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Câu hỏi: Cho hàm số $f\left( x \right)$. Biết $f\left( 0 \right)=4$ và ${f}'\left( x \right)=2{{\cos }^{2}}x+3, \forall x\in \mathbb{R}$, khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}$ bằng
A. $\dfrac{{{\pi }^{2}}+2}{8}$
B. $\dfrac{{{\pi }^{2}}+8\pi +8}{8}$
C. $\dfrac{{{\pi }^{2}}+8\pi +2}{8}$
D. $\dfrac{{{\pi }^{2}}+6\pi +8}{8}$
A. $\dfrac{{{\pi }^{2}}+2}{8}$
B. $\dfrac{{{\pi }^{2}}+8\pi +8}{8}$
C. $\dfrac{{{\pi }^{2}}+8\pi +2}{8}$
D. $\dfrac{{{\pi }^{2}}+6\pi +8}{8}$
Ta có $f\left( x \right)=\int{{f}'\left( x \right)dx}=\int{\left( 2{{\cos }^{2}}x+3 \right)dx}=\int{\left( 2.\dfrac{1+\cos 2x}{2}+3 \right)dx}$
$=\int{\left( \cos 2x+4 \right)dx}=\dfrac{1}{2}\sin 2x+4x+C$ do $f\left( 0 \right)=4\Rightarrow C=4$
Vậy $f\left( x \right)=\dfrac{1}{2}\sin 2x+4x+4$ nên $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{1}{2}\sin 2x+4x+4 \right)dx}$
$=\left( -\dfrac{1}{4}\cos 2x+2{{x}^{2}}+4x \right)\left| \begin{aligned}
& ^{\dfrac{\pi }{4}} \\
& _{0} \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}+8\pi +2}{8}$
$=\int{\left( \cos 2x+4 \right)dx}=\dfrac{1}{2}\sin 2x+4x+C$ do $f\left( 0 \right)=4\Rightarrow C=4$
Vậy $f\left( x \right)=\dfrac{1}{2}\sin 2x+4x+4$ nên $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{1}{2}\sin 2x+4x+4 \right)dx}$
$=\left( -\dfrac{1}{4}\cos 2x+2{{x}^{2}}+4x \right)\left| \begin{aligned}
& ^{\dfrac{\pi }{4}} \\
& _{0} \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}+8\pi +2}{8}$
Đáp án C.