Integrating 2^x: A Practical Guide

In summary, we discussed the process of integrating 2^x and how it can be rewritten as e^{xln2} to make it easier to integrate using the fact that \frac{d}{dx}e^x=e^x. We also explored different approaches to integrating 2^x and found that the simplest way is to use the formula \int a^x = \frac{a^x}{lna} + C.
  • #1
Unto
128
0
Intergrate [tex]2^x[/tex]

...

lol?

How would I even go about this? This is not a homework question, but I'm practising for exams.
 
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  • #2


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

Hint:
[tex]\frac{d}{dx}e^x=e^x[/tex]

You can use that relationship to find the integral of any base exponential.
 
  • #3


Can I use Ln2? Because I don't see how e^x will help when it can't integrate anyway
 
  • #4


Actually that's bollocks, we have not learned how to integrate ln2 yet, I'm stumped lol
 
  • #5


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


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

[tex]\ln{2}[/tex] is a constant.
It is a number.

How would you rewrite [tex]2^x[/tex] to look more like [tex]e^t[/tex] which you do know how to integrate?
 
  • #7


[tex]2^x = e^{(ln2^x)}[/tex]
=
[tex]2^x = e^{(xln2)}[/tex]
 
  • #8


That's right!

Now all you need to know how to do, is integrate [tex]e^{ax}[/tex]
Knowing that [tex]\frac{d}{dx}e^x=e^x[/tex], what can you say about [tex]e^{ax}[/tex] and what can you infer about its integral?
 
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  • #9


What is a?
 
  • #10


The integral is the inverse
 
  • #11


Unto said:
The integral is the inverse

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

And you're right about having to multiply by the reciprocal of [tex]a[/tex]!

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

So to get rid of the [tex]a[/tex] you get when differentiating, you multiply the original function by a constant to "cancel it out," [tex]\frac{1}{a}[/tex]
 
  • #12


[tex]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}[/tex]

Therefore [tex] \int 2^{x}dx = \frac{2^{x}}{ln2}+C[/tex]

What do you think?
 
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  • #13


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 [tex]\int 1\cdot dy[/tex] ?).
I think mine is the simplest approach in this case:

[tex]2^x=e^{x\ln{2}}[/tex]

[tex]\int 2^x = \int e^{x\ln{2}}=\frac{1}{ln{2}} e^{x\ln{2}} + C[/tex]
 
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  • #14


RoyalCat said:
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 [tex]\int 1\cdot dy[/tex] ?).
I think mine is the simplest approach in this case:

[tex]2^x=e^{x\ln{2}}[/tex]

[tex]\int 2^x = \int e^{x\ln{2}}=\frac{1}{ln{2}} e^{x\ln{2}} + C[/tex]
Which equals, of course,
[tex]\frac{1}{ln 2} 2^x+ C[/tex]

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

1. What is the purpose of integrating 2^x?

The purpose of integrating 2^x is to find the area under the curve of the function 2^x. This can be useful in many applications, such as calculating the work done by a variable force or finding the growth rate of a population.

2. What are the steps to integrate 2^x?

The steps to integrate 2^x are as follows: (1) Rewrite the function as e^x, (2) use the power rule of integration which states that the integral of x^n is (x^(n+1))/(n+1), and (3) add a constant of integration to the result.

3. Can you provide an example of integrating 2^x?

Yes, for the function 2^x, the integral would be (2^x)/ln(2) + C, where C is the constant of integration. For example, the integral of 2^x from 0 to 2 would be [(2^2)/ln(2)] - [(2^0)/ln(2)] = (4/ln(2)) - (1/ln(2)) = 3/ln(2).

4. Are there any special cases when integrating 2^x?

Yes, if the upper and lower limits of integration are both 0, the result would be ln(2) instead of a numerical value. This is because when x=0, 2^x equals 1, and the integral of 1 is x. Therefore, the result would be 0 when evaluated at both limits, and the difference would be ln(2).

5. What are some real-life applications of integrating 2^x?

Integrating 2^x can be used in various fields such as physics, economics, and biology. For example, it can be used to calculate the work done by a variable force, the growth rate of a population, or the rate of return on an investment with compound interest.

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