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tandoorichicken
Sep25-04, 04:54 PM
Hello everyone, its been a while.

It's been almost 4 months since I did anything calculus related so I forgot all of my skills. :bugeye:

The problem is:
Use the Fundamental Theorem of Calculus to find the derivative of the function
h(x) = \int_{2}^{\frac{1}{x}} \arctan{t} \,dt

vsage
Sep25-04, 05:55 PM
IIRC (whichI might not) the fundamental theorem of calculus says that given F(x) = S(f(x),x,a,b) F'(x) = f(b)-f(a)

BLaH!
Sep25-04, 05:59 PM
Consider the function F(x) = \int_{a}^{x} f(t) \,dt .

The Fundamental Theorem of Calculus is given by: \frac{dF}{dx} = f(x). In your case the upper integration limit is 1/x. Therefore, you will have to use the chain rule. Let u=1/x \Rightarrow \frac{dh}{dx} = \frac{dh}{du}\frac{du}{dx} = -\frac{1}{x^2}arctan(\frac{1}{x})