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Câu hỏi: Cho hàm số $f\left( x \right)$. Biết $f\left( 0 \right)=4$ và ${f}'\left( x \right)=2{{\sin }^{2}}x+3,\forall \in \mathbb{R}$, khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}$ bằng
A. $\dfrac{{{\pi }^{2}}-2}{8}$
B. $\dfrac{{{\pi }^{2}}+8\pi -8}{8}$
C. $\dfrac{{{\pi }^{2}}+8\pi -2}{8}$
D. $\dfrac{3{{\pi }^{2}}+2\pi -3}{8}$
A. $\dfrac{{{\pi }^{2}}-2}{8}$
B. $\dfrac{{{\pi }^{2}}+8\pi -8}{8}$
C. $\dfrac{{{\pi }^{2}}+8\pi -2}{8}$
D. $\dfrac{3{{\pi }^{2}}+2\pi -3}{8}$
$f\left( x \right)=\int{{f}'\left( x \right)dx}=\int{\left( 2{{\sin }^{2}}x+3 \right)dx=\int{\left( 2.\dfrac{1-\cos 2x}{2}+3 \right)dx}}$
$=\int{\left( 4-\cos 2x \right)dx=4x-\dfrac{\sin 2x}{2}+C}$
Do $f\left( 0 \right)=4\Rightarrow C=4\Rightarrow f\left( x \right)=4x-\dfrac{\sin 2x}{2}+4$
Khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 4x-\dfrac{\sin 2x}{2}+4 \right)dx}=\left. \left( 2{{x}^{2}}+\dfrac{\cos 2x}{4}+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{{{\pi }^{2}}+8\pi -2}{8}$.
$=\int{\left( 4-\cos 2x \right)dx=4x-\dfrac{\sin 2x}{2}+C}$
Do $f\left( 0 \right)=4\Rightarrow C=4\Rightarrow f\left( x \right)=4x-\dfrac{\sin 2x}{2}+4$
Khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 4x-\dfrac{\sin 2x}{2}+4 \right)dx}=\left. \left( 2{{x}^{2}}+\dfrac{\cos 2x}{4}+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{{{\pi }^{2}}+8\pi -2}{8}$.
Đáp án C.