1. exam tomorrow!!! integral

ok..sorry for such an easy qeuestion but i have hit a complete mind blank...

what is the integral of $e^{2x}$

i know the answer it is....e^2x / 2

sorry but i have 2 exams and been studying for both of them forgetting one or the other...

2. Originally Posted by b00yeah05
ok..sorry for such an easy qeuestion but i have hit a complete mind blank...

what is the integral of $e^{2x}$

i know the answer it is....e^2x / 2

sorry but i have 2 exams and been studying for both of them forgetting one or the other...
$\int e^{2x}dx$

-----------------------------
| SUBSTITUTE
| $a=e^x$
| $da=e^x~dx$
| $\frac 1{e^x}~da=dx$
| $\frac 1a~da=dx$
-----------------------------

$=\int \frac 1a*a^2~da$

$=\int a~da$

$=\frac{a^2}2+C$

Anti-Substituite
$=\frac{(e^x)^2}2+C$

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

3. Originally Posted by angel.white
$\int e^{2x}dx$

-----------------------------
| SUBSTITUTE
| $a=e^x$
| $da=e^x~dx$
| $\frac 1{e^x}~da=dx$
| $\frac 1a~da=dx$
-----------------------------

$=\int \frac 1a*a^2~da$

$=\int a+C$

$=\frac{a^2}2+C$

Anti-Substituite
$=\frac{(e^x)^2}2+C$

$=\frac{e^{2x}}2+C$
interesting substitution. i would have done it this way:

(of course, i assume we know $\int e^x~dx = e^x + C$)

We wish to find $\int e^{2x}~dx$

Let $u = 2x$

$\Rightarrow du = 2~dx$

$\Rightarrow \frac 12 ~du = dx$

So our integral becomes:

$\frac 12 \int e^u~du = \frac 12 e^u + C$

............... $= \frac 12e^{2x} + C$

but maybe that's not scary enough for you

"Anti-substitute"?

4. Well (ohhh c'mon, this is only chain rule...)

$\int {e^{2x} \,dx} = \frac{1}
{2}\int {\left( {e^{2x} } \right)'\,dx} = \frac{{e^{2x} }}
{2} + k.$

Originally Posted by Jhevon
"Anti-substitute"?
I saw several times to Soroban saying "back-substitute". I think it's a pretty cool word to express "turn back $u$ into $x$".

5. Originally Posted by Jhevon
but maybe that's not scary enough for you
It's amazing I ever got past pre-algebra.
Originally Posted by Jhevon
"Anti-substitute"?
Is there an official term for this? I've been using anti-substitute, because it makes me happier than "desubstitute" or "unsubstitute" or "remove the substitution" Although I have to agree with Krizalid that Soroban's "back u into x" is quite amusing. Although "Anti-substitute" has a nice fatal ring to it, which compliments that knot of dread in my stomach that comes from the scary way I integrate.
Originally Posted by Krizalid
Well (ohhh c'mon, this is only chain rule...)
I got the impression the OP knew the answer because of this, but needed it to be shown with substitution for their test.

6. Originally Posted by Krizalid
Well (ohhh c'mon, this is only chain rule...)

$\int {e^{2x} \,dx} = \frac{1}
{2}\int {\left( {e^{2x} } \right)'\,dx} = \frac{{e^{2x} }}
{2} + k.$
indeed!

pretty much any substitution problem can be done in this way. after all, integration by substitution was developed to undo the chain rule. (integration by parts matches the product rule etc...)

i assumed the poster wanted the standard/traditional way to integrate this...

Originally Posted by angel.white
It's amazing I ever got past pre-algebra.
hehe

Originally Posted by Krizalid
I saw several times to Soroban saying "back-substitute". I think it's a pretty cool word to express "turn back $u$ into $x$".
Is there an official term for this? I've been using anti-substitute, because it makes me happier than "desubstitute" or "unsubstitute" or "remove the substitution" Although I have to agree with Krizalid that Soroban's "back u into x" is quite amusing.
it is not just cool and amazing, but it is the official term. the text books i have read use "back-substitute" to describe the process.

Originally Posted by angel.white
Although "Anti-substitute" has a nice fatal ring to it, which compliments that knot of dread in my stomach that comes from the scary way I integrate.
ok, fair enough. keep using it. there's no problem.

i found the term to be, well, cool, which is why i said anything

Originally Posted by angel.white
I got the impression the OP knew the answer because of this, but needed it to be shown with substitution for their test.
i did too.

7. Originally Posted by Krizalid
Well (ohhh c'mon, this is only chain rule...)
$\int {e^{2x} \,dx} = \frac{1}
{2}\int {\left( {e^{2x} } \right)'\,dx} = \frac{{e^{2x} }}
{2} + k.$
Or we can do it like,
$\int e^x\cdot e^x~dx$

$e^x~dx = d(e^x)$

$\int e^x~d(e^x)$

$\frac{(e^x)^2}{2} = \frac{e^{2x}}{2}$
I think this is easier for him to see..

8. I have never seen so many different ways to solve $\int {e^{2x} } dx$ in one place before. It is really quite amusing.

9. Originally Posted by xifentoozlerix
I have never seen so many different ways to solve $\int {e^{2x} } dx$ in one place before. It is really quite amusing.
isn't it!

there was a thread once, where we did a lot of methods to find $\int \ln x~dx$

i like seeing things like that...

10. Originally Posted by Jhevon
isn't it!

there was a thread once, where we did a lot of methods to find $\int \ln x~dx$

i like seeing things like that...
Classic example of back substitution...

$u=ln(x)$
$du=\frac{1}{x}dx$

$dx=xdu$
$x=e^u$

$dx=e^u du$

$\int ln(x) dx = \int ue^u du$

Proceed with integration by parts (tabular integration if you are like me).

11. Originally Posted by colby2152
Classic example of back substitution...

$u=ln(x)$
$du=\frac{1}{x}dx$

$dx=xdu$
$x=e^u$

$dx=e^u du$

$\int ln(x) dx = \int ue^u du$

Proceed with integration by parts (tabular integration if you are like me).
ah! the standard way to do this is to do integration by parts on the original, that is, $\int \ln x~dx$, using $u = \ln x$ and $dv = 1$. so that would save you the trouble of your first substitution

the way, Krizalid, i believe, did it, was to, again, reverse the product rule

consider $\frac d{dx} x \ln x$

we know, $\frac d{dx} x \ln x = \ln x + 1$

now integrate both sides and solve for $\int \ln x~dx$, and the result follows.

incidentally, the same reverse-product rule method could be used on $\int ue^u~du$, as opposed to integration by parts or tabular integration (if you're like Krizalid, i guess)

12. Originally Posted by Jhevon
there was a thread once, where we did a lot of methods to find $\int \ln x~dx$

i like seeing things like that...
Yes, having $1\le x\le a$ (where $a$ is a positive integer) or the well known improper integral for $0\le x\le1,$ a double integration trick works nice for both of them. (Also series and gamma function for the last one.)

----

About $\ln x$ without integration by parts, it's like Jhevon did: define $f(x)=x\ln x,$ contemplate its derivative and integrate. (Some integrals can be killed with this method.)