# Integrate 2^x

1. Dec 27, 2009

### Unto

Intergrate $$2^x$$

....

lol?

How would I even go about this? This is not a homework question, but I'm practising for exams.

2. Dec 27, 2009

### RoyalCat

Re: Intergration

What exponential function -DO- you know how to integrate?

Hint:
$$\frac{d}{dx}e^x=e^x$$

You can use that relationship to find the integral of any base exponential.

3. Dec 27, 2009

### Unto

Re: Intergration

Can I use Ln2? Because I don't see how e^x will help when it can't integrate anyway

4. Dec 27, 2009

### Unto

Re: Intergration

Actually that's bollocks, we have not learnt how to integrate ln2 yet, I'm stumped lol

5. Dec 27, 2009

### Unto

Re: Intergration

Ok I'm going to use e^xln2 since that = 2^x
Ok?

How to integrate this? Because it in terms of dx, but what to do with ln2

6. Dec 27, 2009

### RoyalCat

Re: Intergration

$$\ln{2}$$ is a constant.
It is a number.

How would you rewrite $$2^x$$ to look more like $$e^t$$ which you do know how to integrate?

7. Dec 27, 2009

### Unto

Re: Intergration

$$2^x = e^{(ln2^x)}$$
=
$$2^x = e^{(xln2)}$$

8. Dec 27, 2009

### RoyalCat

Re: Intergration

That's right!

Now all you need to know how to do, is integrate $$e^{ax}$$
Knowing that $$\frac{d}{dx}e^x=e^x$$, what can you say about $$e^{ax}$$ and what can you infer about its integral?

Last edited: Dec 27, 2009
9. Dec 27, 2009

### Unto

Re: Intergration

What is a?

10. Dec 27, 2009

### Unto

Re: Intergration

The integral is the inverse

11. Dec 27, 2009

### RoyalCat

Re: Intergration

$$a$$ is just any number.
If you can solve for any number, you can solve for the very specific number, $$\ln{2}$$

And you're right about having to multiply by the reciprocal of $$a$$!

When you differentiate $$e^{ax}$$ you get $$e^{ax}$$ times $$\frac{d}{dx} ax$$ which is just $$a\cdot e^{ax}$$ because of the chain rule.

So to get rid of the $$a$$ you get when differentiating, you multiply the original function by a constant to "cancel it out," $$\frac{1}{a}$$

12. Dec 28, 2009

### yungman

Re: Intergration

$$y=2^{x} \Rightarrow lny = xln2 \Rightarrow \frac{1}{y}\frac{dy}{dx} = ln2 \Rightarrow dy= yln2dx \Rightarrow \int dy = ln2\int 2^{x}dx\Rightarrow \int 2^{x}dx=\frac{y}{ln2} = \frac{2^{x}}{ln2}$$

Therefore $$\int 2^{x}dx = \frac{2^{x}}{ln2}+C$$

What do you think?

Last edited: Dec 28, 2009
13. Dec 28, 2009

### RoyalCat

Re: Intergration

That's true, but it's a bit of a backwards way of approaching it (Though it has its beauty! I mean, what's simpler than taking the integral $$\int 1\cdot dy$$ ?).
I think mine is the simplest approach in this case:

$$2^x=e^{x\ln{2}}$$

$$\int 2^x = \int e^{x\ln{2}}=\frac{1}{ln{2}} e^{x\ln{2}} + C$$

Last edited by a moderator: Dec 28, 2009
14. Dec 28, 2009

### HallsofIvy

Staff Emeritus
Re: Intergration

Which equals, of course,
$$\frac{1}{ln 2} 2^x+ C$$

The derivative of $a^x$ is $ln(a)a^x$ and the integral is $(1/ln(a))a^x+ C$