If ∫_a^b f(x)dx = a+2b, what is ∫_a^b (f(x)+5)dx?

  • Thread starter Thread starter UrbanXrisis
  • Start date Start date
  • Tags Tags
    Integration Stuck
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
30 replies · 4K views
UrbanXrisis
Messages
1,192
Reaction score
1
[tex]\int_a^b f(x)dx = a+2b[/tex]
[tex]\int_a^b (f(x)+5)dx =?[/tex]
[tex]\int_a^b (f(x)+5)dx =\int_a^b f(x)dx+\int_a^b 5dx[/tex]
[tex]a+2b+\int_a^b 5dx[/tex]

I'm stuck, what should be my next step?
 
Physics news on Phys.org
Are you serious? You can't integrate [tex]\int_a^b5dx[/tex]?

Isn't that the same as [tex]5\int_a^b dx= 5(b-a)[/tex]?

Isn't that about the first thing you learned in integration?
 
hehe, thanks

what's the rule for [tex]\int e^x[/tex]?
is it [tex]e^x \int x[/tex]?
 
[tex]\int e^x[/tex] = [tex]\int e^x dx[/tex] ?
 
so if a question was [tex]\int e^{\frac{x}{2}} dx = e^{\frac{x}{2}} \int \frac{x}{2} dt[/tex]
 
[tex]\int e^x dx= e^x+ C[/tex]

because [tex]\frac{d e^x}{dx}= e^x[/tex], of course.
 
UrbanXrisis said:
so if a question was [tex]\int e^{\frac{x}{2}} dx = e^{\frac{x}{2}} \int \frac{x}{2} dt[/tex]


No,if course not.U need to make a substitution

[tex]\frac{x}{2}=u[/tex]

Daniel.

EDIT:BTW,knowledge of integration techniques assumed knowledge of differentiation methodes.
 
[tex]\int e^{\frac{x}{2}} dx[/tex]
[tex]u=x/2[/tex]
[tex]du=1/2dx[/tex]
[tex]\int e^{u} 2du[/tex]
[tex]=2e^{\frac{x}{2}}[/tex]
 
UrbanXrisis said:
so if a question was [tex]\int e^{\frac{x}{2}} dx = e^{\frac{x}{2}} \int \frac{x}{2} dt[/tex]


Let u= x/2. Then 2u= x so 2du= dx.
[tex]\int e^{\frac{x}{2}}dx[/tex] becomes [tex]2\int e^u du= 2e^u+ C= 2e^{\frac{x}{2}}+ C[/tex]

Surely you've learned simple substitutions.

[tex]e^x\int \frac{x}{2}dt[/tex], on the other hand, is [tex]e^x(\frac{x^2}{4}+ C)[/tex].
 
Last edited by a moderator:
is [tex]cos^2(x)=cos(x)cos(x)[/tex]
so..
[tex]\frac{d}{dx}cos^2(x)=-2sin(x)cos(x)[/tex]
 
UrbanXrisis said:
is [tex]cos^2(x)=cos(x)cos(x)[/tex]
so..
[tex]\frac{d}{dx}cos^2(x)=-2sin(x)cos(x)[/tex]


Yes, by golly!

(I think we are posting a cross purposes now!)
 
so if e is raised to any exponet that is not x, then I must use a subsitution when integrating?
 
Yes.Always.Make that depends from case to case.Usually the antiderivatives of exponentials of "weird" arguments are not expressible in terms of "elementary functions".Simples example

[tex]\int e^{-x^{2}} \ dx[/tex]

Daniel.
 
Last edited:
Just out of curiousity, when I did that particular integral on Mathematica's Online Integrator, I got:

[tex]\int e^{-x^{2}} \ dx = \frac{1}{2}\sqrt{\pi} ERF[x] + C[/tex]

Can someone tell me what Erf is?
 
Well, then that makes perfect sense. Is that something that is standardly used in Calculus? The constant seems to be random.
 
It is a very famous function.It is connected to the Gauss' normal distribution (error function).

And that is a normalization constant ...

[tex]\lim_{x\rightarrow +\infty} \mbox{erf} \ (x) =1[/tex]

Daniel.
 
The normalization constant does vary by source. It's basically used to decide what [itex]\lim_{x \rightarrow \infty} \mathop{erf} x[/itex] should be. Dividing by √π makes it an integer. This particular normalization makes it 1, I think. Incidentally, the error function is closely connected with the normal probability distribution.[/size]
 
dextercioby said:
Yes.Always.Make that depends from case to case.Usually the antiderivatives of exponentials of "weird" arguments are not expressible in terms of "elementary functions".Simples example

[tex]\int e^{-x^{2}} \ dx[/tex]

Daniel.

how do you do this problem without a calculator?
[tex]u= -x^{2}[/tex]
[tex]du=-2x dx[/tex]

what next?
 
That was the whole point. You can't express that with elementary functions. You won't be able to do that by hand... unless you know what Hurkyl and Dextercioby were just talking about.
 
You don't. [tex]e^{-x^2}[/tex] is one of the many integrable functions (in fact most of them) whose anti-derivatives, as dextercioby said, cannot be written in terms of "elementary functions".

You can, of course, write that [tex]\int e^{-x^2}dx= \frac{\sqrt{\pi}}{2}Erf(x)- just rewriting what dextercioby also said, because that is the way "Erf(x)", not an elementary function, is defined![/tex]
 
oh, what level calculus in college do I learn this? Or is this for grad school?
 
Which I think is the best way to learn anything I think. When no one is there to really help it kind of forces you to think on a different level, for me at least.
 
basically, would my high school math teach know how to solve it if I asked him that question?