fundamental theorem of calculus

**gracy** · November 23rd 2006, 05:59 AM

compute the following
(a) d/dx[ integrate(x to e^(x^2)) (ln(t))dt]
(b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]

**topsquark** · November 23rd 2006, 06:26 AM

Originally Posted by gracy

compute the following
(a) d/dx[ integrate(x to e^(x^2)) (ln(t))dt]

$\frac{d}{dx} \int_x^{e^{x^2}} dt \, ln(t)$

= $\frac{d}{dx} \left ( t \cdot ln(t) - t \right )|_x^{e^{x^2}}$

= $\frac{d}{dx} \left ( e^{x^2} \cdot x^2 - e^{x^2} - x \cdot ln(x) + x \right )$

= $2xe^{x^2} + 2x^3e^{x^2} - 2xe^{x^2} - ln(x) - x \cdot \frac{1}{x} + 1$

= $2x^3e^{x^2} - ln(x)$

-Dan

**ThePerfectHacker** · November 23rd 2006, 06:29 AM

Originally Posted by gracy

(b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]

Again I am so angry that you do not mention a function on what domain.
But since,
$e^{x^2}>0$ this function is on the entire real domain.

You have,
$f(x)=\int_2^{e^{x^2}}\ln t dt$

You can consider this function as a composition of functions.
$g(x)=\int_2^x \ln tdt,\, x>0$
$h(x)=e^{x^2}$
Then,
$f =g\circ h$
Since, $h$ is differenciable everywhere and $g$ is differenciable for $x>0$ for that is the fundamental theorem. Thus,
$f'=h' g'\circ h$
Thus,
$f'=2xe^{x^2} \ln (e^{x^2})=2x^3e^{x^2}$ for $x>0$ .

**topsquark** · November 23rd 2006, 06:30 AM

Originally Posted by gracy

(b) d/dx[ int (2 to e^(x^2) )(ln(t))dt]

$\frac{d}{dx} \int_2^{e^{x^2}} dt \, ln(t)$

= $\frac{d}{dx}(t \cdot ln(t) - t)|_2^{e^{x^2}}$

= $\frac{d}{dx} \left ( e^{x^2} \cdot x^2 - e^{x^2} - 2 \cdot ln(2) + 2 \right )$

= $2xe^{x^2} + 2x^3e^{x^2} - 2xe^{x^2}$

= $2x^3e^{x^2}$

-Dan

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