Required to prove that ∫f(x)dx[b, a] =∫f(x−c)dx [b+c, a+c]

  • Thread starter Thread starter jodecy
  • Start date Start date
jodecy
Messages
5
Reaction score
0

Homework Statement


required to prove that
∫()[, ] =∫(−) [+, +]

where f is a real valued function integrable over the interval [a, b]

Homework Equations



∫() [, ]=()−()

The Attempt at a Solution




∫() [b, a]=()−()

∫(−) [+, +]=(+−)−(+−)=()−()
∴∫()[, ] =∫(−) [+, +]

right i placed the interval in the [] brackets


is this correct?
 
Physics news on Phys.org


You just assumed that the antiderivative of f(x-c) is F(x-c). Why is that true?
 


i believe it was given in a lecture i had so i assumed is that a wrong assumption?
 


jodecy said:
i believe it was given in a lecture i had so i assumed is that a wrong assumption?

It's not a wrong assumption. It just needs to be proved. If F'(x)=f(x), why is F'(x-c)=f(x-c)? It's easy, but you should say why. Use the chain rule. In other language, they may expect you to prove this using the substitution u=x-c. Why is dx=du?
 
Last edited:
You are probably not aware, but the way you posted makes your post unreadable to at least XP Windows users using Chrome, IE & Opera, attachment shows what they see. It looks little bit better under Vista, but is still barely readable.

I have corrected thread subject.
 

Attachments

  • Untitled-1.png
    Untitled-1.png
    6.6 KB · Views: 501
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top