Integral of \( a^x \)

Step-by-Step Derivation, Exponential Base Change, Formula, and References

Derivation and Evaluation

Evaluate the indefinite integral:

\[ \int a^x \, dx \]

where \( a \) is a constant such that \( a > 0 \) and \( a \neq 1 \).

We first change the base of the exponential expression \( a^x \):

Let \( y = a^x \) and take the natural logarithm (\( \ln \)) of both sides:

\[ \ln y = \ln(a^x) \]

Using the logarithm property \( \ln(a^x) = x \ln a \), we rewrite this as:

\[ \ln y = x \ln a \]

Applying the exponential function to both sides gives:

\[ y = e^{x \ln a} \]

Since \( y = a^x \), we obtain the identity:

Exponential Base Conversion: \[ a^x = e^{x \ln a} \]

Using this identity, the given integral can be rewritten as:

\[ \int a^x \, dx = \int e^{x \ln a} \, dx \qquad (1) \]

We now evaluate the integral on the right side using substitution:

Let \( u = x \ln a \), which gives \( \dfrac{du}{dx} = \ln a \). By definition [1] [2] [3], the differential is \( du = \ln a \, dx \), or \( dx = \dfrac{1}{\ln a} \, du \).

Substitute into integral (1):

\[ \int e^{x \ln a} \, dx = \int e^u \left( \dfrac{1}{\ln a} \right) du = \dfrac{1}{\ln a} \int e^u \, du \]

Evaluate the standard integral \( \int e^u \, du = e^u + c \):

\[ = \dfrac{1}{\ln a} e^u + c \]

where \( c \) is the constant of integration.

Substitute back \( u = x \ln a \):

\[ = \dfrac{1}{\ln a} e^{x \ln a} + c \]

Finally, using the relationship \( e^{x \ln a} = a^x \), we arrive at the standard integration formula:

Integral Formula for \( a^x \): \[ \int a^x \, dx = \dfrac{1}{\ln a} a^x + c \]

More References and Links

  1. Table of Integral Formulas
  2. University Calculus - Early Transcendentals - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13: 978-0134995540
  3. Calculus - Gilbert Strang - MIT - ISBN-13: 978-0961408824
  4. Calculus - Early Transcendentals - James Stewart - ISBN-13: 978-0-495-01166-8