Integrating Exponential Functions
Integrating exponential functions uses \(\displaystyle\int e^{x}\,dx=e^{x}+C\) and \(\displaystyle\int e^{ax+b}\,dx=\dfrac1a e^{ax+b}+C\).
Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Integral calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
Theory
\(e^{x}\) integrates back to \(e^{x}\); \(e^{ax}\) picks up a \(\dfrac1a\). This Year 12 Mathematics Advanced topic (MAV-12-05) integrates exponential functions.
\(\displaystyle\int e^{x}dx=e^{x}+C\) and \(\displaystyle\int e^{ax}dx=\dfrac1a e^{ax}+C\): divide by the coefficient of \(x\) in the exponent. Constant multiples stay out the front.
Method
- Keep the exponential.
- Divide by the coefficient of \(x\) in the exponent.
- Add \(+C\) (or evaluate limits for a definite integral).
| \(\ \) | \(=\) | \(\dfrac12 e^{2x}+C\) |
| \(\ \) | \(=\) | \(e^{x}+C\) |
| \(\ \) | \(=\) | \(-3e^{-x}+C\) |
| \(\ \) | \(=\) | \(\big[e^{x}\big]_0^1=e-1\) |
Common pitfalls
Frequently asked questions
What is the integral of e to the x?
It is e to the x plus C, because e to the x is its own antiderivative.
How do you integrate e to the 2x?
Divide by the coefficient 2: the integral is one half e to the 2x plus C.
What is the integral of e to the minus x?
It is minus e to the minus x plus C, because you divide by minus 1.
What is the area under e to the x from 0 to 1?
It is e to the 1 minus e to the 0, which equals e minus 1.