What is the Use of the Constant in the Integral?

  • Thread starter Thread starter songoku
  • Start date Start date
songoku
Messages
2,470
Reaction score
383
Homework Statement
Find f(x) where ##f(x)=\frac{\cos x}{x}## and f(3) = 4. State the answer in form of ##f(x)=\int_{t=p}^{t=q} (........)##
Relevant Equations
Fundamental Theorem of Calculus
This is my attempt:
$$f(x)=\int_{t=p}^{t=x} \frac{\cos t}{t} dt$$

But I am not sure what ##p## is and what the use of ##f(3)=4##

Thanks
 
Physics news on Phys.org
You forgot about the constant that is added to the integral. If you start the integral at p=3, then you know that the integral part is 0 at x=3. So what constant is added to the integral?
 
  • Like
Likes songoku and Delta2
songoku said:
Homework Statement:: Find f(x) where ##f(x)=\frac{\cos x}{x}## and f(3) = 4. State the answer in form of ##f(x)=\int_{t=p}^{t=q} (...)##
Did you forget to add the prime? Shouldn't it be ##f'(x) = \frac{\cos x}{x}##?
 
  • Like
Likes songoku, Delta2 and berkeman
FactChecker said:
You forgot about the constant that is added to the integral. If you start the integral at p=3, then you know that the integral part is 0 at x=3. So what constant is added to the integral?
I understand

Mark44 said:
Did you forget to add the prime? Shouldn't it be ##f'(x) = \frac{\cos x}{x}##?
Yes, I am sorry

Thank you very much FactChecker and Mark44
 
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