
#1
Sep2104, 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. 



#2
Sep2104, 02:13 AM

P: 406

If you try differentiating e^{x}, it comes to (1)e^{x} which is e^{x}. So it makes sense!




#3
Sep2104, 02:16 AM

P: 86

Yes, but how do u get the integral without knowing thats the answer?




#4
Sep2104, 02:16 AM

P: 86

Integral of e^(x)
Oh crap nevermind, its easy, just use u substitution u = x




#5
Sep2104, 11:24 PM

P: 492

The integral of e^u du = e^u * (du/dx).




#6
Mar2209, 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... 



#7
Mar2309, 02:49 AM

HW Helper
P: 3,436

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



#8
Mar2309, 04:14 AM

P: 34

[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
Mar2309, 06:21 AM

Math
Emeritus
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Thanks
PF Gold
P: 38,895





#10
Mar2309, 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) = e^{u} 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 usubstitution. 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) = e^{u} 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] 



#11
Mar2309, 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] 



#12
Mar2409, 01:22 AM

HW Helper
P: 3,436

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
Mar2409, 05:39 AM

P: 878





#14
Jul710, 06:14 AM

P: 1





#15
Jul710, 08:53 AM

HW Helper
P: 3,436





#16
Jul710, 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.
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 backdifferentiation  when solving something so simple. 



#17
Sep610, 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? 



#18
Sep610, 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.



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