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Integral of exp(x)

Prerequisites

Description

This equation shows that the integral of ex\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} is itself plus some constant (vertical translation)

∫ex dx=ex+C\htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} + \htmlClass{sdt-0000000070}{C}

Symbols Used:

This is a symbol for any generic variable. It can hold any value, whether that be an integer or a real number, or a complex number, or a matrix etc.
e e

This symbol represents Euler's constant. It is approximately 2.7182.718.
∫ \int

This is the symbol for an integral, sometimes referred to as an antiderivative. Graphically, it can be understood as the area between a curve and the axis the integral is taken with respect to.
C C

This symbol represents the constant of integration. It must be added to the result of all definite integrals to encompass all possible solutions that satisfy the integral.
 d \:d

This is the symbol for a differential. It represents an infinitesimally small (infinitely close to zero) change in whatever variable it is with respect to.

Derivation

Consider the Derivative of ex\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}}:

(ex)′=ex(\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}})\htmlClass{sdt-0000000025}{'} = \htmlClass{sdt-0000000035}{e}^{x}

We can now integrate both sides to get:

∫(ex)′ dx+C=∫(ex) dx \htmlClass{sdt-0000000060}{\int} (\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}})\htmlClass{sdt-0000000025}{'} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} + \htmlClass{sdt-0000000070}{C} = \htmlClass{sdt-0000000060}{\int}(\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}}) \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x}

Now also consider the integral of a derivative:

∫f′(x) dx=f(x)+C\htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000065}{f'(x)} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \htmlClass{sdt-0000000096}{f}(\htmlClass{sdt-0000000003}{x}) + \htmlClass{sdt-0000000070}{C}

From here we can simplify the left hand side of our second equation to get:

ex+C=∫ex dx \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} + \htmlClass{sdt-0000000070}{C} = \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x}

By swapping the left hand side and the right hand side, we finally get:

∫ex dx=ex+C \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} + \htmlClass{sdt-0000000070}{C}

as required.

Example

Coming soon...