Integral of -e^(-x)


by CinderBlockFist
Tags: integral
CinderBlockFist
CinderBlockFist is offline
#1
Sep21-04, 02:10 AM
P: 86
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 dont 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.
Phys.Org News Partner Mathematics news on Phys.org
Hyperbolic homogeneous polynomials, oh my!
Researchers help Boston Marathon organizers plan for 2014 race
'Math detective' analyzes odds for suspicious lottery wins
recon
recon is offline
#2
Sep21-04, 02:13 AM
recon's Avatar
P: 406
If you try differentiating -e-x, it comes to (-1)-e-x which is e-x. So it makes sense!
CinderBlockFist
CinderBlockFist is offline
#3
Sep21-04, 02:16 AM
P: 86
Yes, but how do u get the integral without knowing thats the answer?

CinderBlockFist
CinderBlockFist is offline
#4
Sep21-04, 02:16 AM
P: 86

Integral of -e^(-x)


Oh crap nevermind, its easy, just use u substitution u = -x
VinnyCee
VinnyCee is offline
#5
Sep21-04, 11:24 PM
P: 492
The integral of e^u du = e^u * (du/dx).
roseSCC9
roseSCC9 is offline
#6
Mar22-09, 09:03 AM
P: 1
If you have e^(-x), you must divide e^(-x) by (-1), cuz (-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...
Mentallic
Mentallic is offline
#7
Mar23-09, 02:49 AM
HW Helper
P: 3,436
Quote Quote by roseSCC9 View Post
If you have e^(-x), you must divide e^(-x) by (-1), cuz (-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]
ThirstyDog
ThirstyDog is offline
#8
Mar23-09, 04:14 AM
P: 34
Quote Quote by Mentallic View Post
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]
HallsofIvy
HallsofIvy is online now
#9
Mar23-09, 06:21 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,895
Quote Quote by Mentallic View Post
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.
slider142
slider142 is offline
#10
Mar23-09, 05:30 PM
P: 878
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]
deancodemo
deancodemo is offline
#11
Mar23-09, 06:52 PM
P: 20
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]
Mentallic
Mentallic is offline
#12
Mar24-09, 01:22 AM
HW Helper
P: 3,436
Quote Quote by ThirstyDog View Post
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]
Quote Quote by HallsofIvy View Post
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 highschool 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]
slider142
slider142 is offline
#13
Mar24-09, 05:39 AM
P: 878
Quote Quote by Mentallic View Post
...
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.
cookie5
cookie5 is offline
#14
Jul7-10, 06:14 AM
P: 1
Quote Quote by Mentallic View Post
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 :)
Mentallic
Mentallic is offline
#15
Jul7-10, 08:53 AM
HW Helper
P: 3,436
Quote Quote by ThirstyDog View Post
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]
Quote Quote by HallsofIvy View Post
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.
Mentallic
Mentallic is offline
#16
Jul7-10, 08:58 AM
HW Helper
P: 3,436
And after reading the rest of the posts, it seems like my restriction has been mentioned multiple times.

Quote Quote by cookie5 View Post
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
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.
dazzwater
dazzwater is offline
#17
Sep6-10, 12:50 AM
P: 2
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?
Mentallic
Mentallic is offline
#18
Sep6-10, 01:31 AM
HW Helper
P: 3,436
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.


Register to reply

Related Discussions
integral more general then Lebesgue integral? Calculus 7
integral - Double integral Calculus & Beyond Homework 2
Need help w/ an integral Calculus 4
QM Integral and Online Integral Tables Calculus 5
Integral using integral tables Calculus 2