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I have to solve this exercise:

If $k$ is a non-zero constant, determine by inspection the indefinite integral of $\int e^{kx} dx$.

By inspection, I guess it means that it should be solved by elementary methods that do not involve the fundamental theorem of calculus (my professor shown in class some examples of integrals in which the FTC is unnecessary because of some geometric properties of some functions: Perhaps this is one of these cases, but I can't see it). But I've integrated this function on Mathematica and obtained the result:

$$\frac{e^{k x}}{k \log (e)}$$

I guess that without the help of the FTC, I could never guess that answer. So, what would I need to know in order to guess that?

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  • $\begingroup$ Guess based on knowing the derivative and knowing that integration is the inverse procedure. Also, remember $\log e=1$ so that can be omitted. $\endgroup$ – Adam Hughes Jun 25 '15 at 3:17
  • $\begingroup$ Is that $\log(e) = 1$? The natural logarithm? $\endgroup$ – muaddib Jun 25 '15 at 3:17
  • $\begingroup$ So given change of variables, it seems you want to show that $\int e^x dx = e^x + C$ without the FTC. Or as Adam Hughes said, just use the FTC to demonstrate the equality. $\endgroup$ – muaddib Jun 25 '15 at 3:21
  • $\begingroup$ yes, that's right. Log is the natural logarithm. And you're not really supposed to not use the FTC, you're supposed to not use things like u-substitution. $\endgroup$ – Adam Hughes Jun 25 '15 at 3:22
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    $\begingroup$ In my experience, "determine by inspection" means "determine by whatever method you want, just do it, and then show that it's correct". $\endgroup$ – fonini Jun 25 '15 at 3:24
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This is a basic integral and the result is

$$\int e^{kx} dx = \frac1k e^{kx}$$

A basic reasoning is: The derivative is

$$\frac{d}{dx}e^{kx}=k*e^{kx}$$ To avoid k and obtain the same result as above, we should divide by k (constant)

The resolution by subtitusion is:

$$u=k*x; du=k*dx;dx=1/k $$

$$\int e^{u} \frac1kdu = \frac1k\int e^{u} du = \frac1k e^{u} = \frac1k e^{kx}$$

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