0
$\begingroup$

I had this show up in a problem and I went completely blank. How do I compute

$$\int a^t \mathrm{d}t$$

where $a$ is some constant?

  • 2
    $a^t = e^{t \ln a}$2012-11-17
  • 0
    Thanks for the responses but I found the answer here: http://www.math.com/tables/integrals/more/b%5Ex.htm2012-11-17
  • 0
    Why anyone would bother to downvote this. -_-2012-11-18
  • 0
    Perhaps the downvote came from the fact that this is a sufficiently simple integral that it is (as you found) easy to find in almost any table of integrals.2012-11-18
  • 1
    @RickDecker Everything's easy when you know how. I didn't have access to a table of integrals at the time and, even if I did, I wanted to learn how to work it out. What a ridiculous reason to downvote.2013-05-23

5 Answers 5

4

$a^t = e^{t \ln a}$.

From there, $\int a^t dt = \int e^{t \ln a} dt$, which is trivial to integrate. The answer is $$\frac{a^t}{\ln a} + k$$

2

Note that $a^t = (e^{\log(a)})^t = e^{t \log (a)}$ and $$\int e^{bt} dt = \dfrac{e^{bt}}{b} + \text{ constant}$$

1

$$\int a^tdt=1/\log(a) \cdot \int a^t \log(a)dt=a^t/ \log(a)+C$$ This is because the derivative of $a^t$ is $a^t \log(a)$ .

  • 5
    Try `\log(a)` instead of `log(a)`.2012-11-17
1

Since $\dfrac{\text{d}}{\text{dt}} a^t=a^t\ln(a),$ we have $$\int a^t \text{dt}=\dfrac{a^t}{\ln(a)}+C$$

0

Use the fact that $$\frac{\text{d}(a^t)}{\text{d}t}=a^t\ln(a)$$