Cho hàm số $\Large f\left( x \right)$. Biết $\Large f\left( 0 \right)=

Cho hàm số $\Large f\left( x \right)$. Biết $\Large f\left( 0 \right)=4$ và $\Large {f}'\left( x \right)=2{{\sin }^{2}}x+3$, $\Large \forall x\in R$, khi đó $\Large \int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{d}x}$ bằng

• A. $\Large \dfrac{{{\pi }^{2}}-2}{8}$.
• B. . $\Large \dfrac{{{\pi }^{2}}+8\pi -8}{8}$.
• C. $\Large \dfrac{{{\pi }^{2}}+8\pi -2}{8}$.
• D. $\Large \dfrac{3{{\pi }^{2}}+2\pi -3}{8}$.

Chọn C
$\Large \int{{f}'\left( x \right)\text{d}x=}\int{\left( 2{{\sin }^{2}}x+3 \right)}\text{ d}x=\int{\left( 1-\cos 2x+3 \right)}\text{ d}x=\int{\left( 4-\cos 2x \right)}\text{ d}x=4x-\dfrac{1}{2}\sin 2x+C$.
Ta có $\Large f\left( 0 \right)=4$ nên $\Large 4.0-\dfrac{1}{2}\sin 0+C=4\Leftrightarrow C=4$.
Nên $\Large f\left( x \right)=4x-\dfrac{1}{2}\sin 2x+4$.
 $\Large \int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 4x-\dfrac{1}{2}\sin 2x+4 \right)\text{d}x}$$\Large =\left( 2{{x}^{2}}+\dfrac{1}{4}\cos 2x+4x \right)\left| \begin{matrix}
& \dfrac{\pi }{4} \\ 
& 0 \\ 
\end{matrix} \right.=$$\Large \dfrac{{{\pi }^{2}}+8\pi -2}{8}$