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Câu hỏi: Cho hàm số $f\left( x \right)$. Biết $f\left( 0 \right)=4$ và ${f}'\left( x \right)=2{{\sin }^{2}}x+1,\forall x\in \mathbb{R}$, khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{d}x}$ bằng
A. $\dfrac{{{\pi }^{2}}-4}{16}.$
B. $\dfrac{{{\pi }^{2}}+15\pi }{16}.$
C. $\dfrac{{{\pi }^{2}}+16\pi -16}{16}.$
D. $\dfrac{{{\pi }^{2}}+16\pi -4}{16}.$
A. $\dfrac{{{\pi }^{2}}-4}{16}.$
B. $\dfrac{{{\pi }^{2}}+15\pi }{16}.$
C. $\dfrac{{{\pi }^{2}}+16\pi -16}{16}.$
D. $\dfrac{{{\pi }^{2}}+16\pi -4}{16}.$
Ta có $\int{{f}'\left( x \right)}\text{d}x=\int{\left( 2{{\sin }^{2}}x+1 \right)\text{d}x=\int{\left( 2-\cos 2x \right)}}\text{d}x=2x-\dfrac{1}{2}\sin 2x+C.$
Suy ra $f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+C.$
Vì $f\left( 0 \right)=4\Rightarrow C=4$ hay $f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+4.$
Khi đó: $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x-\dfrac{1}{2}\sin 2x+4 \right)\text{d}x}$
$=\left( {{x}^{2}}+\dfrac{1}{4}\cos 2x+4x \right)\left| \begin{aligned}
& \dfrac{\pi }{4} \\
& 0 \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}}{16}+\pi -\dfrac{1}{4}=\dfrac{{{\pi }^{2}}+16\pi -4}{16}.$
Suy ra $f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+C.$
Vì $f\left( 0 \right)=4\Rightarrow C=4$ hay $f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+4.$
Khi đó: $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x-\dfrac{1}{2}\sin 2x+4 \right)\text{d}x}$
$=\left( {{x}^{2}}+\dfrac{1}{4}\cos 2x+4x \right)\left| \begin{aligned}
& \dfrac{\pi }{4} \\
& 0 \\
\end{aligned} \right.=\dfrac{{{\pi }^{2}}}{16}+\pi -\dfrac{1}{4}=\dfrac{{{\pi }^{2}}+16\pi -4}{16}.$
Đáp án D.