Question

Let R be the first quadrant region enclosed by the graph of `y= 2e^-x` and the line x=k, how od you find the area of R in terms of k?

Answer 1

Integrate `y` from `x=0` to `x=k` to get `R=2-2e^(-k)`.

Explanation:

We know that the area under a curve can be found by integrating it; so the area under `y=2e^(-x)` is given by: `int2e^(-x)dx`. The only problem is what `x`-values we need to integrate.

The first quadrant starts at `x=0`, and continues for all positive values of `x`. The problem states that we're finding the area under `y` in the first quadrant (which begins at `x=0`) to the line `x=k`. So we'll be integrating from `x=0` to `x=k`:
`R=int_0^k2e^(-x)dx`
`R=2int_0^ke^(-x)dx`
`R=2[-e^(-x)]_0^k`
`R=2(-e^((-k))-(-e^((-0)))`
`R=2(-e^(-k)+1)=2-2e^(-k)`

Note that `2e^(-k)` will never be greater than `2`, which means the area will always be greater than `0`.