What is the Integral of -e^(-x)?

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In summary, the discussion revolved around the integral of e^(-x)dx and how to arrive at the answer -e^(-x). The participants discussed the use of u substitution and the chain rule in solving this type of integral. They also mentioned a general rule for integrating exponentials, which involves dividing by the derivative of the power. However, the correctness of one of the participants' result was questioned and the correct solution was provided by another participant.
  • #1
CinderBlockFist
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Hi guys, I am looking in my calculus book, and it tells me in the example problem that the Integral of e^(-x)dx = -e^(-x)


I don't see how u get this answer. I know the integral of e^x = e^x but how does that negative sign come out to the left? Can someone explain this step by step, thanks.
 
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  • #2
If you try differentiating -e-x, it comes to (-1)-e-x which is e-x. So it makes sense!
 
  • #3
Yes, but how do u get the integral without knowing that's the answer?
 
  • #4
Oh crap nevermind, its easy, just use u substitution u = -x
 
  • #5
This is a theorem.

The integral of e^u du = e^u * (du/dx).
 
  • #6
If you have e^(-x), you must divide e^(-x) by (-1), because (-1) is the coefficient of (-x)

that's the way you integrate

when you have e^(-3x) you must divide e^(-3x) by (-3)... so on...
 
  • #7
roseSCC9 said:
If you have e^(-x), you must divide e^(-x) by (-1), because (-1) is the coefficient of (-x)

that's the way you integrate

when you have e^(-3x) you must divide e^(-3x) by (-3)... so on...

or more simply, divide by the derivative of the power.

[tex]\int_a^b{e^{f(x)}}=\left [\frac{e^{f(x)}}{f'(x)} + C \right ]^b_a[/tex]
 
  • #8
Mentallic said:
or more simply, divide by the derivative of the power.

[tex]\int_a^b{e^{f(x)}}=\left [\frac{e^{f(x)}}{f'(x)} + C \right ]^b_a[/tex]

I don't think that your last bit is true in general... I think that
[tex]\int_a^b{ f'(x) e^{f(x)}}dx=\left [e^{f(x)} + C \right ]^b_a[/tex]
is what you mean and you have divided through. This is only correct when the f'(x) is constant... you might have known this, I just found it confusing the way you stated. Also for definite integrals you can omit the constant.

For example
[tex]\int_a^b{e^{-x^{2}}}=\left [\frac{\sqrt{\pi}}{2}erf(x) \right ]^b_a[/tex]
 
  • #9
Mentallic said:
or more simply, divide by the derivative of the power.

[tex]\int_a^b{e^{f(x)}}=\left [\frac{e^{f(x)}}{f'(x)} + C \right ]^b_a[/tex]
That is completely wrong. In order to use "substitution" with u= f(x), f'(x), unless it is a constant, must already be in the integral.
 
  • #10
You have been given:
[tex]\int e^u du = e^u + C[/tex]
where C is an undetermined constant of integration.
You have come upon the integral:
[tex]\int f(g(x)) dx[/tex]
where you know the indefinite integral of f(u), but not f(g(x)). In your case f(u) = eu and g(x) = -x. If we go back to derivatives, we may remember a related rule called the chain rule:
(f(g(x))' = f'(g(x))*g'(x)
In other words:
[tex]\int f'(g(x))*g'(x) dx = f(g(x)) + C[/tex]
In some texts, they will use shorthand Leibnitz notation: Let u = g(x). Then du = g'(x) dx. The above integral is then written:
[tex]\int f'(u) du = f(u) + C[/tex]
This is usually called u-substitution. Note that you have to have both f'(g(x)) = f'(u) and du = g'(x) dx in the integrand. Sometimes you will have to multiply the integrand by a creative version of 1 in order to make this happen. In your example, let f'(u) = eu since we already know how to integrate that and of course u = -x. Then du = -1 dx. Then we need our integrand to be f'(u) du = -e-x dx. Unfortunately, our integrand is actually -(e^-x dx) = -f'(u) du. Luckily, the -1 can be factored out of the integrand as it is a constant, so we have
[tex]-\int f'(u) du = - f(u) + C = -e^u + C[/tex]
 
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  • #11
Here is a simple rule you can use:
[tex]
\int f(ax + b) \, dx = \frac{F(ax + b)}{a} + C
[/tex]

where F(x) is a primitive function of f(x). We can use this like so:
[tex]
\int e^{-x} \, dx = \frac{e^{-x}}{-1} + C
[/tex]
 
  • #12
ThirstyDog said:
I don't think that your last bit is true in general... I think that
[tex]\int_a^b{ f'(x) e^{f(x)}}dx=\left [e^{f(x)} + C \right ]^b_a[/tex]
is what you mean and you have divided through. This is only correct when the f'(x) is constant... you might have known this, I just found it confusing the way you stated. Also for definite integrals you can omit the constant.

For example
[tex]\int_a^b{e^{-x^{2}}}=\left [\frac{\sqrt{\pi}}{2}erf(x) \right ]^b_a[/tex]

HallsofIvy said:
That is completely wrong. In order to use "substitution" with u= f(x), f'(x), unless it is a constant, must already be in the integral.

Thanks for correcting me. It seems high school doesn't expect any more than the elementary integrals of ef(x), f(x) being a linear function.
Hallsofivy, I'm not exactly sure what you're saying. This result has been given to us in class.

Maybe, [tex]\int f'(x)e^{f(x)}=\frac{f'(x)e^{f(x)}}{f'(x)}+C=e^{f(x)}+C[/tex]
 
  • #13
Mentallic said:
...
Hallsofivy, I'm not exactly sure what you're saying. This result has been given to us in class.

Maybe, [tex]\int f'(x)e^{f(x)}=\frac{f'(x)e^{f(x)}}{f'(x)}+C=e^{f(x)}+C[/tex]

Halls is correct; your previous result makes no sense unless f(x) is linear, in which case f'(x) is just a constant, from which you can derive the result yourself from the chain rule. The far left and far right hand sides of the above quoted integral are correct, and is a straightforward result of the chain rule; your middle step is unjustified.
 
  • #14
Mentallic said:
or more simply, divide by the derivative of the power.

[tex]\int_a^b{e^{f(x)}}=\left [\frac{e^{f(x)}}{f'(x)} + C \right ]^b_a[/tex]

listen to this guy k.. the substitution method is sooooo not necessary in this case.. it just complicates something that is actually very very simple.. above is the general method of integrating exponentials.. it might be easier if u understand integration as the opposite of differentiation.. when differentiating exponentials you multiply by the derivative of function x.. and integrating is the opposite so you divide by the derivative of function x.. anyway good luck :)
 
  • #15
ThirstyDog said:
I don't think that your last bit is true in general... I think that
[tex]\int_a^b{ f'(x) e^{f(x)}}dx=\left [e^{f(x)} + C \right ]^b_a[/tex]
is what you mean and you have divided through. This is only correct when the f'(x) is constant... you might have known this, I just found it confusing the way you stated. Also for definite integrals you can omit the constant.

For example
[tex]\int_a^b{e^{-x^{2}}}=\left [\frac{\sqrt{\pi}}{2}erf(x) \right ]^b_a[/tex]

HallsofIvy said:
That is completely wrong. In order to use "substitution" with u= f(x), f'(x), unless it is a constant, must already be in the integral.

Yes sorry, I was only thinking of functions f(x) that were linear y=ax+b.
 
  • #16
And after reading the rest of the posts, it seems like my restriction has been mentioned multiple times.

cookie5 said:
listen to this guy k.. the substitution method is sooooo not necessary in this case.. it just complicates something that is actually very very simple.. above is the general method of integrating exponentials.. it might be easier if u understand integration as the opposite of differentiation.. when differentiating exponentials you multiply by the derivative of function x.. and integrating is the opposite so you divide by the derivative of function x.. anyway good luck :)

The only quarrel they had was that the result I gave is only correct if f(x)=ax+b, [itex]a\neq[/itex]0. It was my mistake. We should instead stick to the proper formula :smile:
And yes I agree with you. In this problem substitution should be avoided and that's what I try to get students to do - think of integrals as back-differentiation - when solving something so simple.
 
  • #17
Question:

The integral of 2x.e^(x^2) is e^(x^2), isn't it?

Surely the integral of f'(x).e^f(x) is always e^f(x), regardless of whether f'(x) is a constant?
 
  • #18
Yes, but what I did was rather than saying [tex]\int{f'(x)e^{f(x)}dx}=e^{f(x)}+c[/tex] I mistakenly wrote [tex]\int{e^{f(x)}dx}=\frac{e^{f(x)}}{f'(x)}+c[/tex] which is only true for f(x)=ax+b since then f'(x) is a constant.
 
  • #19
u = -x
du = -1dx
dx = -du

so you subsitute e-x for eu and then substitute dx for -du, pull out the negative infront of the integral and you get -eu + C, subsituting back in and get:

-e-x + C
 
  • #20
CinderBlockFist said:
Yes, but how do u get the integral without knowing that's the answer?

A well kept secret - you practically can't get any integral without knowing that's the answer.
 
  • #21
Mentallic said:
Yes, but what I did was rather than saying [tex]\int{f'(x)e^{f(x)}dx}=e^{f(x)}+c[/tex] I mistakenly wrote [tex]\int{e^{f(x)}dx}=\frac{e^{f(x)}}{f'(x)}+c[/tex] which is only true for f(x)=ax+b since then f'(x) is a constant.

You've not done any mistake, you're right in both cases... it's 100% correct and if somebody is interested in fighting, i suggest them to visit university classes... It is correct and it's the same.
 
  • #22
Naii said:
You've not done any mistake, you're right in both cases... it's 100% correct and if somebody is interested in fighting, i suggest them to visit university classes... It is correct and it's the same.

But it isn't, if you take the derivative of both sides, they won't be equal except in the special case that f'(x)=c for some constant [tex]c\neq 0[/tex]
 
  • #23
Mentallic said:
But it isn't, if you take the derivative of both sides, they won't be equal except in the special case that f'(x)=c for some constant [tex]c\neq 0[/tex]

*sigh*
if you take it simple way, it is correct... if you want to digg in and know the answer for any e^f(x) ... you have a problem :) I've got here one problem I'm trying to figure out right now too, it's something like this:
[tex]\int{e^{\int{\frac{2x}{x^{2}+1}}dx}\int{4xe^{2x^{2}}dx}}dx[/tex]
and hopefully it will show up correctly.. O.O it's quiet simple, I've got it done already but you can torture yourself too if you are interested hehe
 
  • #24
Naii said:
*sigh*
if you take it simple way, it is correct... if you want to digg in and know the answer for any e^f(x) ... you have a problem :)
Well isn't that what "100% correct" implies? :tongue:

Naii said:
i've got here one problem I'm trying to figure out right now too, it's something like this:
[tex]\int{e^{\int{\frac{2x}{x^{2}+1}}dx}\int{4xe^{2x^{2}}dx}}dx[/tex]
and hopefully it will show up correctly.. O.O it's quiet simple, I've got it done already but you can torture yourself too if you are interested hehe

Ahh well I haven't studied double integration yet but yeah if I were to take a stab at it, I would guess the answer is... 27
 
  • #25
Mentallic said:
Ahh well I haven't studied double integration yet but yeah if I were to take a stab at it, I would guess the answer is... 27

:D well... I'm just fighting this:
[tex]e^{-ln|x^{2}+1|}=\frac{e^{0}}{e^{ln|x^{2}+1|}}=\frac{1}{|x^{2}+1|}[/tex] any if this is correct? o_O I'm not sure, that minus is uggly... whatever, i'll continue and see the result check if it is correct :)
 
  • #26
Please take this in a new thread. Do not clutter an already cluttered thread.

This is an easy calculus problem.

Firstly you have integrated the first part incorrectly. Strictly speaking you can omitt the absolute value since this is an indefinite integral.

You should therefore get

[tex] \int (x^2+1)e^{2x^2} \, \text{dx} [/tex]

Which does not have a elementary solution, I think you meant to state the problem:

[tex]\int{e^{\int{\frac{2x}{x^{2}+1}}dx} dx \cdot \int{4xe^{2x^{2}}dx}}[/tex]

Which does have a "nice" solution.
 
  • #27
Naii said:
:D well... I'm just fighting this:
[tex]e^{-ln|x^{2}+1|}=\frac{e^{0}}{e^{ln|x^{2}+1|}}=\frac{1}{|x^{2}+1|}[/tex] any if this is correct? o_O I'm not sure, that minus is uggly... whatever, i'll continue and see the result check if it is correct :)

I agree with Nebuchadnezza.

And the absolute value signs are unnecessary. [tex]x^2+1>0[/tex] for all [tex]x\in\Re[/tex]
 
  • #28
let's close this with the correct answer which solves my equation i had, and it works like i thought with that -ln thing... as i said right at the begining, i knew the answer already and i did it correctly...
and i know that these absolute braclets are useless, but it was torn out of the context so ... they are not really uselss if you put it back to the equation :)
[tex]Y(x)=\frac{x^{4}+2x^{2}+C}{x^{2}+1}[/tex]
solves the thing
 
  • #29
Naii said:
*sigh*
if you take it simple way, it is correct... if you want to digg in and know the answer for any e^f(x) ... you have a problem :) I've got here one problem I'm trying to figure out right now too, it's something like this:
[tex]\int{e^{\int{\frac{2x}{x^{2}+1}}dx}\int{4xe^{2x^{2}}dx}}dx[/tex]
and hopefully it will show up correctly.. O.O it's quiet simple, I've got it done already but you can torture yourself too if you are interested hehe
Do it one step at a time. Let u= x^2+ 1 so that du= 2x and the integral in the exponential becomes integral 1/u du= ln u+ C= ln(x^2+ 1)+ C. Where C can be any constant. And then e^{ln(x^2+ 1)+ C= e^c (x^2+ 1)= C'(x^2+ 1). So the first integral is just C'((1/3)x^3+ x+ D) where D is another constant.

To do the second integral let u= 2x^2 so that du= 4x dx and the integral becomes integral e^u du= e^u+ E= e^(2x^2)+ E.

The product of integrals is (C'((1/3)x^3+ x+ D)(e^(2x^2)+ E)

That is certainly NOT what you give.
 
  • #30
integrals were only one side of the equation
 
  • #31
Reading through this thread I can't help thinking to myself, "How many Mathematicians does it take to change a light bulb?"
 

1. What is the formula for the integral of -e^(-x)?

The formula for the integral of -e^(-x) is ∫-e^(-x) dx = e^(-x) + C, where C is the constant of integration.

2. How do you solve the integral of -e^(-x)?

To solve the integral of -e^(-x), you can use the substitution method. Let u = -x, then du = -dx. This will transform the integral into ∫e^u du, which can be easily solved by using the power rule for integration.

3. What is the significance of the constant of integration in the integral of -e^(-x)?

The constant of integration, denoted by C, represents the family of solutions to the indefinite integral. It is added to the result of the integral to account for all possible solutions.

4. Can the integral of -e^(-x) be solved using other methods?

Yes, besides the substitution method, the integral of -e^(-x) can also be solved using integration by parts or the u-substitution method.

5. Is there a graphical representation of the integral of -e^(-x)?

Yes, the integral of -e^(-x) represents the area under the curve of the function -e^(-x) on the x-axis. This area can be visualized using a graphing calculator or by plotting the function and shading the area below it.

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