Question

find the area y=e^x y=xe^x x=0

Answer 1

Can you be more precise?

Answer 2

oh. sorry. find the area enclosed by y=e^x, y=xe^x and x=0

Answer 3

my guess is you are looking for
\[\int_0^1e^x-xe^xdx\]

Answer 4

It can also be between 0 and 10

Answer 5

that is just a guess
the only part that requires much work is finding where they intersect

Answer 6

but, how do you know is not integral from xe^x-e^x,

Answer 7

@eliassaab i don't think it can be 0 to ten because they intersect at \(x=1\)

Answer 8

oh that is a good point
you need to know which one is larger on the interval \([0,1]\) don't you?

Answer 9

Why not?

Answer 10

but if \(0<x<1\) it should be fairly obvious that \(e^x>xe^x\)

Answer 11

Yes, I agree. But in general you can find the area between 0 and b

Answer 12

reedited question included the phrase "find the area enclosed by"

Answer 13

oh, got it!!! thanks it should be from 0 to 1. also, my second question. how to get anti of xe^x?

Answer 14

Integrate by parts

Answer 15

i took "enclosed" to mean between the curves

Answer 16

enclosed---be bounded by, i think. by part?? yo mean substitution rule?

Answer 17

no not substitution, "parts"

Answer 18

can you be more specific?

Answer 19

\[\int udv=uv-\int vdu\]

Answer 20

oh, i have not been introduced to iintegration by parts

Answer 21

then i have no idea how you would do this otherwise

Answer 22

i can show you how to do it in this example if you like
it is a good example to start with

Answer 23

oh,, yes, please, I would love to love. It is strange. We finished 6.2--volumes. my professor put this problem as final review problem. But we did not learn what you said--by parts??

Answer 24

yes, "parts" is the product rule in reverse, just like "u-sub" is the chain rule in reverse

Answer 25

you have
\[\int xe^xdx\] and you can see that is it very easy to find the derivative of \(x\) which is 1, so it will be gone in the second step, and also easy to find the anti derivative of \(e^x\) which is just \(e^x\)
so starting with
\[\int xe^xdx \] and looking at \[\int udv=uv-\int vdu\]you want to make \(u=x, dv=e^xdx\) which gives \(du=dx\) and also \(v=e^x\)

Answer 26

therefore
\[\int xe^xdx=xe^x-\int e^xdx\]

Answer 27

the now it is easy, because \[\int e^xdx=e^x\] so you have as a final result
\[\int xe^xdx=xe^x-e^x\]
you can check that this is right, and also why "parts" works, by taking the derivative of
\[xe^x-e^x\] and seeing that indeed you get \[xe^x\]

Answer 28

the product rule makes it work
\[\frac{d}{dx}[xe^x-e^x]=xe^x+e^x-e^x=xe^x\] tada!

Answer 29

I need a little time to think about what is "part". And I don't this is the method professor is intended. But, I think your product rule seems more straightforward.

Answer 30

i am not sure what your professor intended, but i am very sure that you are not going to find the integral of \(xe^x\) without using parts

Answer 31

OK, haha, Got you!!THank you so much!!

Answer 32

the product rule say s
\[(fg)'=fg'+gf'\] integrating both sides gives
\[fg=\int fg'+\int gf'\] or
\[\int fg'=fg-\int gf'\] this is usually written as
\[\int udv=uv-\int vdu\]

Answer 33

yw

Answer 34

Yeah!! I completely got it !! Thanks a lot!! satellite73

Answer 35

yw (again)

Answer 36

haha, one more question, How do you get all the integral symbols

Answer 37

latex

Answer 38

if you want to see the code, right click on the equation or expression then select "show math as" then "lex commands"

try it

Answer 39

I just tried .And I guess I am using iPad,so it didn.t come up as you told me. But, I will try it on my laptop , now, I need to practice other problems before tomorrow's final .Thanks lot again