Integrating 2^x: A Practical Guide

[Unto]

Intergrate 2^x

...

lol?

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

 

[RoyalCat]

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.

 

[Unto]

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

 

[Unto]

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

 

[Unto]

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

 

[RoyalCat]

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

\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?

 

[Unto]

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

 

[RoyalCat]

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?

 

[Unto]

What is a?

 

[Unto]

The integral is the inverse

 

[RoyalCat]

The integral is the inverse

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}

 

[yungman]

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?

 

[RoyalCat]

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

 

[HallsofIvy]

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

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