# Let f and g be the functions given by f(x)=e^x and g(x)=1/x. What is the area of the region enclosed by the graphs of f and g between x=1 and x=2?

Apr 15, 2016

$A = {e}^{2} - e - \ln 2$ units

#### Explanation:

If we take a peek at the graphs of these functions, we'll see this:

Of course, we won't see that monstrous yellow blob - that's me. But what we need to concentrate on are the red graph ($g \left(x\right) = \frac{1}{x}$) and the blue graph ($f \left(x\right) = {e}^{x}$). The area between these two functions is the yellow region - and that's what we're calculating.

To find it, all we need to do is find the area under $\frac{1}{x}$ and subtract it from the area under ${e}^{x}$. We do this because the yellow region is really the area under ${e}^{x}$, without the area under $\frac{1}{x}$. Basically, take the area under $\frac{1}{x}$ away and you have the yellow region.

Mathematically, we express the area of the region like this:
$A = {\int}_{1}^{2} f \left(x\right) \mathrm{dx} - {\int}_{1}^{2} g \left(x\right) \mathrm{dx}$

Because $f \left(x\right) = {e}^{x}$ and $g \left(x\right) = \frac{1}{x}$,
$A = {\int}_{1}^{2} {e}^{x} \mathrm{dx} - {\int}_{1}^{2} \frac{1}{x} \mathrm{dx}$

Lucky for us, these two integrals are quite simple. Because the derivative of ${e}^{x}$ is ${e}^{x}$, the antiderivative is ${e}^{x}$; and because the derivative of $\ln x$ is $\frac{1}{x}$, the antiderivative is $\ln x$:
$A = {\left[{e}^{x}\right]}_{1}^{2} - {\left[\ln x\right]}_{1}^{2}$

Now it's on to evaluation:
$A = \left({e}^{2} - {e}^{1}\right) - \left(\ln 2 - \ln 1\right)$
$A = {e}^{2} - e - \ln 2$ units

An approximation to three decimal places is $A \approx 3.978$ units.