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Math Help - generalized version of Therorem 4

  1. #1
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    generalized version of Therorem 4

    F(x) = \int_1^{\ln ({x^2})} {\frac{1}<br />
{t}dt} <br />
    generalized theorem: F(x) = \int_{h(x)}^{g(x)} {f(t)dt} <br />
F'(x) = f(g(x))*g'(x) = f(h(x))*h'(x)<br />

    I tired:
    \frac{1}<br />
{{\ln ({x^2})}}*\frac{{2x}}<br />
{{{x^2}}} = \frac{{2x}}<br />
{{{x^2}\ln ({x^2})}}<br />
    F'(e) = \frac{{2e}}<br />
{{{e^2}\ln {{(e)}^2}}} = \frac{{2e}}<br />
{{2{e^2}}} = .368<br />
    is this correct?
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  2. #2
    Super Member redsoxfan325's Avatar
    Joined
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    Swampscott, MA
    Posts
    943
    Quote Originally Posted by genlovesmusic09 View Post
    F(x) = \int_1^{\ln ({x^2})} {\frac{1}<br />
{t}dt} <br />
    generalized theorem: F(x) = \int_{h(x)}^{g(x)} {f(t)dt} <br />
F'(x) = f(g(x))*g'(x) = f(h(x))*h'(x)<br />

    I tired:
    \frac{1}<br />
{{\ln ({x^2})}}*\frac{{2x}}<br />
{{{x^2}}} = \frac{{2x}}<br />
{{{x^2}\ln ({x^2})}}<br />
    F'(e) = \frac{{2e}}<br />
{{{e^2}\ln {{(e)}^2}}} = \frac{{2e}}<br />
{{2{e^2}}} = .368<br />
    is this correct?
    Yes, but your professor will probably want it written as \frac{1}{e} as opposed to a rounded decimal.
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