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Câu hỏi: Cho $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=5$. Tính $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ f\left( x \right)+2\sin x \right]\text{d}x}$.
A. $I=5$
B. $I=5+\dfrac{\pi }{2}$
C. $I=3$
D. $I=7.$
A. $I=5$
B. $I=5+\dfrac{\pi }{2}$
C. $I=3$
D. $I=7.$
Ta có
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ f\left( x \right) +2\sin x \right] \text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right) \text{d}x \text{+2}\int\limits_{0}^{\tfrac{\pi }{2}}{\sin x \text{d}x}}}$ $=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)} \text{d}x -\left. 2\cos x \right|_{0}^{\dfrac{\pi }{2}}=5-2\left( 0-1 \right)=7$.
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ f\left( x \right) +2\sin x \right] \text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right) \text{d}x \text{+2}\int\limits_{0}^{\tfrac{\pi }{2}}{\sin x \text{d}x}}}$ $=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)} \text{d}x -\left. 2\cos x \right|_{0}^{\dfrac{\pi }{2}}=5-2\left( 0-1 \right)=7$.
Đáp án D.