Fundamental Theorem of Calculus?

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. Thank You.

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  • 1 decade ago
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    Well, you have:

     3

     ∫ u³ / (1 - u² ) du

    1-3x

    You can't use the theorem until the x part is in the top limit of integration, so first you have to:

     1-3x

    - ∫ u³ / (1 - u² ) du

     3

    Remember, if you switch the two limits of integration, you need to add a negative sign. So then you remember the rule, which is:

           g(x)

    (d/dx) ∫ f(u) du = f(g(x)) g'(x)

             a

    for any functions f and g, and any constant u. The g'(x) that comes out there is essentially the chain rule in effect here.

    Applying it to our case, you have:

    g(x) = 1-3x

    g'(x) = -3

    f(g(x)) = (1-3x)³ / ( 1 - (1-3x)² )

    So your answer is:

    - f(g(x)) g'(x)

    = 3 (1-3x)³ / (1 - (1-3x)² )

    = 3 (1-3x)³ / ( 1 - (1 - 6x + 9x² ) )

    = 3 (1-3x)³ / ( 6x - 9x² )

    = (1-3x)³ / ( 2x - 3x² )

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