Definite IntegrationHard
Question
If f(π) = 2 and 
sin x dx = 5, then f(0) is equal to :
(it is given that f(x) is continuous in [0, π])
sin x dx = 5, then f(0) is equal to : (it is given that f(x) is continuous in [0, π])
Options
A.7
B.3
C.5
D.1
Solution
5 = 
f(x) . sin x dx + 
sin x . f″(x) dx
= (f(x) . cos x)p0 +
f′(x) cos x dx + (sinx . f′(x)0p) - 
cos x f″(x) dx
5 = - f(n) . (-1) + f(0) + 0
f(0) = 5 - 2 = 3
f(x) . sin x dx + sin x . f″(x) dx= (f(x) . cos x)p0 +
f′(x) cos x dx + (sinx . f′(x)0p) - cos x f″(x) dx5 = - f(n) . (-1) + f(0) + 0
f(0) = 5 - 2 = 3
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