Definite integrals of the exponential

Differentiated or integrated, $e^x$ stays $e^x$ ¹. We compute its integral.

\int_0^1 e^x\,dx = \left[ e^x \right]_0^1 = e - 1 \approx 1.71828

Since the antiderivative is the function itself, the area comes from the difference of the end values alone.

The area function has the same shape

Writing the area from $0$ to $x$ as $F(x) = e^x - 1$ , the graph of $F$ is that of $e^x$ moved down by $1$ . Usually an area function has a shape different from the original; for $e^x$ that does not happen.

The area equals the height

Consider the area from $-\infty$ to $x$ .

\int_{-\infty}^{x} e^t\,dt = \left[ e^t \right]_{-\infty}^{x} = e^x

It is exactly the height $e^x$ at the point $x$ . The infinite strip stretching to the left has a finite area because $e^t$ falls to $0$ so quickly.

The main integrals

Integral	Result
$\int_0^1 e^x dx$	$e - 1 \approx 1.718$
$\int_{-\infty}^{x} e^t dt$	$e^x$
$\int_0^{\infty} e^{-x} dx$	$1$
$\int_0^1 2^x dx$	$\dfrac{1}{\log 2} \approx 1.443$

The decay of an exponential confines an infinite interval to a finite area. When the base is not $e$ a coefficient appears: differentiation multiplied by $\log 2$ , so integration divides by it.

Applications

In applications this integral appears as a total of some quantity.

The amount left after the decay of a radioactive substance
The charge flowing into a capacitor
The sum accumulated by compound interest

In each the rate of change is proportional to the quantity, and integrating brings out an exponential.

The steeply rising curve on the graph is $y = e^x$ , the curve moved down by $1$ representing the area is $y = e^x - 1$ , and the large dots are $(0, 1)$ and $(1, e)$ .

Exponential function, Wikipedia