# 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?

Apr 30, 2016

Integrate $y$ from $x = 0$ to $x = k$ to get $R = 2 - 2 {e}^{- k}$.

#### Explanation:

We know that the area under a curve can be found by integrating it; so the area under $y = 2 {e}^{- x}$ is given by: $\int 2 {e}^{- x} \mathrm{dx}$. 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 $x = k$:
$R = {\int}_{0}^{k} 2 {e}^{- x} \mathrm{dx}$
$R = 2 {\int}_{0}^{k} {e}^{- x} \mathrm{dx}$
$R = 2 {\left[- {e}^{- x}\right]}_{0}^{k}$
R=2(-e^((-k))-(-e^((-0)))
$R = 2 \left(- {e}^{- k} + 1\right) = 2 - 2 {e}^{- k}$

Note that $2 {e}^{- k}$ will never be greater than $2$, which means the area will always be greater than $0$.