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roadjester
 5 years ago
integral(f(x))dx from 0 to pi where f(x)= sinx if 0<=x<pi/2 and cosx if pi/2<=x<=pi
roadjester
 5 years ago
integral(f(x))dx from 0 to pi where f(x)= sinx if 0<=x<pi/2 and cosx if pi/2<=x<=pi

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roadjester
 5 years ago
Best ResponseYou've already chosen the best response.0that's what i got too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would you not just add the integral of sin(x) from 0 to pi/2 to the integral of cos(x) from pi/2 to pi? which is 1 + (1) = 0, this is a piecewise defined function

roadjester
 5 years ago
Best ResponseYou've already chosen the best response.0ok, the answer is 0, can u elaborate please?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0note the integral of sinx is cosx

roadjester
 5 years ago
Best ResponseYou've already chosen the best response.0ok, i can follow that

roadjester
 5 years ago
Best ResponseYou've already chosen the best response.0i also know that the integral of cosx is sinx

roadjester
 5 years ago
Best ResponseYou've already chosen the best response.0it's the evaluation idk how to do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x) = sin(x) for 0< x </2 and f(x) = cos(x) for pi/2< x <pi so in evaluating the integral of sin(x), which is cos(x) you get: cos(pi/2)  (cos(0)) which is 0  (1) which is 1. then evaluate the integral of cos(x) which is sin(x): sin(pi)  sin(pi/2) which is 1  0 = 1 add the values and you will get 0

roadjester
 5 years ago
Best ResponseYou've already chosen the best response.0but aren't the 0, pi/2, and pi, restrictions of the graph? I see your logic but the question was integral of f(x)dx where f(x)=sin(x) if 0< = x < pi/2 and f(x)=cos(x) if pi/2< = x <= pi or is that irrelevant?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this says, f(x) = sin(x) where x is greater than or equal to x and less than pi/2, and f(x) = cos(x) where x is greater than or equal to pi/2 and less than or equal to pi. all values you mentioned are defined in the graph (0,pi/2,pi) you are finding area under curve from [0,pi/2) and [pi/2,pi]

roadjester
 5 years ago
Best ResponseYou've already chosen the best response.0COOL! THANKS MAN (or girl/woman/miss, don't know your gender sorry) YOU ROCK!
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