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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 0 \right)=1$ và ${f}'\left( x \right)=2{{\sin }^{2}}x-3,\forall x\in \mathbb{R}.$ Tích phân $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)}dx$ bằng
A. $\dfrac{{{\pi }^{2}}-4\pi +4}{16}.$
B. $-\dfrac{{{\pi }^{2}}-4\pi +4}{16}.$
C. $\dfrac{{{\pi }^{2}}+4\pi -4}{16}.$
D. $-\dfrac{{{\pi }^{2}}+4\pi -4}{16}.$
A. $\dfrac{{{\pi }^{2}}-4\pi +4}{16}.$
B. $-\dfrac{{{\pi }^{2}}-4\pi +4}{16}.$
C. $\dfrac{{{\pi }^{2}}+4\pi -4}{16}.$
D. $-\dfrac{{{\pi }^{2}}+4\pi -4}{16}.$
Ta có $f\left( x \right)=\int{\left( 2{{\sin }^{2}}x-3 \right)dx}=\int{\left( 1-\cos 2x-3 \right)dx}=-\dfrac{\sin 2x}{2}-2x+C.$
Mà $f\left( 0 \right)=1\Rightarrow C=1\Rightarrow f\left( x \right)=-\dfrac{1}{2}\sin 2x-2x+1.$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( -\dfrac{1}{2}\sin 2x-2x+1 \right)}dx=\left( \dfrac{1}{4}\cos 2x-{{x}^{2}}+x \right)\left| _{\begin{smallmatrix}
\\
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
\dfrac{\pi }{4} \\
\end{smallmatrix}} \right.$
$=-\dfrac{{{\pi }^{2}}}{16}+\dfrac{\pi }{4}-\dfrac{1}{4}=-\dfrac{{{\pi }^{2}}-4\pi +4}{16}.$
Mà $f\left( 0 \right)=1\Rightarrow C=1\Rightarrow f\left( x \right)=-\dfrac{1}{2}\sin 2x-2x+1.$
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( -\dfrac{1}{2}\sin 2x-2x+1 \right)}dx=\left( \dfrac{1}{4}\cos 2x-{{x}^{2}}+x \right)\left| _{\begin{smallmatrix}
\\
\\
0
\end{smallmatrix}}^{\begin{smallmatrix}
\dfrac{\pi }{4} \\
\end{smallmatrix}} \right.$
$=-\dfrac{{{\pi }^{2}}}{16}+\dfrac{\pi }{4}-\dfrac{1}{4}=-\dfrac{{{\pi }^{2}}-4\pi +4}{16}.$
Đáp án B.