Question

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?

Answer 1

`A=e^2-e-ln2` 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(x)=1/x`) and the blue graph (`f(x)=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 `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 `1/x`. Basically, take the area under `1/x` away and you have the yellow region.

Mathematically, we express the area of the region like this:
`A=int_1^2f(x)dx-int_1^2g(x)dx`

Because `f(x)=e^x` and `g(x)=1/x`,
`A=int_1^2e^xdx-int_1^2 1/xdx`

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 `lnx` is `1/x`, the antiderivative is `lnx`:
`A=[e^x]_1^2-[lnx]_1^2`

Now it's on to evaluation:
`A=(e^(2)-e^(1))-(ln2-ln1)`
`A=e^2-e-ln2` units

An approximation to three decimal places is `A~~3.978` units.