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