The integral of xe^-(x^2/2) is quite simple. just put -x^2/2 = t , we get -xdx = dt
The given integral becomes - e^t dt which is -e^t
Now you can proceed further.
Hi, I have been having trouble finding areas of region bounded by graphs.
Like for example:
y=xe^{-(x^2/2)}, y=0, x=0, x=√2
and
y=3^{x}, y=0, x=0, and x=3
So for example, the second one.
First I would graph it.
Then I would find out the axis it gets cut off right? From which x it starts, and the x it ends.
So then it would be [0, 3]
Then I would use the y=3^{x} right? As the integral.
Then it would be (1/(3-0))∫3^{x}
=1/3∫3^{x}
Then I would find the antiderivative?
Which would equal 1/ln3 * 3^{x} = 3^{x}/ln3
Then I would plug in the points 3 and 0 right and subtract it from each other?
So 1/3[(3^{(3)}/ln3)-3^{(0)}/ln3]
Which is kind of not a pretty number... but is this the right process?
Just making sure before I do the first one... because the antiderivative of xe^{-(x^2/2)} looks a bit difficult.
So can anyone make sure if these steps are right? Thanks!
And if it really is wrong, explain the process or some advices as well? Thanks!