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Please check my answer and tell me where I am wrong? The answer according to the answer key for thi question is 2 0 ∫ pi/2 (1+sin x)1/2dx = 0 ∫ pi/2 (sin x/2 + cos x/2)dx = [-2cos x/2 + 2sin x/2] 0 pi/2 = -2cos pi/4 + 2sin pi/4 = -√2 + √2 = 0

Please check my answer and tell me where I am wrong? The answer according to the answer key for thi question is 2


0pi/2(1+sin x)1/2dx = 0pi/2(sin x/2 + cos x/2)dx = [-2cos x/2 + 2sin x/2]0pi/2  = -2cos pi/4 + 2sin pi/4 = -√2 + √2 = 0

Grade:upto college level

2 Answers

akshay akshay
17 Points
9 years ago

you have not substituted the lower limit

Har Simrat Singh
42 Points
9 years ago

its right upto  the step of subs the limits 

ie -2cospi/4+ 2sin pi/4 -(-2 cos 0 + 2sin0)

= 0 - (-2) = 2

remember cos 0 is 1 

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