generalized version of Therorem 4

**genlovesmusic09** · October 4th 2009, 10:17 AM

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

I tired:
$\frac{1} {{\ln ({x^2})}}*\frac{{2x}} {{{x^2}}} = \frac{{2x}} {{{x^2}\ln ({x^2})}} $
$F'(e) = \frac{{2e}} {{{e^2}\ln {{(e)}^2}}} = \frac{{2e}} {{2{e^2}}} = .368 $
is this correct?

**redsoxfan325** · October 4th 2009, 10:39 AM

Originally Posted by genlovesmusic09

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

I tired:
$\frac{1} {{\ln ({x^2})}}*\frac{{2x}} {{{x^2}}} = \frac{{2x}} {{{x^2}\ln ({x^2})}} $
$F'(e) = \frac{{2e}} {{{e^2}\ln {{(e)}^2}}} = \frac{{2e}} {{2{e^2}}} = .368 $
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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