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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

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  1. anonymous
    • 5 years ago
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    -2

  2. roadjester
    • 5 years ago
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    not -2

  3. roadjester
    • 5 years ago
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    that's what i got too

  4. anonymous
    • 5 years ago
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    would 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

  5. roadjester
    • 5 years ago
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    ok, the answer is 0, can u elaborate please?

  6. anonymous
    • 5 years ago
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    note the integral of sinx is -cosx

  7. roadjester
    • 5 years ago
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    ok, i can follow that

  8. roadjester
    • 5 years ago
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    i also know that the integral of cosx is sinx

  9. roadjester
    • 5 years ago
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    it's the evaluation idk how to do

  10. anonymous
    • 5 years ago
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    f(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

  11. roadjester
    • 5 years ago
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    but 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?

  12. anonymous
    • 5 years ago
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    so 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]

  13. roadjester
    • 5 years ago
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    COOL! THANKS MAN (or girl/woman/miss, don't know your gender sorry) YOU ROCK!

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