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Integrate $$\int{2^{2x}} dx$$

How do I do it, at first, I thought I treat 2 as $e$ and I will get something like $\dfrac{1}{2} 2^{2x}$, but according to WolframAlpha its supposed to be $\dfrac{4^x}{\lg{4}}$

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Why would you treat $2$ as $e$? At most, you can think of $2^x=e^{x \log 2}$ from where you can solve the given integral using $$\int e^{ax} dx=\frac{e^{ax}}{a}$$ Also, note that $2^{2x} = 4^x = e^{x \log 4}$ –  Pedro Tamaroff Apr 24 '12 at 14:31
    
@PeterTamaroff, I was thinking both are constants so maybe they have the same properties –  Jiew Meng Apr 25 '12 at 1:34
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It isn't the case. $e$ is a very special constant. –  Pedro Tamaroff Apr 25 '12 at 1:36
    
If you just divide your answer by $\ln(2)$ then you'll get Wolfram's answer. And you're right, $2$ and $e$ are both constants and they almost share the same properties, at least relative to integration of the corresponding exponentials. $e^x$ is a bit cleaner though since dividing by $\ln(e)$ amounts to dividing by $1$.. –  Patrick May 8 '12 at 1:54
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3 Answers

up vote 7 down vote accepted

Well, $$\frac{d}{dx}a^x = a^x \log a.$$ Hence $$\int a^x \, dx = \frac{1}{\log a}a^x+C.$$

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Hmm, somehow ... I don't remember $\frac{d}{dx} a^x = a^x \log{a}$ ... could you explain how do I get there, chain rule etc? Just to aid my understanding, I could of course add this into my formula sheet –  Jiew Meng Apr 24 '12 at 14:27
    
Alternatively, you have $\int 2^{2x}\mathrm dx=\int \exp(x\log\,4)\mathrm dx$. Let $u=x\log\,4$ and remember how to integrate the natural exponential function... –  J. M. Apr 24 '12 at 14:28
    
@JiewMeng: simply write $a^x$ as $e^{x\log a}$ and use the chain rule. –  Peter Apr 24 '12 at 14:32
    
These things are always a bit "circular". You need to know that $$\lim_{x \to 0} \frac{a^x-1}{x}=\log a,$$ otherwise you are out of luck. But this is the definition of the derivative of $x \mapsto a^x$ at $x=0$. Exponential and logarithms should be either defined correctly, or taken as granted to some extent. –  Siminore May 8 '12 at 7:48
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First, rewrite $$2^{2x} = (2^2)^x = 4^x.$$ This eliminates the need to do a substitution. Then, use the rule that $$\frac{d}{dx} a^x = a^x \ln a,$$ so that $$\int a^x \,dx = \frac{a^x}{\ln a} + C.$$ If you don't see where this comes from, use logarithmic differentiation. That is, let $y = a^x$. Then, $\ln y = \ln a^x = x \ln a$. Taking the derivative of both sides gives $$\frac{y'}{y} = \ln a$$ so that $$y' = y \ln a = a^x \ln a.$$ Now, it is simple to see that $$\int 2^{2x} \,dx = \int 4^x \,dx = \frac{4^x}{\ln 4} + C.$$

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$$ I = \int 2^{2x} \mathrm {d}x \tag{1}$$ Let

$$ \begin{align*} y &=2^{2x} \hspace{5pt} \\ \Rightarrow \ln y &= 2x \ln 2 \end{align*} $$

Differentiate both sides

$$ \begin{align*} \frac{1}{y} \frac{dy}{dx} &= 2 \ln 2 \\ \frac{dy}{y}&= 2 \ln 2 \hspace{4pt}dx\\ &= \ln 2^2 \hspace{4pt} dx \end{align*} $$

$$ \begin{align*} dx &= \frac{dy}{y \hspace{4pt} \ln 2^2} \end{align*} $$

Substitute for $x$ and $dx$ in $(1)$

$$ \begin{align*} I &= \int y \frac{dy}{y \hspace{4pt} \ln 2^2}\\ &= \int \frac{1}{\ln 2^2} \mathrm{d}y = \frac{1}{\ln 2^2} \int \mathrm{d}y = \frac{1}{\ln 4} \int \mathrm{d}y \\ &= \frac{y}{\ln 4} + C \hspace{14pt} (\textit{But } y=2^{2x})\\ &= \frac{2^{2x}}{\ln 4} + C \end{align*} $$

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