Indefinite Integral Question; e^x

In summary, the homework statement is to find the indefinite integral (preferably using u-substitution), but I'm slightly confused as to which path I'm supposed to take. I'm also having trouble with the +∫2 dx. Somebody please tell me if I have the right answer or if I'm on the right track.
  • #1
Celadon
11
0

Homework Statement


Find the indefinite integral (preferably using u-substitution):
∫(ex-e-x)2 dx


Homework Equations


N/A


The Attempt at a Solution


To be honest, I'm slightly confused as to which path I'm supposed to take with this, especially since I'm not sure what I should be using u to substitute here. This is one of my attempts at solving the question:

∫[(ex-e-x)2 dx]
= ∫[e2x dx] - ∫[e-2x dx]
= (e2x / 2) + (1 / 2e2x) + C

Could someone please tell me if I have the right answer / I'm on the right track / I'm completely wrong? If the latter, I'd appreciate it if you could help get me on track!

Thanks in advance!
 
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  • #2
You almost got it. But you've forgotten the cross-terms when you did the square.

There is a good substitution you could use instead. But your method is just as good.

Edit: p.s. welcome to physicsforums, your name is good. reminds me of the town in pokemon
 
  • #3
If you need the u-substitution, try just putting u=ex.

(Remember that 1/a is the same as a-1)


EDIT: Also in your attempt you should also have


∫e2x dx + ∫e-2x dx +∫2 dx
 
  • #4
Before I continue asking questions, thanks for all the help! I really do appreciate it and I'm sorry if I'm a bit slow, haha.

Now then...
@BruceW: Thanks for the help, the welcome, and the name compliment! I'm actually a fan of the colour itself and the city in Pokemon, heh.
@rock.freak667: Ugh, how could I forget 1/a = a-1? Thanks for the reminder. I have a question about the +∫2 dx, though. How exactly did it get there?

I think I'm actually a bit more confused than I initially thought I was. Let's say I went with this method:

u=ex
du=exdx ; dx=du/ex

...and then what? This is what I'm leaning towards:
=∫(u2 dx) - ∫(1/u2 dx)

Do I substitute dx for du/ex or am I over-complicating things? Oh man, my head's going to burst. @@
 
  • #5
(ex-e-x)2 = (ex-e-x)(ex-e-x)

Multiply it out. (Use FOIL, if that's what you're used to.)
 
  • #6
(e^x-e^(-x))^2=(e^x-e^(-x))*(e^x-e^(-x)). Multiply that out and you'll see where the '2' came from. rock.freak667 had the sign wrong on it.
 
  • #7
Okay, do I tried expanding it and then went through this:

(ex-e-x)2 = (ex-e-x)(ex-e-x) = e2x + e-2x

∫(e2x + e-2x) dx
u = e2x
du = (e2x / 2) dx

Final solution:
(u2/e2x) + (2 / e2x)ln|e2x|
 
Last edited:
  • #8
Celadon said:
Okay, do I tried expanding it and then went through this:

(ex-e-x)2 = (ex-e-x)(ex-e-x) = e2x + e-2x
No, this is incorrect, as Sammy and Dick pointed out. (a - b)2 ≠ a2 - b2, which is essentially what you are doing.
When you expand (ex - e-x)2 you should get three terms, not two.

Also, there is an error below - you shouldn't get any log terms.
Celadon said:
∫(e2x + e-2x) dx
u = e2x
du = (e2x / 2) dx

Final solution:
(u2/e2x) + (2 / e2x)ln|e2x|
 
  • #9
Oh I see it now.
(ex - e-x)(ex - e-x) = e2x - 2 + e-2x

I somehow wrote "+ e0 - e0" at some point, oops. Thanks!
Time to try again. (:
 
  • #10
For you last error, I think this is what you did:

∫dx/e2x = ln|e2x|

That is incorrect. Don't write the integrand as a fraction - use the negative exponents.
 
  • #11
Okay, another attempt. Hope I fixed more than I broke, this time. :/
Thanks again everyone, I'm sorry for bothering you all. D:

∫(ex-e-x)2dx = ∫(e2x-2+e-2x)dx
= ∫e2xdx + ∫e-2xdx - ∫2dx

Let u = ex
du = exdx

∫e2xdx + ∫e-2xdx - ∫2dx = (1/ex)(u3/3)+(1/ex)(u-1/-1)-2x+C
= (1/ex)[(u3/3)+(u-1/-1)]-2x+C
= (1/ex)[(e3x/3) - (1/ex)]-2x+C
= (1/ex)[(e4x-3)/(3ex)]-2x+C
= [(e4x-3)/(3e2x)]-2x+C
 
Last edited:
  • #12
Celadon said:
Okay, another attempt. Hope I fixed more than I broke, this time. :/
Thanks again everyone, I'm sorry for bothering you all. D:

∫(ex-e-x)2dx = ∫(e2x-2+e-2x)dx
= ∫e2xdx + ∫e-2xdx - ∫2dx

Let u = ex
du = exdx

∫e2xdx + ∫e-2xdx - ∫2dx = (1/ex)(u3/3)+(1/ex)(u-1/-1)-2x+C
= (1/ex)[(u3/3)+(u-1/-1)]-2x+C
= (1/ex)[(e3x/3) - (1/ex)]-2x+C
= (1/ex)[(e4x-3)/(3ex)]-2x+C
= [(e4x-3)/(3e2x)]-2x+C
What is [itex]\displaystyle\int du \,?[/itex]
 
  • #13
SammyS said:
What is [itex]\displaystyle\int du \,?[/itex]

ex+C?
 
  • #14
Celadon said:
ex+C?
Ignore my last post.

If u = ex and du = exdx,

then [itex]\displaystyle\int e^{2x}dx[/itex] becomes what integral in terms of u ?
 
  • #15
SammyS said:
Ignore my last post.

If u = ex and du = exdx,

then [itex]\displaystyle\int e^{2x}dx[/itex] becomes what integral in terms of u ?

∫u2dx? :x

[EDIT] Another attempt:
2e2x-2x-2e-2x+C

x_x
 
Last edited:
  • #16
Celadon said:
∫u2dx? :x

I think I'm confused as to what happens with du. ._.

If du = exdx, then dx = du(1/ex), so:
(1/ex)∫u2du?
Yes. You appear to be confused.

You can't take 1/ex out of the integral. In fact you should leave it inside the integral as 1/u .

What is u2/u ?
 
  • #17
SammyS said:
Yes. You appear to be confused.

You can't take 1/ex out of the integral. In fact you should leave it inside the integral as 1/u .

What is u2/u ?

u1 or just u? xD

(also, I edited my previous post with another answer in case you missed it.)
 
  • #18
Celadon said:
u1 or just u? xD

(also, I edited my previous post with another answer in case you missed it.)

OK, so what is [itex]\displaystyle\int u \,du \,?[/itex]
 
  • #19
(u2)/2? :x
 
  • #20
Celadon said:
(u2)/2? :x

What does the ":x" mean ?
 
  • #21
SammyS said:
What does the ":x" mean ?

Just an emoticon; I add them to a lot of my sentences out of habit, haha... Sorry.
Okay so, I tried it again like this:

u=ex
du=exdx
dx=du(1/ex)

∫u2dx-2x+∫u-2dx = ∫u2(1/u)du-2x+∫u-2(1/u)du
=∫udu - 2x + ∫u-3du
=u2/2 - 2x + (u-2/-2) + C
=u2/2 - 2x - 1/(2u2) + C
=e2x/2 - 2x - 1/(2e2x) + C
 
  • #22
Celadon said:
Just an emoticon; I add them to a lot of my sentences out of habit, haha... Sorry.
Okay so, I tried it again like this:

u=ex
du=exdx
dx=du(1/ex)

∫u2dx-2x+∫u-2dx = ∫u2(1/u)du-2x+∫u-2(1/u)du
=∫udu - 2x + ∫u-3du
=u2/2 - 2x + (u-2/-2) + C
=u2/2 - 2x - 1/(2u2) + C
=e2x/2 - 2x - 1/(2e2x) + C
Correct.

It would likely be expressed in a manner closer to the given problem.

e2x/2 - 2x - e-2x/2 + C

or

(e2x - e-2x)/2 - 2x + C
 
  • #23
SammyS said:
Correct.

It would likely be expressed in a manner closer to the given problem.

e2x/2 - 2x - e-2x/2 + C

or

(e2x - e-2x)/2 - 2x + C

Thank you very much for all your help! I can finally move on and not worry about this one question anymore. The sense of relief from when you solve a math problem.. words can't describe it.

I really do appreciate your help. :D
 
  • #24
just to add one last thing: I think the substitution that was hinted was the hyperbolic function, because that makes the integral nice and simple. But your way is good as well.

Edit: actually, its not much more simple. It just depends if you are more comfortable with sinusoids or exponentials.
 
Last edited:
  • #25
Celadon said:
(u2)/2? :x

SammyS said:
What does the ":x" mean ?

Celadon said:
Just an emoticon; I add them to a lot of my sentences out of habit, haha... Sorry.
Celadon, that's not a good idea in a mathematics context, especially when x already has a meaning. You don't want to put anything in that prevents someone from understanding what you're trying to say.
 

Related to Indefinite Integral Question; e^x

1. What is an indefinite integral?

An indefinite integral is an expression that represents the antiderivative of a function. It is a mathematical operation that reverses the process of differentiation, and it is commonly used in calculus to find the original function from its derivative.

2. What is the indefinite integral of e^x?

The indefinite integral of e^x is e^x + C, where C is a constant. This can be written as ∫e^x dx = e^x + C. In other words, the integral of e^x is equal to the original function e^x plus a constant value.

3. How do you solve an indefinite integral of e^x?

To solve an indefinite integral of e^x, you can use the basic rules of integration, such as the power rule and constant multiple rule. In the case of e^x, you can simply use the fact that the integral of e^x is e^x + C, where C is a constant. You can also use substitution or integration by parts, depending on the complexity of the function.

4. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, while an indefinite integral does not. This means that a definite integral gives a numerical value, while an indefinite integral gives an expression that represents a family of functions. In other words, the definite integral is a number, while the indefinite integral is a function.

5. What are some real-life applications of indefinite integrals?

Indefinite integrals have many real-life applications, especially in physics and engineering. For example, they are used to calculate the total distance traveled by an object with varying speed, the total work done by a variable force, and the total charge accumulated in a capacitor. They are also used in economics to calculate total revenue and total cost functions.

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