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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
11 years ago

you have not substituted the lower limit

Har Simrat Singh
42 Points
11 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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