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anonymous

  • 4 years ago

If the average rate of change of F on [1,3] is k, find ∫sin(t^2) dt [1,3] in terms of k

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  1. JamesJ
    • 4 years ago
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    ...you mean where \[ F(t) = \int \sin(t^2) \ dt \]

  2. anonymous
    • 4 years ago
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    yes i think this is a trap rule Q

  3. JamesJ
    • 4 years ago
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    or where \( F(t) = \sin(t^2) \), or what? What's your definition of F?

  4. anonymous
    • 4 years ago
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    the first...the integral of

  5. anonymous
    • 4 years ago
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    F(t)=∫sin(t^2) dt

  6. JamesJ
    • 4 years ago
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    Well, by the Fundamental Theorem of Calculus, the rate of change of F, is the derivative dF/dt is given by \[ dF/dt = \sin(t^2) \] Now integrals find averages of things. The average of a function \( f(t) \) over an interval [a,b] is given by \[ \frac{1}{b-a} \int_a^b f(t) \ dt \] You're told that \[ \frac{1}{3-1} \int_1^3 \sin(t^2) \ dt = k \] Hence ...

  7. JamesJ
    • 4 years ago
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    Hence what must \[ \int_1^3 \sin(t^2) \ dt \] be equal to?

  8. anonymous
    • 4 years ago
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    1/2 of the approximation of the integral?

  9. JamesJ
    • 4 years ago
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    No need to approximate, none at all.

  10. anonymous
    • 4 years ago
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    k

  11. anonymous
    • 4 years ago
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    they ask for it in terms of k

  12. JamesJ
    • 4 years ago
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    Yes ... read again the last equations I wrote up there for you.

  13. anonymous
    • 4 years ago
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    1/2*k

  14. JamesJ
    • 4 years ago
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    No

  15. anonymous
    • 4 years ago
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    2K

  16. JamesJ
    • 4 years ago
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    Yes

  17. anonymous
    • 4 years ago
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    ok....thank you so much for walking me thru it

  18. JamesJ
    • 4 years ago
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    ok

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