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Integral of (a exp(ax)) wrt x

Prerequisites

Power Rule for Integration | \(\int (a x^{k}) \:d x = \frac{a}{k + 1}x^ {k + 1} + C\)
Integral of exp(ax) wrt x | \(\int e^{a x} \:d x = \frac{e^{a x}}{a} + C, a \neq 0\)
Integration by Parts | \(\int u \:d v = u v - \int v \:d u\)

Description

This equation shows the result of the integral:

\[\htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000003}{x} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \frac{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x} - 1}{\htmlClass{sdt-0000000121}{a}^{2}} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} + \htmlClass{sdt-0000000070}{C}\]

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.
\( e \)

This symbol represents Euler's constant. It is approximately \(2.718\).
\( \int \)

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.
\( C \)

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.
\( \:d \)

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.
\( a \)

This is a symbol for any generic constant. It can hold any numerical value

Derivation

Consider our integral:

\[ \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000003}{x} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} \]

We will use integration by parts:

\[\htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000008}{u} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000019}{v} = \htmlClass{sdt-0000000008}{u} \htmlClass{sdt-0000000019}{v} - \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000019}{v} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000008}{u}\]

Let us take \(\htmlClass{sdt-0000000008}{u} = \htmlClass{sdt-0000000003}{x}\) and \(\htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000019}{v} = \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x}\)

From here it follows that:

\(\htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000008}{u} = \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x}\)

We can also integrate both sides of the equation:

\[ 1 \cdot \htmlClass{sdt-0000000102}{\:d}\htmlClass{sdt-0000000019}{v} = \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} \]

(\(\htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000019}{v}\) is multiplied by 1 to make the next step more clear)

\[ \htmlClass{sdt-0000000060}{\int} 1 \cdot \htmlClass{sdt-0000000102}{\:d}\htmlClass{sdt-0000000019}{v} = \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} \]

We can now use the power rule to simplify the left hand side. The power rule is:

\[\htmlClass{sdt-0000000060}{\int} (\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000015}{k}}) \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \frac{\htmlClass{sdt-0000000121}{a}}{\htmlClass{sdt-0000000015}{k} + 1}\htmlClass{sdt-0000000003}{x}^ {\htmlClass{sdt-0000000015}{k} + 1} + \htmlClass{sdt-0000000070}{C}\]

This makes the left hand side just \( \htmlClass{sdt-0000000019}{v} \)

We also know the structure of the right hand side:

\[\htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \frac{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}} + \htmlClass{sdt-0000000070}{C}, \htmlClass{sdt-0000000121}{a} \neq 0\]

This simplifies that whole equation to:

\[ \htmlClass{sdt-0000000019}{v} = \frac{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}} \]

We can now apply integration by parts: The equation becomes:

\[ \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000003}{x} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = (\htmlClass{sdt-0000000003}{x} \cdot \frac{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}}) - \htmlClass{sdt-0000000060}{\int} (\frac{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x}) \]

Our right hand side now:

\[ = \frac{\htmlClass{sdt-0000000003}{x} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}} - \frac{1}{\htmlClass{sdt-0000000121}{a}} \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000118}{t}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000118}{t} \]

Again, we know the value of the integral, yielding:

\[ = \frac{\htmlClass{sdt-0000000003}{x} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}} - \frac{1}{\htmlClass{sdt-0000000121}{a}} (\frac{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}}) \]

From here it follows that the right hand side:

\[ = \frac{\htmlClass{sdt-0000000003}{x} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}} - \frac{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}^{2}} + \htmlClass{sdt-0000000070}{C} \]

Finally, we can simplify the fraction, to get our final result of:

\[ \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000003}{x} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \frac{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x} - 1}{\htmlClass{sdt-0000000121}{a}^{2}} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} + \htmlClass{sdt-0000000070}{C} \]

as required.

Example

Coming soon...