Question

Someone help find the integral of y/e^2y from 0 to 1

Answer 1

@Luigi0210

Answer 2

@zepdrix

Answer 3

@agent0smith

Answer 4

@jim_thompson5910

Answer 5

have you tried integration by parts?

Answer 6

I'm not good at it :(

Answer 7

Do you agree that
\[\LARGE \frac{y}{e^{2y}} = y*e^{-2y}\]
??

Answer 8

Yes

Answer 9

Y would be e^2y would be easier to integrate

Answer 10

I wanted to get it into u*dv form

Answer 11

So u is y? Right

Answer 12

it might be easier to make u = e^(-2y) and dv = y

I'm not 100% sure which path is easier

Answer 13

Oh okay :)

Answer 14

let's try it out though

u = e^(-2y)
du = -2e^(-2y)dy

dv = y
v = (1/2)*y^2

agree with everything so far?

Answer 15

Yes

Answer 16

Uv-integral vdu

Answer 17

this you mean?
\[\LARGE \int u*dv = u*v - \int v*du\]

Answer 18

Correct

Answer 19

So we have
\[\large \int u*dv = u*v - \int v*du\]

\[\large \int e^{-2y}*y = e^{-2y}*\left(\frac{1}{2}\right)*y^2 - \int \left(\frac{1}{2}\right)*y^2*(-2e^{-2y})dy\]

Answer 20

hmm the problem looks like it's getting more complicated, let me try another way

Answer 21

Okay

Answer 22

Take your time I got a lot of time

Answer 23

You were right in making u = y. That's much simpler


u = y
du = dy


dv = e^(-2y)dy
v = (-1/2)*e^(-2y)



\[\Large \int u*dv = u*v - \int v*du\]

\[\Large \int y*e^{-2y}dy = y*\left(-\frac{1}{2}*e^{-2y}\right) - \int \frac{-1}{2}*e^{-2y}dy\]

\[\Large \int y*e^{-2y}dy = -\frac{1}{2}*y*e^{-2y} + \frac{1}{2}\int e^{-2y}dy\]


Do you see how to finish up?

Answer 24

Is it -y2(e^-2y)-1/4e^-2y?

Answer 25

-y/2

Answer 26

I think you mean this right?
\[\Large -\frac{y}{2}e^{-2y}-\frac{1}{4}e^{-2y}+C\]

Answer 27

Yeah except it's a definite integral

Answer 28

ok so you'll let

\[\Large g(y) = -\frac{y}{2}e^{-2y}-\frac{1}{4}e^{-2y}+C\]

Answer 29

then you'll compute the value of `g(1) - g(0)`

Answer 30

the constant C will cancel when computing `g(1) - g(0)`

Answer 31

Here give me a few minutes and I'll let you know what I get :)

Answer 32

Okay -3/4e^-2-1/4?

Answer 33

Yes the answer is \[\Large \frac{1}{4}-\frac{3}{4}e^{-2}\]

Answer 34

I'm getting +1/4 not -1/4

Answer 35

Hmmmm

Answer 36

plug in y = 0
\[\Large g(y) = -\frac{y}{2}e^{-2y}-\frac{1}{4}e^{-2y}+C\]
\[\Large g(0) = -\frac{0}{2}e^{-2*0}-\frac{1}{4}e^{-2*0}+C\]
\[\Large g(0) = -\frac{1}{4}+C\]


plug in y = 1
\[\Large g(y) = -\frac{y}{2}e^{-2y}-\frac{1}{4}e^{-2y}+C\]
\[\Large g(1) = -\frac{1}{2}e^{-2*1}-\frac{1}{4}e^{-2*1}+C\]
\[\Large g(1) = -\frac{1}{2}e^{-2}-\frac{1}{4}e^{-2}+C\]


now subtract
\[\Large g(1) - g(0) = \left(-\frac{1}{2}e^{-2}-\frac{1}{4}e^{-2}+C\right) - \left(-\frac{1}{4}+C\right)\]

\[\Large g(1) - g(0) = -\frac{1}{2}e^{-2}-\frac{1}{4}e^{-2}+C +\frac{1}{4}-C\]

\[\Large g(1) - g(0) = -\frac{3}{4}e^{-2}+\frac{1}{4}\]

\[\Large g(1) - g(0) = \frac{1}{4}-\frac{3}{4}e^{-2}\]

Answer 37

you might have forgotten about a negative somewhere

Answer 38

Oh yeah it's minus a negative oops!

Answer 39

other than that mistake, you did good

Answer 40

Thank you, and thank you for the help :)