( {{x^3} + 5{x^2} + 3x} \right)} \right|_0^1 = 9\].

\[\begin{array}{l}f(x) = \int {(2{{\cos }^2}} x + 1){\rm{d}}x = \int {\left( {2\left( {\frac{{1 + \cos 2x}}{2}} \right) + 1} \right)} {\rm{d}}x = \int {\left( {\cos 2x + 2} \right)} {\rm{d}}x\\ = \int {\cos 2x{\rm{d}}x + \int {2{\rm{d}}x} } = \frac{{\sin 2x}}{2} + 2x + C.\end{array}\]

Lại có \(f(0) = 4 \Leftrightarrow C = 4 \Rightarrow f(x) = \frac{{\sin 2x}}{2} + 2x + 4.\)

\(\begin{array}{l} \Rightarrow \int\limits_0^{\frac{\pi }{4}} {f(x){\rm{d}}x = } \int\limits_0^{\frac{\pi }{4}} {\left( {\frac{{\sin 2x}}{2} + 2x + 4} \right){\rm{d}}x = } \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {\sin 2x{\rm{d}}(2x) + \int\limits_0^{\frac{\pi }{4}} {2x{\rm{d}}x + \int\limits_0^{\frac{\pi }{4}} {4{\rm{d}}x} } } \\ = \frac{{ - \cos 2x}}{4}\left| \begin{array}{l}\frac{\pi }{4}\\0\end{array} \right. + ({x^2} + 4x)\left| \begin{array}{l}\frac{\pi }{4}\\0\end{array} \right. = \frac{{{\pi ^2} + 16\pi + 4}}{{16}}.\end{array}\).