Author Topic: exponential int (Read 1851 times) Tweet Share

bananna · « **on:** March 01, 2017, 03:39:12 pm »

hello

can someone pls help me with this q
I'm used to just dealing with linear exponential functions
so have trouble with this

thank you!!

jakesilove · « **Reply #1 on:** March 01, 2017, 03:51:20 pm »

Quote from: bananna on March 01, 2017, 03:39:12 pm

hello

can someone pls help me with this q
I'm used to just dealing with linear exponential functions
so have trouble with this

thank you!!

Hey!

$\frac{d}{dx}e^{x^2+3}=2xe^{x^2+3}$

This is because

$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$

So, it's clear that

$\int 2xe^{x^2+3} = e^{x^2+3} + C$

ie. If we differentiate something, then integrate it again, it because the original function. However, because we have completed an indefinite integral, we have to remember to plus C. Does that make sense?

bananna · « **Reply #2 on:** March 02, 2017, 11:32:08 am »

Quote from: jakesilove on March 01, 2017, 03:51:20 pm

Hey!

$\frac{d}{dx}e^{x^2+3}=2xe^{x^2+3}$

This is because

$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$

So, it's clear that

$\int 2xe^{x^2+3} = e^{x^2+3} + C$

ie. If we differentiate something, then integrate it again, it because the original function. However, because we have completed an indefinite integral, we have to remember to plus C. Does that make sense?

yes it makes sense

thank you!!

bananna · « **Reply #3 on:** March 02, 2017, 11:41:10 am »

Quote from: jakesilove on March 01, 2017, 03:51:20 pm

Hey!

$\frac{d}{dx}e^{x^2+3}=2xe^{x^2+3}$

This is because

$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$

So, it's clear that

$\int 2xe^{x^2+3} = e^{x^2+3} + C$

ie. If we differentiate something, then integrate it again, it because the original function. However, because we have completed an indefinite integral, we have to remember to plus C. Does that make sense?

just clarifying--

2xe^x^2+3
is divided by the differential of x^2 = (2x)

and that cancels with 2x on the top to give the integrated form, which is =e^x^2+3 + C ?

thnx

jamonwindeyer · « **Reply #4 on:** March 02, 2017, 11:44:14 am »

Quote from: bananna on March 02, 2017, 11:41:10 am

just clarifying--

2xe^x^2+3
is divided by the differential of x^2 = (2x)

and that cancels with 2x on the top to give the integrated form, which is =e^x^2+3 + C ?

thnx

Pretty much! It is kind of like integrating with the chain rule; you have to be really careful when you do it, and they'll usually lead you into it like they have here!

RuiAce · « **Reply #5 on:** March 02, 2017, 11:46:16 am »

You don't really get told how the chain rule is reversed as far as 2U goes. It's taught a bit more properly in 3U.

For 2U purposes, you may take that method of reversing the chain rule for granted

bananna · « **Reply #6 on:** March 02, 2017, 11:55:19 am »

Quote from: jamonwindeyer on March 02, 2017, 11:44:14 am

Pretty much! It is kind of like integrating with the chain rule; you have to be really careful when you do it, and they'll usually lead you into it like they have here!

oh ok..is there a formula for integrating with the chain rule or is that a method you use just for working out

thnx

Quote from: RuiAce on March 02, 2017, 11:46:16 am

You don't really get told how the chain rule is reversed as far as 2U goes. It's taught a bit more properly in 3U.

For 2U purposes, you may take that method of reversing the chain rule for granted

oh okay answered my question haha

thanks

bananna · « **Reply #7 on:** March 02, 2017, 11:58:49 am »

sorry, one more q :

I was able to differentiate it successfully, but I'm not able to integrate it/reverse it.

thanks

jamonwindeyer · « **Reply #8 on:** March 02, 2017, 12:27:24 pm »

Quote from: bananna on March 02, 2017, 11:58:49 am

sorry, one more q :

I was able to differentiate it successfully, but I'm not able to integrate it/reverse it.

thanks

YOur result for the differentiation should have been:

$(2x-2)e^{x^2-2x+3}$

Notice that the integral we are asked to find is half of this expression - We've just lost the factor of 2 in the bracket out the front. So the answer is just:

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

Does that help?

bananna · « **Reply #9 on:** March 02, 2017, 07:55:51 pm »

Quote from: jamonwindeyer on March 02, 2017, 12:27:24 pm

YOur result for the differentiation should have been:

$(2x-2)e^{x^2-2x+3}$

Notice that the integral we are asked to find is half of this expression - We've just lost the factor of 2 in the bracket out the front. So the answer is just:

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

Does that help?

sorry, in the final product, where did the

(2x-2) go?

thanks

RuiAce · « **Reply #10 on:** March 02, 2017, 08:16:09 pm »

Quote from: bananna on March 02, 2017, 07:55:51 pm

sorry, in the final product, where did the

(2x-2) go?

thanks

$\text{Put it this way}\\ \begin{align*}(2x-2)e^{x^2-2x+3}&= \frac{d}{dx}e^{x^2-2x+3}\\ \implies \int (2x-2)e^{x^2-2x+3dx}&=e^{x^2-2x+3}\\ \implies \frac12 \int (2x-2)e^{x^2-2x+3}dx&=\frac12e^{x^2-2x+3}\\ \implies \int \frac12 (2x-2)e^{x^2-2x+3}dx&=e^{x^2-2x+3}\\ \implies \int (x-1)e^{x^2-2x+3}&= e^{x^2-2x+3} \end{align*}\\ \text{Your integration UNDO's the differentiation}\\ \text{The differentiation brought in the }2x-2\text{, so the integration eliminates it}\\ \text{much like in the previous examples}$

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