Google
Câu hỏi: Cho hàm số $f\left( x \right)$. Biết $f\left( 0 \right)=4$ và $f'\left( x \right)=2{{\sin }^{2}}x+3,\forall x\in \mathbb{R}$, 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. $\dfrac{3{{\pi }^{2}}+2\pi -3}{8}.$
Ta có:
$\int{f'\left( x \right)dx}=\int{(2{{\sin }^{2}}x+3)dx}=\int{(1-\cos 2x+3)dx}=\int{(4-\cos 2x)dx}=4x-\dfrac{1}{2}\sin 2x+C$
Ta có $f\left( 0 \right)=4$ nên $4.0-\dfrac{1}{2}\sin 0+C=4\Leftrightarrow C=4$. Nên $f\left( x \right)=4x-\dfrac{1}{2}\sin 2x+4$
$\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 4x-\dfrac{1}{2}\sin 2x+4 \right)dx=\left. \left( 2{{x}^{2}}+\dfrac{1}{4}\cos 2x+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{{{\pi }^{2}}+8\pi -2}{8}}$.
$\int{f'\left( x \right)dx}=\int{(2{{\sin }^{2}}x+3)dx}=\int{(1-\cos 2x+3)dx}=\int{(4-\cos 2x)dx}=4x-\dfrac{1}{2}\sin 2x+C$
Ta có $f\left( 0 \right)=4$ nên $4.0-\dfrac{1}{2}\sin 0+C=4\Leftrightarrow C=4$. Nên $f\left( x \right)=4x-\dfrac{1}{2}\sin 2x+4$
$\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 4x-\dfrac{1}{2}\sin 2x+4 \right)dx=\left. \left( 2{{x}^{2}}+\dfrac{1}{4}\cos 2x+4x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{{{\pi }^{2}}+8\pi -2}{8}}$.
Đáp án C.