1. ## Double Integral Problem

Hello, I am having difficulty setting up this double integral.

$\displaystyle f(s,t)=e^slnt$ over the region in the first quadrant of the st plane that lies above the curve $\displaystyle s=lnt$ from $\displaystyle t=1$ to $\displaystyle t=2$

so far I have converted the $\displaystyle s=lnt$ into the format: $\displaystyle t=e^s$ Then I graphed it, and for my bound on my integral, I got: $\displaystyle \int_1^2\int_o^{e^s}{e^slnt}$ $\displaystyle dtds$

Nevermind....after trying numerous different things, I think I have finally found the right answer. The x bounds were really getting me. I knew the y bounds had to be 1 and 2, since that was given, but for some reason, the x bounds were giving me trouble.

$\displaystyle \int_1^2\int_o^{lny}{e^slnt}$ $\displaystyle dtds$

2. Originally Posted by BCHurricane89
$\displaystyle \int_1^2\int_o^{e^s}{e^slnt}$ $\displaystyle dtds$

It is wrong ( just my view )

I think the correct boundary is

$\displaystyle 1 \leq t \leq 2 , 0 \leq s \leq ln(t)$

so what we have to calculate is

$\displaystyle \int_1^2\int_0^{\ln(t)} e^s \ln(t)~dsdt$

integrate w.r.t. $\displaystyle s$ first .

$\displaystyle = \int_1^2 \ln(t) [ e^s ]_0^{\ln(t)} ~dt$

$\displaystyle = \int_1^2 (t-1)\ln(t)~dt = [(t-1)^2\ln(t)/2 - t^2/4 + t - \ln(t)/2 ]_1^2$

$\displaystyle = 2 - 1 + 1/4 -1 = 1/4$

3. - Wolfram|Alpha

It tells us the answer but no steps are shown ( pity ) .

I think it is easy to input ... so why not you check your exercises online

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# integreate e^s. lnt over the region in the first quadrant of the ST plane that lies above the curve s=lnt from t=1 to t=2

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