# integral question

• Nov 28th 2012, 11:07 AM
kingsolomonsgrave
integral question
h(x)= integral from 0 to e^x of ln(t) dt

solution

let u = e^x then du/dx =e^x

also dh/dx=dh/du du/dx

h'(x)=d/dx integral from 1 to e^x of ln(t) dt = d/du integral from 1 to u ln(t) dt times du/dx = ln u du/dx = (lne^x)*e^x = xe^x

for "d/du integral from 1 to u ln(t) dt times du/dx" above, why is the integral multiplied by the derivative of u WRT x
• Nov 28th 2012, 11:16 AM
Plato
Re: integral question
Quote:

Originally Posted by kingsolomonsgrave
[FONT=Times New Roman][SIZE=2][COLOR=#0e0e0e][FONT=Times New Roman][SIZE=2][COLOR=#0e0e0e]h(x)= integral from 0 to e^x of ln(t) dt

Do you really mean $\int_0^{e^x } {\ln (t)dt}$ or was that a typo and you want $\int_1^{e^x } {\ln (t)dt}~?$
• Nov 28th 2012, 11:19 AM
kingsolomonsgrave
Re: integral question
the second one, sorry
• Nov 28th 2012, 11:23 AM
skeeter
Re: integral question
$\frac{d}{dx} \int_a^u f(t) \, dt = f(u) \cdot \frac{du}{dx}$

$\frac{d}{dx} \int_1^{e^x} \ln(t) \, dt = \ln(e^x) \cdot e^x = x \cdot e^x$
• Nov 28th 2012, 11:28 AM
Plato
Re: integral question
Quote:

Originally Posted by kingsolomonsgrave
the second one, sorry

If $h(x)=\int_1^{e^x } {\ln (t)dt}$ then $h'(x)=xe^x~.$

Because $\ln(e^x)=x$ and the derivative of $e^x=e^x~.$