Integration of e power x, e power negative x, e power ax

The integral of the exponential function , $e^x$, can be found using basic calculus principles. The integral of $e^x$ with respect to (x) is:

\[
\int e^x \, dx = e^x + C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant. This result is derived from the fact that the derivative of $e^x$ is $e^x$.

The Exponential Function: A Primer

The exponential function $ e^{x} $, where ( e ) is Euler’s number (approximately 2.71828), is known for its unique property of being its own derivative. This implies that the rate of change of $ e^{x} $ with respect to ( x ) is proportional to its current value, a property that makes it crucial in modeling growth processes, decay, and many natural phenomena.

Integration of e power x by Principal of Derivatives

we can use the fundamental theorem of calculus, which states that if a function f is the derivative of another function F, then the integral of f is F plus a constant.

Now we know that

$\frac {d}{dx} e^x = e^x$
or
$\frac {d}{dx} (e^x+ C) = e^x$

Where C is the constant

Now from fundamental theorem of calculus stated above, we can say that

\[
\int e^x \, dx = e^x + C
\]

We can derive this using expansion series of e to the power x also

Definite Integration of e to the power x

Definite integral will be represent as

$\int _{0}^{1} e^x \, dx $

To solve this, we use the antiderivative of $ e^{x} $ , which is $ e^{x} $ . So, the integral becomes:

\[
\left[ e^{x} \right]_{0}^{1}
\]

Now, we evaluate this expression at the upper and lower limits of the integral:

\[
= e^1 -e^0
\]

\[
= e -1
\]

Integration of e power negative x

The integral of the exponential function , $e^{-x}$ is give as

\[
\int e^{-x} \, dx = -e^{-x} + C
\]

Proof

Let -x = t
then -dx = dt
therefore

\[
\int e^{-x} \, dx =- \int e^{t} \, dt = -e^{t} + C
\]

Substituting back,

\[
\int e^{-x} \, dx = -e^{-x} + C
\]

Integration of e power ax

The integral of ( e^{ax} ) with respect to ( x ) is straightforward due to its simple derivative. The integral is given by:

\[
\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C
\]

Proof

Let ax = t
then adx = dt
therefore

\[
\int e^{ax} \, dx =\frac {1}{a} \int e^{t} \, dt = \frac {1}{a} e^{t} + C
\]

Substituting back,

\[
\int e^{ax} \, dx = \frac {1}{a} e^{ax} + C
\]

Special Integral based on exponential function

$\int e^x( f(x) + f^{‘} (x) ) \; dx = e^x f(x) + C$
Proof
$\int e^x( f(x) + f^{‘} (x) ) dx= \int e^x f(x) dx + \int e^x f^{‘} (x) dx$
Now lets calculate $\int e^x f(x) dx$ using integration by parts by taking f(x) and $e^x$ as the first function and second function
$\int e^x f(x) dx = f(x) e^x – \int e^x f^{‘} (x) dx$
Substituting in Above we get
$\int e^x{ f(x) + f^{‘} (x) } dx=f(x) e^x – \int e^x f^{‘} (x) dx + \int e^x f^{‘} (x) dx=e^x f(x) + C $

Solved examples

Question 1
$$
\int e^{3x} \, dx
$$

Solution:
$$
\int e^{3x} \, dx = \frac{1}{3} e^{3x} + C
$$
where $ C $ is the constant of integration.

Question 2

$$
\int e^{x} + 4 e^{2x} \, dx
$$

Solution:
$$
\int e^{x} + 4 e^{2x} \, dx = e^x + 2e^{2x} + C
$$
where $ C $ is the constant of integration.

Question 3

A particle’s velocity $ v(t) $ increases exponentially with time as $ v(t) = e^{2t} $ m/s. Find the displacement of the particle from time $ t = 0 $ to $ t = T $.

Solution:
Displacement is the integral of velocity. So,
$$
\text{Displacement} = \int_0^T e^{2t} \, dt = \left[\frac{1}{2} e^{2t}\right]_0^T = \frac{1}{2} e^{2T} – \frac{1}{2}
$$