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Câu hỏi: Cho hàm số $f(x)$.Biết $f(0)=4$ và ${f}'(x)=2{{\cos }^{2}}x+3, \forall x\in \mathbb{R}$, khi đó $\int\limits_{0}^{\dfrac{\pi }{4}}{f(x)dx}$ bằng?
A. $\dfrac{{{\pi }^{2}}+2}{8}$.
B. $\dfrac{{{\pi }^{2}}+8\pi +8}{8}$.
C. $\dfrac{{{\pi }^{2}}+8\pi +2}{8}$.
D. $\dfrac{{{\pi }^{2}}+6\pi +8}{8}$.
A. $\dfrac{{{\pi }^{2}}+2}{8}$.
B. $\dfrac{{{\pi }^{2}}+8\pi +8}{8}$.
C. $\dfrac{{{\pi }^{2}}+8\pi +2}{8}$.
D. $\dfrac{{{\pi }^{2}}+6\pi +8}{8}$.
Ta có $f(x)=\int{{{f}^{^{,}}}(x)dx}=\int{(2{{\cos }^{2}}x+3)dx}$ $=\int{(2.\dfrac{1+\cos 2x}{2}+3)dx}$.
$=\int{(\cos 2x+4)dx}$ = $\dfrac{1}{2}\sin 2x+4x+C$ do $f(0)=4\Rightarrow C=4$.
Vậy $f(x)=\dfrac{1}{2}\sin 2x+4x+4$ nên $\int\limits_{0}^{\dfrac{\pi }{4}}{f(x)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{(\dfrac{1}{2}\sin 2x+4x+4})dx$.
$=\left. (-\dfrac{1}{4}\cos 2x+2{{x}^{2}}+4x) \right|_{0}^{\dfrac{\pi }{4}}$ $=\dfrac{{{\pi }^{2}}+8\pi +2}{8}$.
$=\int{(\cos 2x+4)dx}$ = $\dfrac{1}{2}\sin 2x+4x+C$ do $f(0)=4\Rightarrow C=4$.
Vậy $f(x)=\dfrac{1}{2}\sin 2x+4x+4$ nên $\int\limits_{0}^{\dfrac{\pi }{4}}{f(x)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{(\dfrac{1}{2}\sin 2x+4x+4})dx$.
$=\left. (-\dfrac{1}{4}\cos 2x+2{{x}^{2}}+4x) \right|_{0}^{\dfrac{\pi }{4}}$ $=\dfrac{{{\pi }^{2}}+8\pi +2}{8}$.
Đáp án C.