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Math Help - Derivative and Definite Integral question

  1. #1
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    Derivative and Definite Integral question

    Derivative:
    <br />
p(t) = t^-^3 - 3t\sqrt{t^3}<br />
    Is the square root of t^3 the same as t^1.5?

    Integrals(Express the following as limits, don't evaluate):
    As 0 goes to 4
    <br />
\int {\( (2x+1)^1^/^3} dx = <br />

    As 4 goes to 7
    <br />
\int {\(\sqrt{2t} dt = <br />
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  2. #2
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    Quote Originally Posted by nirva
    Derivative:
    <br />
p(t) = t^-^3 - 3t\sqrt{t^3}<br />
    Is the square root of t^3 the same as t^1.5?
    square root t^3 is pretty much just t^3*.5 (when a power is raised to a power, you multiply the powers). So yes. If nobody has helped you by morning, I'll do what I can. Until then, nap time!
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  3. #3
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    Quote Originally Posted by nirva
    Integrals(Express the following as limits, don't evaluate):
    As 0 goes to 4
    <br />
\int {\( (2x+1)^1^/^3} dx = <br />
    Use substitution rule. Let u=2x+1 then, the integral becomes,
    \frac{1}{2} \int^9_1 u^3 du the reason for the limits is because when x=0 then u=1 and when x=4 then u=9
    Quote Originally Posted by Nirva
    As 4 goes to 7
    <br />
\int {\(\sqrt{2t} dt = <br />
    Use subsitution x=2t then using the rule properly you have,
    \frac{1}{2}\int^{14}_{8}\sqrt{x}dx
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