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Câu hỏi: Cho hàm số $f\left( x \right)$ có $f\left( 0 \right)=4$ và ${f}'\left( x \right)=2{{\cos }^{2}}x+1$, $\forall x\in \mathbb{R}$. Khi đó $\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{d}x}$ bằng
A. $\frac{{{\pi }^{2}}+16\pi +16}{16}$.
B. $\frac{{{\pi }^{2}}+4}{16}$.
C. $\frac{{{\pi }^{2}}+14\pi }{16}$.
D. $\frac{{{\pi }^{2}}+16\pi +4}{16}$.
A. $\frac{{{\pi }^{2}}+16\pi +16}{16}$.
B. $\frac{{{\pi }^{2}}+4}{16}$.
C. $\frac{{{\pi }^{2}}+14\pi }{16}$.
D. $\frac{{{\pi }^{2}}+16\pi +4}{16}$.
Ta có $\int{{f}'\left( x \right)\text{d}x}=\int{\left( 2{{\cos }^{2}}x+1 \right)\text{d}x}$
$=\int{\left( 2+\cos 2x \right)\text{d}x}=2x+\frac{1}{2}\sin 2x+C$.
Do đó $f(x)=2x+\frac{1}{2}\sin 2x+C$.
Vì $f\left( 0 \right)=4$ nên $C=4$. Do đó $f\left( x \right)=2x+\frac{1}{2}\sin 2x+4$.
Vậy $\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{d}x}$ $=\int\limits_{0}^{\frac{\pi }{4}}{\left( 2x+\frac{1}{2}\sin 2x+4 \right)\text{d}x}$ $=\left. \left( {{x}^{2}}-\frac{1}{4}\cos 2x+4x \right) \right|_{0}^{\frac{\pi }{4}}$ $=\frac{{{\pi }^{2}}+16\pi +4}{16}$.
$=\int{\left( 2+\cos 2x \right)\text{d}x}=2x+\frac{1}{2}\sin 2x+C$.
Do đó $f(x)=2x+\frac{1}{2}\sin 2x+C$.
Vì $f\left( 0 \right)=4$ nên $C=4$. Do đó $f\left( x \right)=2x+\frac{1}{2}\sin 2x+4$.
Vậy $\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{d}x}$ $=\int\limits_{0}^{\frac{\pi }{4}}{\left( 2x+\frac{1}{2}\sin 2x+4 \right)\text{d}x}$ $=\left. \left( {{x}^{2}}-\frac{1}{4}\cos 2x+4x \right) \right|_{0}^{\frac{\pi }{4}}$ $=\frac{{{\pi }^{2}}+16\pi +4}{16}$.
Đáp án D.