Question
∫π/2−π/2cos xdx
Answer: Option B
:
B
I=∫π/2−π/2cosxdx=sinx|π/2−π/2=sinπ2−sin(−π2)=1+1=2∵∫baf(x)dx=I(x)∣∣∣ba=I(b)−I(a)
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:
B
I=∫π/2−π/2cosxdx=sinx|π/2−π/2=sinπ2−sin(−π2)=1+1=2∵∫baf(x)dx=I(x)∣∣∣ba=I(b)−I(a)
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