Your idea matters-5

$2\int_{0}^{\infty} e^{-x} dx$ = $-2e^{-x}$ $\Big|$ $^{\infty}_{0}$

= $-2e^{-\infty} - (-2e^{0})$

Let's set the $2$ to the side and let $A = -e^{-\infty}$ and $B = -e^{0}$ .

With this being true, our solution $= 2A - 2B$

We can't find a finite value for $e^{-\infty}$ , but we can plug $x$ values into $e^{-x}$ that approach infinity $(\infty)$ .

Example of values of x that approach $\infty$ :

$\Big| x_1 = 1 \Big| x_2 = 10 \Big| x_3 = 100 \Big| x_4 = 1000 \Big|$

From the table, we can see that as $x$ approaches $\infty$ , $-e^{-x}$ approaches zero.

So, it's safe to say that $A = -e^{-\infty} = 0$ .

Now, we just need $B = -e^{0}$ , and we know from exponential laws that $-e^{0}=-1$ .

So, knowing that $A = 0$ and $B = -1$ , we can solve:

$2\int_{0}^{\infty} e^{-x} dx$ = $-2e^{-\infty} - (-2e^{0})$ = $2A - 2B$

$2A - 2B = 2(0) - 2(-1)$ $= 0 + 2 = 2$

So, our solution is 2 .