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cander3

  • one year ago

If the definite integral between [0,x] of f(t)dt=e^(3x)cos(x)+c, what is the value of c?

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  1. xkrromania
    • one year ago
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    If f(t)dt=e^(3x)cos(x)|0 to x then using the Fundamental Theorem of calculus you get f(t)dt=[e^(3x)cos(x)+c]-[e^(0)*cos(0)+c]=e^3x*cos(x)-1+(c-c)=e^3x*cos(x)-1. The definite integral doesn't have a constant simply because the c cancels when you subtract.

  2. dan815
    • one year ago
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    ^

  3. Ahmad1
    • one year ago
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    I think you can simply find the answer by setting x=0 , you will find c=-1

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