Fundemental theorem of Calc. question?

  • Thread starter Thread starter Integral8850
  • Start date Start date
  • Tags Tags
    Theorem
Integral8850
Messages
15
Reaction score
0

Homework Statement


Find function f(x) and constant a such that,

6+ integral f(t)/t^2 dt = 2x
limits are a to x

Homework Equations


I think I am on the right track could someone check this?

The Attempt at a Solution


if x = a then the integral will = 0,
6 = 2a, a = 3 limits now are 3 to x,
now,
6 + integral f(x)/x^2 dx = 2x, fundamental theorem
6 + f(x)/x^2 = 2 => 6 + f(x)= 2x^2 => f(x)= 2x^2 - 6
Thanks!
 
Physics news on Phys.org
try to take the derivative of both sides and see what happens?
on your last line, when you took the derivative, you missed something. what is the derivative of 6?
 
Ok thanks that was the part I was not sure about so the derivative of 6 goes to 0.
Thanks!
 
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