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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: $f\left( x \right)=\int{{f}'\left( x \right)dx}=\int{\left( 2{{\cos }^{2}}x+1 \right)dx}=\int{\left( 2+\cos 2x \right)}dx=2x+\dfrac{1}{2}\sin 2x+C$.
Theo bài: $f\left( 0 \right)=4\Leftrightarrow 2.0+\dfrac{1}{2}.\sin 0+C=4\Leftrightarrow C=4$. Suy ra $f\left( x \right)=2x+\dfrac{1}{2}\sin 2x+4$.
Vậy:
$\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x+\dfrac{1}{2}\sin 2x+4 \right)dx}=\left. \left( {{x}^{2}}-\dfrac{\cos 2x}{4}+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\left( \dfrac{{{\pi }^{2}}}{16}+\pi \right)-\left( -\dfrac{1}{4} \right)=\dfrac{{{\pi }^{2}}+16\pi +4}{16}$.
Theo bài: $f\left( 0 \right)=4\Leftrightarrow 2.0+\dfrac{1}{2}.\sin 0+C=4\Leftrightarrow C=4$. Suy ra $f\left( x \right)=2x+\dfrac{1}{2}\sin 2x+4$.
Vậy:
$\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x+\dfrac{1}{2}\sin 2x+4 \right)dx}=\left. \left( {{x}^{2}}-\dfrac{\cos 2x}{4}+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\left( \dfrac{{{\pi }^{2}}}{16}+\pi \right)-\left( -\dfrac{1}{4} \right)=\dfrac{{{\pi }^{2}}+16\pi +4}{16}$.
Đáp án C.