Solution to Problem 12.3-2

The problem is to find the area bounded by the y-axis, the curve, y = e^x, and the curve, y = 2e^-x. Note that with this problem, we don't have to solve for one of the limits of the integral. One of the limits is given by the requirement that the area is bounded on one side by the y-axis. So we only have to look for one intersection point between the two curves. We do this by solving for when e^x - 2e^-x = 0.

There are several ways to solve this. One is the way I suggested in the main text -- set the function on the left of minus sign to the function on the right, and take the natural log of both sides.

  e^x = 2e^-x

  x  =  ln(2e^-x)

Now applying the log identity, ln(ab) = ln(a) + ln(b):

  x  =  ln(2) - x

  2x  =  ln(2)

x =

ln(2) 2

An alternative method to solve an equation like this one is to let u = e^x. Then 1/u = e^-x, and you have

u -

2 = 0 u

  u²  =  2

  u  =  ± √2

Since exponentials can never be negative, we have to take the + of the ±. So u = √2 = e^x, hence x = ln(√2), which, by another log identity, is the same as what we got using the first method.

Now we know that we take the integral of the difference between the two curves from zero to ln(2)/2 (this time I'm looking at the graph to make sure I take the difference the right way in order to avoid getting a negative area).

A =

ln(2)/2 0

2e^-x - e^x dx =

-2e^-x - e^x

ln(2)/2 0

  A  =  -(√2 + √2) + 3  =  3 - 2√2  =  0.171572875253810 ± ξ,  0 ≤ ξ ≤ 5×10^-16

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