| \( x \) | 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 |
This uses integration by u substitution (no page for this yet).
We are tasked with the integration:
\[ \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} \]
let
\[ u = \htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x} \].
we will now use the power rule:
\[\frac{\htmlClass{sdt-0000000102}{\:d}}{\htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x}}(\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000015}{k}}) = \htmlClass{sdt-0000000015}{k} \htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}^{\htmlClass{sdt-0000000015}{k} - 1}\]
to take the derivative of both sides, getting:
\[ \frac{\htmlClass{sdt-0000000102}{\:d} u}{\htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x}} = \htmlClass{sdt-0000000121}{a} \]
From here, it follows that:
\[ \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \frac{1}{\htmlClass{sdt-0000000121}{a}} \htmlClass{sdt-0000000102}{\:d} u \]
We can now simplify our original integral to get:
\[ \htmlClass{sdt-0000000060}{\int} \frac{1}{\htmlClass{sdt-0000000121}{a} } \htmlClass{sdt-0000000035}{e}^{u} \htmlClass{sdt-0000000102}{\:d} u \]
Our \(\frac{1}{\htmlClass{sdt-0000000121}{a}}\) term is unrelated to \(u\), so we can take that outside our integral yielding:
\[ \frac{1}{\htmlClass{sdt-0000000121}{a}}\htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{u} \htmlClass{sdt-0000000102}{\:d} u \]
We now have an integral we know the value of, \(\htmlClass{sdt-0000000035}{e}^{u} \). We can use the Integral of exp(x):
\[\htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000003}{x}} + \htmlClass{sdt-0000000070}{C}\]
yielding:
\[ \frac{1}{\htmlClass{sdt-0000000121}{a}}(\htmlClass{sdt-0000000035}{e}^{u} + \htmlClass{sdt-0000000070}{C}) \]
We can now multiply both terms in our binomial ( \(\htmlClass{sdt-0000000035}{e}^{u} + \htmlClass{sdt-0000000070}{C}\) ) with our constant \( (\frac{1}{\htmlClass{sdt-0000000121}{a}})\) to get:
\[ \htmlClass{sdt-0000000060}{\int} \htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}} \htmlClass{sdt-0000000102}{\:d} \htmlClass{sdt-0000000003}{x} = \frac{\htmlClass{sdt-0000000035}{e}^{\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}}}{\htmlClass{sdt-0000000121}{a}} + \htmlClass{sdt-0000000070}{C}, \htmlClass{sdt-0000000121}{a} \neq 0 \]
(because \(\frac{\htmlClass{sdt-0000000070}{C}}{\htmlClass{sdt-0000000121}{a}} = \htmlClass{sdt-0000000070}{C}\) as both our constants anyway, and we don't really care what the exact value is because it changes in every case)
Coming soon...