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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm trên $\mathbb{R}$ là ${f}'\left( x \right)=\text{sin}x\text{cos}x$ và $f\left( 0 \right)=1$. Tính tích phân $I=\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)}dx$
A. $I=\dfrac{\pi -4}{2}$.
B. $I=\dfrac{3\pi -4}{8}$.
C. $I=\dfrac{3\pi +2}{16}$.
D. $I=\dfrac{5\pi -2}{16}$.
A. $I=\dfrac{\pi -4}{2}$.
B. $I=\dfrac{3\pi -4}{8}$.
C. $I=\dfrac{3\pi +2}{16}$.
D. $I=\dfrac{5\pi -2}{16}$.
Ta có: $f\left( x \right)=\int{\left( {f}'\left( x \right) \right)}dx=\int{\text{sin}x\text{cos}x}dx=\int{\sin x}d\left( \sin x \right)=\dfrac{{{\sin }^{2}}x}{2}+C$.
Mà $f\left( 0 \right)=1 \Rightarrow \dfrac{{{\sin }^{2}}0}{2}+C=1 \Rightarrow C=1$.
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)}dx=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{{{\sin }^{2}}x}{2}+1 \right)}dx=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{5}{4}-\dfrac{\cos 2x}{4} \right)}dx=\dfrac{5}{4}\left. x \right|_{0}^{\dfrac{\pi }{4}}-\dfrac{1}{8}\left. \sin 2x \right|_{0}^{\dfrac{\pi }{4}}-=\dfrac{5\pi -2}{16}$.
Mà $f\left( 0 \right)=1 \Rightarrow \dfrac{{{\sin }^{2}}0}{2}+C=1 \Rightarrow C=1$.
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)}dx=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{{{\sin }^{2}}x}{2}+1 \right)}dx=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \dfrac{5}{4}-\dfrac{\cos 2x}{4} \right)}dx=\dfrac{5}{4}\left. x \right|_{0}^{\dfrac{\pi }{4}}-\dfrac{1}{8}\left. \sin 2x \right|_{0}^{\dfrac{\pi }{4}}-=\dfrac{5\pi -2}{16}$.
Đáp án D.