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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 $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}$ bằng
A. $\dfrac{{{\pi }^{2}}+15\pi }{16}$
B. $\dfrac{{{\pi }^{2}}+16\pi -16}{16}$
C. $\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
D. $\dfrac{{{\pi }^{2}}-4}{16}$
A. $\dfrac{{{\pi }^{2}}+15\pi }{16}$
B. $\dfrac{{{\pi }^{2}}+16\pi -16}{16}$
C. $\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
D. $\dfrac{{{\pi }^{2}}-4}{16}$
Ta có $f\left( x \right)=\int{\left( 2{{\sin }^{2}}x+1 \right)dx=\int{\left( 2-2\cos x \right)dx=2x-\dfrac{1}{2}\sin 2x+C}}$
Vì $f\left( 0 \right)=4\Rightarrow C=4$ do đó $f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+4$
Suy ra $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x-\dfrac{1}{2}\sin 2x+4 \right)}}dx$
$={{x}^{2}}+\dfrac{1}{4}cos2x+4x\left| _{_{_{0}}}^{^{^{\dfrac{\pi }{4}}}} \right.=\dfrac{{{\pi }^{2}}}{16}+\pi -\dfrac{1}{4}=\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
Vì $f\left( 0 \right)=4\Rightarrow C=4$ do đó $f\left( x \right)=2x-\dfrac{1}{2}\sin 2x+4$
Suy ra $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 2x-\dfrac{1}{2}\sin 2x+4 \right)}}dx$
$={{x}^{2}}+\dfrac{1}{4}cos2x+4x\left| _{_{_{0}}}^{^{^{\dfrac{\pi }{4}}}} \right.=\dfrac{{{\pi }^{2}}}{16}+\pi -\dfrac{1}{4}=\dfrac{{{\pi }^{2}}+16\pi -4}{16}$
Đáp án C.