I'm trying to solve the definite integral $\int_0^n\pi^{ex}dx$

Wolfram says that the answer is $\frac{\pi^{en}-1}{e \ln(\pi)}$, but I got $\frac{\pi^{en}-1}{\ln(\pi)}$. Can anyone help me figure out where I went wrong? These are my steps:

$$ \int_0^n\pi^{ex}dx $$

$$ = \frac{\pi^{ex}}{\ln \pi}\Bigg|_0^n $$

$$ =\frac{\pi^{en}}{\ln \pi} - \frac{1}{\ln \pi} $$

$$ = \frac{\pi^{en}-1}{\ln \pi} $$

Thanks in advance.

  • 3
    $\begingroup$ Your antiderivative for $\Large e^{(\ln\pi)\color{Red}{e}x}$ is incorrect. See where Wolfram's $\color{Red}{e}$ comes from? $\endgroup$ – anon Dec 22 '14 at 7:20
  • $\begingroup$ Your answer should have a factor of $\frac{1}{e}$ since you're integrating $\pi^{ex}$ and not $\pi^{x}$. $\endgroup$ – sayantankhan Dec 22 '14 at 7:21
  • $\begingroup$ @Anon Ah okay, I see it now. Thanks $\endgroup$ – David Dec 22 '14 at 7:23

$$\int_{0}^{n}\pi ^{ex}dx=\int_{0}^{e\cdot n}\frac{\pi^{t}}{e}dt=\frac{1}{e\cdot \ln \pi}\pi^{t}\Bigg|_0^{en}=\frac{1}{e \ln \pi}\left ( \pi^{en}-1 \right )$$ I introduced the substitution $ex=t$, and thus $e dx=dt$

  • 1
    $\begingroup$ Oh interesting! the question is rather old :-) $\endgroup$ – Math-fun Nov 2 '15 at 20:19

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