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Câu hỏi: Cho $\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x=5$. Tính $I=\int_0^{\dfrac{\pi}{2}}[f(x)+2 \sin x] \mathrm{d} x$.
A. $I=5+\pi$.
B. $I=7$.
C. $I=3$.
D. $I=5+\dfrac{\pi}{2}$.
A. $I=5+\pi$.
B. $I=7$.
C. $I=3$.
D. $I=5+\dfrac{\pi}{2}$.
Ta có:
$I=\int_0^{\dfrac{\pi}{2}}[f(x)+2 \sin x] \mathrm{d} x=\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x+2 \int_0^{\dfrac{\pi}{2}} \sin x \mathrm{~d} x=\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x-\left.2 \cos x\right|_0 ^{\dfrac{\pi}{2}}=5-$ 2. $(0-1)=7$
$I=\int_0^{\dfrac{\pi}{2}}[f(x)+2 \sin x] \mathrm{d} x=\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x+2 \int_0^{\dfrac{\pi}{2}} \sin x \mathrm{~d} x=\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x-\left.2 \cos x\right|_0 ^{\dfrac{\pi}{2}}=5-$ 2. $(0-1)=7$
Đáp án B.