how do I find the 1st quadrant area bounded by the curve y=x*e^(-x)?

Question

i think the answer is one but i dont know how to get it....thanks

Answer 1

The domain of your integration is from 0 to infinity. Now you just need to inegrate x*e^(-x).

You can't do a straight integration for this function because the function is too complex, thus you must use integration by parts.

Integration by parts is defined by:

---------integral(u*dv) = u*v - integral(v*du)--------

When performing integration by parts, you need to define 'u' and 'dv' of your initial integral {integral(x*e^-x*dx)}

For the components of integration by parts, you need to take the derivative of 'u to get 'du' and the integral of 'dv' to get 'v'. When performing integration by parts on an integral with an 'e^x' term, you usually make that the 'dv' term because the integral has an 'e^x' term also.

Thus, we make u = x and dv = e^-x*dx

Therefore, du = dx and v = -e^-x

Now you have all the parts for the integral:

integral(x*e^-x) = -xe^-x[evaluated from 0 to infinity] - integral(-e^-x*dx)

evaluating the first part, you get 0.
The 2nd part results in e^-x (evaluated from 0 to infinity)

Thus, integral = 0-(-1) = 1

Hope this helps

Answer 2

The curve will not intersect any other point except at origin. therefore we have to find the are in first quadrant from 0 to infinity.
Solve the given integral by parts
Area =-e^-x{x+1} where limits are 0 to infinity
Area = 1=absolute one =1 unit
Note e^0=1 and e^-infinity =0

Answer 3

Your're right. the answer is one. And you intergrate the function y(x) from x=0 to x=infinitiy. I took calculus before and learned some techniques of integration, but I'm a bit rusty right now. So I can't integrate this for you. the only reason I know it is 1 is that I used Mathcad

http://i2.photobucket.com/albums/y26/CBasedLifeform/plot-1.jpg

Answer 4

to integrate first find limits i.e, where it cuts x axis. substitute y=o in given equation u get x=0 & x=infinity . so integrate over limits 0 to infinity.

int(x e^-x) should be evaluated by parts using formula

int(u.v dx)=u.int(V dx)-int{ d(u)/dx . int(v dx) dx}

here u=x& v= x.e^-x

in final ans substitute limits 0 & infinity.

u get ans as 1.

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Answer 5

integral(xe^-x dx) from 0 to infinity =
-(x+1)e^-x from 0 to infinity. =
0-(-1) = 1

(-(x+1)/e^x | ∞ = lim|x→∞(-1/e^x) = -1)

Answer 6

Integrate by parts
int(xe^(-x)) dx =[ -xe^(-x)] - int(-e^(-x)) dx
=[-xe^(-x)]+int(e^(-x)) dx
=-xe^(-x)-e^(-x)
If you substitute in the limits 0 and infinity, you will get 1 because xe^(-x) tends to 0 as x tends to infinity.