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Câu hỏi: Cho hàm số $f\left( x \right)$. Biết $f\left( 0 \right)=4$ và ${f}'\left( x \right)=2-\cos 2x,\forall x\in \mathbb{R}$, khi đó: $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}$ bằng
A. $\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
B. $\dfrac{{{\pi }^{2}}-4}{16}$
C. $\dfrac{{{\pi }^{2}}+15\pi }{16}$
D. $\dfrac{{{\pi }^{2}}+16\pi -16}{16}$
A. $\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
B. $\dfrac{{{\pi }^{2}}-4}{16}$
C. $\dfrac{{{\pi }^{2}}+15\pi }{16}$
D. $\dfrac{{{\pi }^{2}}+16\pi -16}{16}$
Ta có: $\int{{f}'\left( x \right)dx=\int{\left( 2-\cos 2x \right)dx=2x-\dfrac{1}{2}\sin 2x+{{C}_{1}}}}$. Do $f\left( 0 \right)=4$ nên: ${{C}_{1}}=4$
$\Rightarrow f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+4\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x-\dfrac{1}{2}\sin 2x+4 \right)dx=}\left. \left( {{x}^{2}}+\dfrac{1}{4}\cos 2x+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
$\Rightarrow f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+4\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x-\dfrac{1}{2}\sin 2x+4 \right)dx=}\left. \left( {{x}^{2}}+\dfrac{1}{4}\cos 2x+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
Đáp án A.