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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm ${f}'\left( x \right)=cosx+1, \forall x\in \mathbb{R}$. Biết $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}\text{d}x=\dfrac{{{\pi }^{2}}}{8}+1$. Khi đó $f\left( \dfrac{\pi }{2} \right)$ bằng
A. $\dfrac{\pi }{2}$.
B. $\dfrac{\pi }{2}+1$.
C. $\dfrac{\pi }{2}-1$.
D. $1$.
A. $\dfrac{\pi }{2}$.
B. $\dfrac{\pi }{2}+1$.
C. $\dfrac{\pi }{2}-1$.
D. $1$.
Ta có: $f\left( x \right)=\int{{f}'\left( x \right)}\text{d}x=\int{\left( \text{cos}x+1 \right)}\text{d}x=\sin x+x+C$.
$\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}\text{d}x=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin x+x+C \right)}\text{d}x=\left. \left( -\cos x+\dfrac{{{x}^{2}}}{2}+Cx \right) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{8}+1$
$\Leftrightarrow \dfrac{{{\pi }^{2}}}{8}+\dfrac{\pi }{2}C+1=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow C=0$.
Vậy $f\left( x \right)=\sin x+x\Rightarrow f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}+1$.
$\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)}\text{d}x=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin x+x+C \right)}\text{d}x=\left. \left( -\cos x+\dfrac{{{x}^{2}}}{2}+Cx \right) \right|_{0}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{8}+1$
$\Leftrightarrow \dfrac{{{\pi }^{2}}}{8}+\dfrac{\pi }{2}C+1=\dfrac{{{\pi }^{2}}}{8}+1\Leftrightarrow C=0$.
Vậy $f\left( x \right)=\sin x+x\Rightarrow f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}+1$.
Đáp án B.