Integrate 2^x (1 Viewer)

SunnyScience · Apr 25, 2012

Hi,

Can someone show me a quick and easy (?) way of integrating y = 2^x, by noticing that d(2^x)/dx = ln2.2^x

Ty

umm what · Apr 25, 2012

integration of a^x = a^x / log e a
That's an integral

Edit: so integration of 2^x = 2^x / Ln 2 + C

SunnyScience · Apr 25, 2012

raufmalik said:
integration of a^x = a^x / log e a
That's an integral

Edit: so integration of 2^x = 2^x / Ln 2 + C

Thanks for the reply

But how do you know the integration of a^x?

umm what · Apr 25, 2012

thats an integral (its in my 4U Maths Book (Patel))
so u just memorize this integral

SunnyScience said:
Thanks for the reply

But how do you know the integration of a^x?

funnytomato · Apr 25, 2012

SunnyScience said:
Hi,

Can someone show me a quick and easy (?) way of integrating y = 2^x, by noticing that
d(2^x)/dx = ln2.2^x
Ty

what would you get when you integrate both sides ?

Carrotsticks · Apr 25, 2012

SunnyScience said:
Thanks for the reply

But how do you know the integration of a^x?

Good to see you're being inquisitive and not just using a formula blindly.

raufmalik said:
thats an integral (its in my 4U Maths Book (Patel))
so u just memorize this integral

NO.

$\\ $Recall that $ e^x $ and $ \ln(x) $ are inverse functions such that $ x = e^{\ln(x)} \\\\ $Replace $ x $ with $ a^x: \\\\ a^x = e^{\ln(a^x)} = e^{x \ln(a)} $ using log laws.$ \\\\ $Differentiate both sides with respect to $ x: \\\\ \frac{d}{dx} \left (a^x \right ) = \frac{d}{dx} \left ( e^{x \ln(a)} \right ) = \ln(a) e^{x \ln(a)} = \ln(a) \times a^x \\\\ $Moving the $ \ln(a) $ to the other side yields:$ \\\\ \frac{1}{\ln(a)} \times \frac{d}{dx} \left ( a^x \right ) = a^x \\\\ $Integrate both sides with respect to $ x $ yields:$ \\\\ \int a^x dx = \frac{1}{\ln(a)} \times a^x + C$

umm what · Apr 25, 2012

do we have to show all this working man??

Carrotsticks said:
Good to see you're being inquisitive and not just using a formula blindly.

NO.

$\\ $Recall that $ e^x $ and $ \ln(x) $ are inverse functions such that $ x = e^{\ln(x)} \\\\ $Replace $ x $ with $ a^x: \\\\ a^x = e^{\ln(a^x)} = e^{x \ln(a)} $ using log laws.$ \\\\ $Differentiate both sides with respect to $ x: \\\\ \frac{d}{dx} \left (a^x \right ) = \frac{d}{dx} \left ( e^{x \ln(a)} \right ) = \ln(a) e^{x \ln(a)} = \ln(a) \times a^x \\\\ $Moving the $ \ln(a) $ to the other side yields:$ \\\\ \frac{1}{\ln(a)} \times \frac{d}{dx} \left ( a^x \right ) = a^x \\\\ $Integrate both sides with respect to $ x $ yields:$ \\\\ \int a^x dx = \frac{1}{\ln(a)} \times a^x + C$

D94 · Apr 25, 2012

SunnyScience said:
Hi,

Can someone show me a quick and easy (?) way of integrating y = 2^x, by noticing that d(2^x)/dx = ln2.2^x

Ty

I'm not sure if I've interpreted this correctly, but assuming this means "given", then:
$\\\frac{d}{dx}(2^{x}) = ln(2).2^{x}\\\\ \int \frac{d}{dx}(2^{x}) \ dx = \int ln(2).2^{x} \ dx\\\\ 2^{x} = ln(2)\int 2^{x}\ dx\\\\ \frac{2^{x}}{ln(2)} + C = \int 2^{x} \ dx$

SunnyScience · Apr 25, 2012

Carrotsticks said:
Good to see you're being inquisitive and not just using a formula blindly.

NO.

$\\ $Recall that $ e^x $ and $ \ln(x) $ are inverse functions such that $ x = e^{\ln(x)} \\\\ $Replace $ x $ with $ a^x: \\\\ a^x = e^{\ln(a^x)} = e^{x \ln(a)} $ using log laws.$ \\\\ $Differentiate both sides with respect to $ x: \\\\ \frac{d}{dx} \left (a^x \right ) = \frac{d}{dx} \left ( e^{x \ln(a)} \right ) = \ln(a) e^{x \ln(a)} = \ln(a) \times a^x \\\\ $Moving the $ \ln(a) $ to the other side yields:$ \\\\ \frac{1}{\ln(a)} \times \frac{d}{dx} \left ( a^x \right ) = a^x \\\\ $Integrate both sides with respect to $ x $ yields:$ \\\\ \int a^x dx = \frac{1}{\ln(a)} \times a^x + C$

Thanks Carrot.

I'm just not convinced with...

$$Moving the $ \ln(a) $ to the other side yields:$ \\\\ \frac{1}{\ln(a)} \times \frac{d}{dx} \left ( a^x \right ) = a^x \\\\ $Integrate both sides with respect to $ x $ yields:$ \\\\ \int a^x dx = \frac{1}{\ln(a)} \times a^x + C$

are you allowed to just "move" 1/lna to the other side???

funnytomato · Apr 25, 2012

SunnyScience said:
are you allowed to just "move" 1/lna to the other side???

yes

[a simpler example] would you agree:
d/dx(x^2)=2x
and (1/2)*d/dx(x^2)=x ?

in your question, a is just a constant(a=2,in this particular case) , so is ln(a)
d/dx(a^x)= ln(a)*(a^x)
then (1/ln(a))*d/dx(a^x)=a^x

Carrotsticks · Apr 25, 2012

Yes you can.

Here's another numerical example.

$\\\ \frac{d}{dx} x^2 = 2x \\\\ $Move the 2 to the other side:$ \\\\ \frac{1}{2} \frac{d}{dx} x^2 = x \qquad $which is true.$$

asianese · Apr 25, 2012

You don't need to show the working - ie you can go from $\int 2^x \,dx = \frac{2^x}{\ln2}+C$ but carrotsticks is stressing the knowledge of the process involved in coming to the integral - ie. that you can only do calculus with base e.

D94 · Apr 25, 2012

SunnyScience said:
I'm just not convinced with...
[...]
are you allowed to just "move" 1/lna to the other side???

1/ln(a) is just a constant. Like the number 1 or 5 or 100, it can be moved over to the other side.

Drongoski · Apr 26, 2012

Interesting discussion. I'd also do it more-or-less the Carrot way.

The above exchanges reveal a quite-common lack of grasp of the following principle:

1) $\frac {d}{dx} kf(x) = k\frac {d}{dx} f(x)$

2) $\int kg(x)dx = k\int g(x)dx$

where k is a constant, like ln 2 or ln a above.

Thus:

$\frac {d}{dx} \pi x^5 = \pi\frac {d}{dx} x^5$

$\int-e^3 (1+\sqrt{x^2-1})^7 dx = -e^3\int(1+\sqrt{x^2 -1})^7dx$

mirakon · Apr 26, 2012

raufmalik said:
do we have to show all this working man??

Not every time, but it's certainly important to know and understand the process at least, rather than simply memorising it

Drongoski · Apr 26, 2012

mirakon said:
Not every time, but it's certainly important to know and understand the process at least, rather than simply memorising it

Agreed. Esp in 3U and 4U: being able to derive a result is more important than being able to memorise it. That said, it often helps to be able to remember a result without having to derive it.

Carrotsticks · Apr 26, 2012

Drongoski said:
Agreed. Esp in 3U and 4U: being able to derive a result is more important than being able to memorise it. That said, it often helps to be able to remember a result without having to derive it.

Yes, I too agree.

For example, I have TERRIBLE memory and I can NEVER remember the Pythagorean Identities such as:

$\\ \sec^2(x)=1+\tan^2(x) \\\\ \csc^2(x) = 1+\cot^2(x)$

And when I need to use it, I just start with:

$\sin^2(x) + \cos^2(x) = 1$

And depending on which identity I need, I divide by either sin squared or cos squared.

Drongoski · Apr 26, 2012

Carrot: that'd be too slow. Must have all those imprinted in your head.

It's like some of my students insisting it is unnecessary to remember all the exact values of sin, cos and tan of 0, 30, 45, 60 & 90 degrees. They say you can always derive them from the well-known triangles. Do that and another 1 or more precious minutes of your time is gone during your exam. Besides it then often does not dawn on you, when this is crucial, that 1/sqrt(3) is tan 30 or that sqrt(3)/2 is sin 60.

asianese · Apr 26, 2012

LOL trig identities...CALCULATOR =D=D and I agree carrotsticks... its too easy to get those mixed up! start from sin2+cos2 = 1!!

nightweaver066 · Apr 26, 2012

Carrotsticks said:
Yes, I too agree.

For example, I have TERRIBLE memory and I can NEVER remember the Pythagorean Identities such as:

$\\ \sec^2(x)=1+\tan^2(x) \\\\ \csc^2(x) = 1+\cot^2(x)$

And when I need to use it, I just start with:

$\sin^2(x) + \cos^2(x) = 1$

And depending on which identity I need, I divide by either sin squared or cos squared.

I remember that the term containing 'sec' is always on the LHS, and change the sec to a tan and you get the trig term on the RHS + 1.

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