Question

can u double integral e^y/x dxdy with limpets 4 to 0 and x^2 to 0

Answer 1

\[\int\limits_{0}^{4}\int\limits_{0}^{x^2} e^{y/x} dx dy \]

Answer 2

changing the order may help

Answer 3

can u try and show me
@rational

Answer 4

Your double integral is not a valid region. You must have written it incorrectly.

Look what happens when you integrate the exponential function. You will substitute a function of \(x\) into the antiderivative of the exponential, but then you have to integrate with respect to \(y\).

Answer 5

thats what i have but lets say if u integral respect to y first then to x is it will be work ??

Answer 6

Yes, integrating with respect to y first, and then x yields a finite result. You wouldn't even be able to integrate e^(y/x) with respect to x.

Answer 7

okay so can u show me what is the integral of e^(y/x) respect to y

Answer 8

you need to change the bounds also

Answer 9

start by sketching the region for area of integration

Answer 10

sketch below curves :

y = 0
y = 4
x = 0
x = y^2 ???? (check this bound again)

Answer 11

no no its x^2

Answer 12

your bounds and the differential are not agreeing

x = x^2 as a function is nonsense

Answer 13

\[\large \int\limits_{0}^{4}\int\limits_{\color{red}{0}}^{\color{Red}{x^2}} e^{y/x} \color{Red}{dx} dy\]

Answer 14

the bounds must show x as function of y since the differential is dx

Answer 15

\[\int\limits_{0}^{4}\int\limits_{0}^{x^2} e^{y/x} dy dx\]

Answer 16

how about this ??

Answer 17

that looks fine

Answer 18

\[\large \int\limits_{0}^{4}\int\limits_{\color{red}{0}}^{\color{Red}{x^2}} e^{y/x} \color{Red}{dy} dx\]

like this ?

Answer 19

\[\large \int\limits_{0}^{4}\left(\int\limits_{\color{red}{0}}^{\color{Red}{x^2}} e^{y/x} \color{Red}{dy}\right) dx\]

Answer 20

first integrate the integral inside parenthesis
treat "x" as constant

Answer 21

okay :)

Answer 22

show me plz @rational
i wanna know whats the integral of e^y/x

Answer 23

x is a constant
so it not much exciting as you are expecting

Answer 24

\[\large \int\limits_{0}^{4}\left(\int\limits_{\color{red}{0}}^{\color{Red}{x^2}} e^{y/x} \color{Red}{dy}\right) dx =\int\limits_{0}^{4}\left(\dfrac{e^{y/x}}{1/x} \Bigg|_{\color{red}{0}}^{\color{Red}{x^2}} \right) dx\]

Answer 25

than ??

Answer 26

evaluate the bounds and you will be left with a single integral

Answer 27

can u complete it

Answer 28

\[\large \begin{align}\\ \int\limits_{0}^{4}\left(\int\limits_{\color{red}{0}}^{\color{Red}{x^2}} e^{y/x} \color{Red}{dy}\right) dx &=\int\limits_{0}^{4}\left(\dfrac{e^{y/x}}{1/x} \Bigg|_{\color{red}{0}}^{\color{Red}{x^2}} \right) dx \\~\\ &=\int\limits_{0}^{4}\left(x[e^{\color{Red}{x^2}/x} - e^{\color{red}{0}}] \right) dx \\~\\&= \int\limits_{0}^{4}\left(x[e^{x} - 1] \right) dx\end{align}\]

Answer 29

okay than

Answer 30

thats a single integral
evaluate it

Answer 31

okay just do it i wanna make sure looolz

Answer 32

http://www.wolframalpha.com/input/?i=%5Cint%5Climits_0%5E4%5Cint%5Climits_%7B0%7D%5E%7Bx%5E2%7D+e%5E%7By%2Fx%7D++dydx

Answer 33

you should end up with that answer after evaluating the outer integral