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Math Help - Fundamental Theorem of Caluculus

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    Fundamental Theorem of Caluculus

    x
    G(x)=∫ t((3t^2+4)^1/2)dt
    4

    Use the Fundamental Theorem of Calculus to find an algebraic expression for G(x)
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    Re: Fundamental Theorem of Caluculus

    Quote Originally Posted by irock3337 View Post
    x
    G(x)=∫ t((3t^2+4)^1/2)dt
    4
    Use the Fundamental Theorem of Calculus to find an algebraic expression for G(x)

    What is the derivative of \tfrac{1}{9}(3t^2+4)^{3/2}~?
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    Re: Fundamental Theorem of Caluculus

    t(3t^2+4)^2 or is it t(3t^2+4)^1/2
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    Re: Fundamental Theorem of Caluculus

    Quote Originally Posted by irock3337 View Post
    t(3t^2+4)^2 or is it t(3t^2+4)^1/2
    It is t(3t^2+4)^{1/2}
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    Re: Fundamental Theorem of Caluculus

    So then would I just plug x and 4 into the equation and subtract them from one another to get the answer?
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    Re: Fundamental Theorem of Caluculus

    Quote Originally Posted by irock3337 View Post
    x
    G(x)=∫ t((3t^2+4)^1/2)dt
    4

    Use the Fundamental Theorem of Calculus to find an algebraic expression for G(x)
    If you were going to integrate this directly, first rewrite this integral as \displaystyle \begin{align*} \int_4^x{t\left( 3t^2 + 4 \right)^{\frac{1}{2}}\,dt} = \frac{1}{6}\int_4^x{6t\left( 3t^2 + 4 \right)^{\frac{1}{2}}\,dt} \end{align*} and then use a substitution \displaystyle \begin{align*} u = 3t^2 + 4 \implies du = 6t\,dt \end{align*}, noting that when \displaystyle \begin{align*} t = 4, u = 52 \end{align*} and when \displaystyle \begin{align*} t = x, u = 3x^2 + 4 \end{align*}, and the integral becomes

    \displaystyle \begin{align*} \frac{1}{6}\int_4^x{6t \left( 3t^2 + 4 \right)^{\frac{1}{2}}\,dt} &= \frac{1}{6}\int_{52}^{3x^2 + 4}{u^{\frac{1}{2}}\,du} \\ &= \frac{1}{6}\left[ \frac{2}{3} u^{\frac{3}{2}} \right]_{52}^{3x^2 + 4} \\ &= \frac{1}{9} \left[ \left( 3x^2 + 4 \right)^{\frac{3}{2}} - 52^{\frac{3}{2}} \right] \end{align*}
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