Question

I had a question about using the U substitution for finding antiderivatives:

Answer 1

\[\int\limits_{}^{?}(e^x+e ^{-x})/2\]

Answer 2

That 2 is supposed to be inside the parenthesis, but I didn't know how to do that. So anyways, would a good approach to this be using the e^-x as your U?

Answer 3

What do you mean by... inside the parenthesis? :O I don't think that really changes the problem :D

Answer 4

Oh! Really? I didn't know, sorry. I thought that would've changed the problem entirely.

Answer 5

\[\large \int\limits \frac{e^x+e^{-x}}{2}dx\]Pull the constant outside, then write it as 2 separate integrals, maybe that will make it easier to work with :)\[\large \frac{1}{2}\int\limits e^x+e^{-x}dx\qquad \rightarrow \qquad \frac{1}{2}\int\limits e^x dx+\frac{1}{2}\int\limits e^{-x} dx\]Now you should only really worry about doing a U sub for the second integral. The first one is pretty straight forward :D

Answer 6

Right, okay lemme see if I get this.

Answer 7

Oh wait...okay so basically you don't need to use the U sub. then, correct?

Answer 8

If you are familiar with taking the derivative of an exponential, then you'll very quickly get familiar with the anti-derivative of an exponential. Unless you already understand the process? :)

You certainly don't need a U-sub, unless you still need to learn this concept with what happens to the constant on the exponent. the negative one in this case.

Answer 9

Can you show me that concept, please? I think I may have forgotten. Is the answer for that just -e^-x?

Answer 10

Yes. Since it's just a negative, it's pretty straight forward. But lemme give a quick example with a different constant so it will sink in. :D

Answer 11

Thank you so much, I appreciate this.

Answer 12

Sorry it takes me a little bit to type stuff :C the equation tool CONSTANTLY crashes my browser...

Answer 13

\[\large \int\limits e^{2x}dx\]Hmmm let's say we don't know this integral.
Let's start by taking the derivative of this term and see what happens.\[\huge (e^{2x})'\quad =\quad e^{2x}(2x)'\quad=\quad2e^{2x}\]Hmm ok. Let's take the integral of the thing we just differentiated.

\[\huge \int\limits\limits 2e^{2x}dx \quad = \quad e^{2x}\]How did we end up solving that integral? We actually DIVIDED by 2 didn't we? See how the 2 disappeared?

Answer 14

Oooh! Okay that makes sense.

Answer 15

So we divided by 2... now let's return to the integral we started with.\[\huge \int\limits e^{2x}dx\]The fact that there isn't a 2 in front shouldn't affect the integration process at all. So that's basically the rule for exponentials, you DIVIDE by the coefficient on the exponent.

Answer 16

If the exponent is MORE than just x, let's say x^2, then that wouldn't apply :O If it's a constant attached to X though, then we're ok :D

Answer 17

\[\huge \int\limits\limits e^{2x}dx=\frac{1}{2}e^{2x}\]Just in case I didn't make it clear :D heh

Answer 18

Ah, okay. Thank you, I think I understand that now. So, what if it was like...say e^ln2? Would that apply? Sorry, I'm just throwing a random question in case I see it on my exam.

Answer 19

Well try to be careful, sometimes you'll get things thrown at you that LOOK confusing, but are really just CONSTANTS.

I like to call them fancy constants. They have their tuxedo on, they're really for the ball. But it's just an impostor, posing as something else.

Example:\[\large (e^{\ln2})'=0\]See how it contains no X's?

When we integrate, it will be a tiny bit different.

Answer 20

It's similar to taking the integral of .... 3. An x pops back into the problem right?\[\large \int\limits 3 dx=3x\]\[\large \int\limits e^{\ln2}dx=e^{\ln2}x\]

Answer 21

Oooh!!! Ha! That's really tricky, I guess that ln just messes me up sometimes. Wow :D

Answer 22

If we had say,\[\large \int\limits e^{\ln2 x}dx\]Then yes, this rule that we went over earlier would still apply :)\[\frac{1}{\ln 2}e^{\ln 2 x}\]

Answer 23

woops, the x is suppose to be outside of the natural log, i didn't write that very clearly.

Answer 24

Okay so, basically, for that problem, you do the exact same thing as the other log problem so, you just divide that with the ln2? Right? I Think I'm getting it.

Answer 25

Yah, when you DIFFERENTIATE, you end up MULTIPLYING by the coefficient on x.

When you INTEGRATE, you DIVIDE by the coefficient on x.

Answer 26

Here's a tricky one to try on for size :D\[\huge \int\limits e^{3x+1}dx\]

Answer 27

Okay okay lemme see.

Answer 28

...okay this is what I got...soooo would it be \[1\div3x+1\times e ^{3x+1}\] ?

Answer 29

So the issue with this problem is that I threw that nasty +1 into the exponent! :O

Answer 30

we gotta deal with that before we can integrate XD I so sneaky! lol

Answer 31

Haha! WEll that's good! I have to learn this anyways :P

Answer 32

Remember your rules of exponents?\[\huge e^{a+b}=e^a\cdot e^b\]Does this look familiar? :D

Answer 33

Is it one of those log rules? It does look familiar

Answer 34

Ooh! Or those rules when working with exponents.

Answer 35

It's an exponent rule, but yah you prolly learned them at the same time you learned log rules :D

Answer 36

Ooh! So...I'm guessing I had to split the problem first..and then integrate it?

Answer 37

Yah :D

Answer 38

Gah, that changes everything, okay okay lemme see if I got this now.

Answer 39

Okay so I got \[\int\limits_{?}^{?} 1\div3x \times \int\limits_{?}^{?}e^1\]

Answer 40

Then...it should just be 1/3x (e^3x) and then times that with e^1 which is....zero? Sorry I'm a bit shaky with those logs.

Answer 41

Oh wait nope! It's 1!

Answer 42

Oooh!!! no no, my fault. That would be 1/3 (e^3x) times that by e^1 which is just 1/3(e^3x)

Answer 43

So we're not writing it as 2 integrals. Just rewrite the exponential using exponent rules.\[\huge \int\limits e^{3x+1}dx \quad =\quad \int\limits e^{3x}\cdot e^1dx\]
See how I just rewrote the exponential term within the integral? :O

Answer 44

That extra E is throwing you off I think :D hah

Answer 45

Ugh probably!

Answer 46

It's just a constant, so let's pull it outside of the integral.\[\huge e \int\limits e^{3x}dx\]

Answer 47

Now apply the rule for integrating exponentials at this point :D Ignore the e on the outside, it will just be put back in after you integrate.

Answer 48

Soo....when we integrate the 3x that would just be 1/4(x^4) right? As its' antiderivative?

Answer 49

Ugh! Wait! I think I messed up! ....Crap....Wait, so if we pulled out the e constant, where did that other e come from?

Answer 50

Hmmm :O

Answer 51

From the Rule of Exponents :O
We have a SUM within the exponent.
So we can rewrite it as the PRODUCT of 2 different exponentials with the same base.
Confused about this part?\[\huge e^{3x+1} \quad = \quad e^{3x}\cdot e^1\]That's where the extra E is coming from

Answer 52

Ahhh...okay, so after pulling out that e we put it outside the integral. Okay...and then we just integrate the e^3x which would be....1/3 (3^3x) ....Correct?

Answer 53

Crap i meant e^3x

Answer 54

Yah looks good :) then we can just bring the E back inside after that.

We could even put it back into the exponent if we wanted to.\[\large e\int\limits e^{3x}dx \quad = \quad e\frac{1}{3}e^{3x}\quad = \quad \frac{1}{3}e^{3x+1}\]

Answer 55

Great! Thank you so much!! I think I'm getting this now. Strange how they never showed us this technique in school .

Answer 56

Actually for something that, with the +1.. THAT is probably a good candidate for a U-sub. Doing these silly little exponent rules doesn't really seem worth it :D But I dunno...

Answer 57

Well I'll try it later with the U sub and see if I get the same answer. That was a good practice though, thank you.

Answer 58

Do you mind if I ask this one last question, I apologize if I'm taking up so much of your time.

Answer 59

sure :D

Answer 60

Okay so I have this problem...I think I may know how to solve this...but I Just wanna make sure I"m doing it correctly...So we have