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Câu hỏi: Cho hàm số $f(x)$ có $f\left( \dfrac{\pi }{2} \right)=4$ và ${f}'\left( x \right)=\dfrac{2}{{{\sin }^{2}}x}+1,\forall x\in \left( 0;\pi \right)$ khi đó $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{f\left( x \right)\text{d}x}$ bằng
A. $\dfrac{{{\pi }^{2}}}{2}$.
B. $2\pi $.
C. $\dfrac{8\pi -{{\pi }^{2}}}{2}$.
D. $\dfrac{{{\pi }^{2}}+2\pi }{2}$.
A. $\dfrac{{{\pi }^{2}}}{2}$.
B. $2\pi $.
C. $\dfrac{8\pi -{{\pi }^{2}}}{2}$.
D. $\dfrac{{{\pi }^{2}}+2\pi }{2}$.
Ta có: $\int{{f}'\left( x \right)\text{d}x}=\int{\left( \dfrac{2}{{{\sin }^{2}}x}+1 \right)}dx=-2\cot x+x+C=f(x)$.
$f\left( \dfrac{\pi }{2} \right)=4\Leftrightarrow C=4-\dfrac{\pi }{2}\Rightarrow f\left( x \right)=-2\cot x+x+4-\dfrac{\pi }{2}$.
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\left( -2\cot x+x+4-\dfrac{\pi }{2} \right)\text{d}x}==\left. \left[ -2\ln \left( \sin x \right)+\dfrac{{{x}^{2}}}{2}+4x-\dfrac{\pi }{2}x \right] \right|_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}=2\pi $.
$f\left( \dfrac{\pi }{2} \right)=4\Leftrightarrow C=4-\dfrac{\pi }{2}\Rightarrow f\left( x \right)=-2\cot x+x+4-\dfrac{\pi }{2}$.
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{f\left( x \right)\text{d}x}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}{\left( -2\cot x+x+4-\dfrac{\pi }{2} \right)\text{d}x}==\left. \left[ -2\ln \left( \sin x \right)+\dfrac{{{x}^{2}}}{2}+4x-\dfrac{\pi }{2}x \right] \right|_{\dfrac{\pi }{4}}^{\dfrac{3\pi }{4}}=2\pi $.
Đáp án B.