a)
 = ke^{-kx} - k^2xe^{-kx})
Anti-differentiate both sides now, to yield:
\,dx)
Cancelling out a
:
\, dx)
Splitting the integral term by term:
Rearranging to obtain the desired integral:
)
b)
Mean:
Let
Using part (a):
\right]^\infty_0)
Evaluating the integral: lots of terms go to zero, except 2. One is complicated and one is a constant:
 + \frac{1}{k})
Fact:  = 0)
This fact holds because the exponential function decreases far faster than the linear function, hence as
goes to infinity, this becomes zero. That's not a solid proof, but it is a general rule of thumb that works -- exponentials change faster than polynomials. This isn't something you need to know for Methods - so it's the exercise's problem for having that ambigious term there.
Since
c)
Graph-sketching problem involving exponential functions. Shouldn't be too difficult. Just make sure you show the effects of a different size of
.
d)
Larger values of
will result in smaller
, which decrease the vertical dilation of the graph, so the curves with larger values of
will be flatter. The horizontal dilation of the graph will increase with larger values of
, which also contributes to a flatter-looking graph.
This is consistent with the definition found in part (b),
. Since the graph starts at
, and it diminishes rapidly (being a decaying exponential function), a flatter graph that takes longer to decay (less steep) is required to have a larger mean. Hence, a larger mean (larger value of
) is consistent with a flatter graph.
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