Again, to find the local max/min, you need to find when the slope is equal to zero.
In this integral, you need to use the product rule.
f(x)=xe^(-3x)
The product rule states:
If f(x)=g(x)*h(x), then f'(x)=g(x)*h'(x)+g'(x)*h(x)
Let g(x) in this case be x and h(x) be e^(-3x)
Thus g'(x)=1 and h'(x)=-3e^(-3x)
Thus f'(x)=-3xe^(-3x)+e^(-3x)
Since we want slope to be zero, let's set f'(x)=0
0=-3xe^(-3x)+e^(-3x)
0=e^(-3x)(-3x+1)
since e^(-3x) can't equal zero(since it has a asymptote at y=0), we can cancel it to get
0=(-3x+1)
Thus at x=1/3 is an extrema
Now we plug that into f(x) to get f(1/3)=(1/3)e^(-3/3)=1/(3e)
Now we have to check if it's a max or a min.
To do this, we plug a value greater than the max x value into f'(x) and see if it's positive or negative.
f'(0)=-3(0)e^(-3(0))+e^(-3(0))=1, which is positive.
If the result is positive, it's a minimum.
Thus (1/3, 1/(3e)) is a minimum