Google
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. $\dfrac{{{\pi }^{2}}+6{{\pi }^{2}}+8}{8}$
A. $\dfrac{{{\pi }^{2}}+2}{8}$
B. $\dfrac{{{\pi }^{2}}+8\pi +8}{8}$
C. $\dfrac{{{\pi }^{2}}+8\pi +2}{8}$
D. $\dfrac{{{\pi }^{2}}+6{{\pi }^{2}}+8}{8}$
Ta có ${f}'\left( x \right)=2{{\cos }^{2}}x+3=\cos 2x+4\Rightarrow f\left( x \right)=\int{{f}'\left( x \right)dx}=\dfrac{1}{2}\sin 2x+4x+C$
Mà $f\left( 0 \right)=4\Rightarrow C=4\Rightarrow f\left( x \right)=\dfrac{1}{2}\sin 2x+4x+4$
Do đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{1}{2}\sin 2x+4x+4 \right)dx}=\left( -\dfrac{1}{4}\cos 2x+2{{x}^{2}}+4x \right)\left| \begin{aligned}
& ^{\dfrac{\pi }{4}} \\
& _{0} \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}+8\pi +2}{8}$.
Mà $f\left( 0 \right)=4\Rightarrow C=4\Rightarrow f\left( x \right)=\dfrac{1}{2}\sin 2x+4x+4$
Do đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{1}{2}\sin 2x+4x+4 \right)dx}=\left( -\dfrac{1}{4}\cos 2x+2{{x}^{2}}+4x \right)\left| \begin{aligned}
& ^{\dfrac{\pi }{4}} \\
& _{0} \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}+8\pi +2}{8}$.
Đáp án C.