Question

int_{1}^{e} \frac{ 1+e^x }{ lnx } lnx

Answer 1

\[\int\limits_{1}^{e} \frac{ 1+e^x }{ lnx } lnx\]

Answer 2

i kinda need help with the simplification of this, i see how it goes to
\[1+\frac{ e^x lnx}{ lnx }\]

Answer 3

check your statement

Answer 4

@surjithayer what do you mean. becaus ei forgot to put in the dx?

Answer 5

is there like a u sub you need to do or is it just a fact that e^xlnx/lnx is just equal to e

Answer 6

\[\int\limits_{1}^{e}\frac{ 1+e^x }{ \ln x }\ln x~dx=\int\limits_{1}^{e}\left( 1+e^x \right)dx= \left[ x+e^x \right],~x~from~1\rightarrow~e\]

Answer 7

yea i see that, but for simplification of e^xlnx/lnx, how does that spit out e^x

Answer 8

they just cancel out.

Answer 9

oh, alright.

Answer 10

so there isnt a u sub method to get rid of em its just that they cancel out?

Answer 11

consider the solid region W situated above the region \[0 \le x \le 2, 0 \le y \le x, \], also looks like \[0 \le y \le 2, y \le x \le 2\]

Answer 12

and bounded above by the surface \[z=e ^{x ^{2}}\]

Answer 13

how would i go about drawing this graph so that i can set up a double integral

Answer 14

Write an integral that evaluates the area of each cross-section of W with vertical planes y = a, where a is a constant in [0, 2]. Can you evaluate this integral (as an expression depending on a)?

Answer 15

@surjithayer

Answer 16

sorry i am not in touch with that,because i studied way back in 1974

Answer 17

oh no problem, but @SithsAndGiggles

Answer 18

The projection of region given by
\[W:=\{(x,y,z)~:~0 \le x \le 2,~0 \le y \le x,~\color{red}{0\le z\le e^{x^2}}\}\]
in the \(x-y\) plane looks like this:

(I'm not sure about the red part. It's not explicit whether \(z\) is bounded below by \(z=0\) or not. I'll assume it's correct for now.)

The cross-sections are perpendicular to the \(y\) axis. To help visualize just one of these, here's the solid: