What are F(1) and F'(1) for F(x) = \int_{-x}^{x} \frac{dt}{1+t^2}?

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SUMMARY

For the function F(x) defined as F(x) = ∫_{-x}^{x} (dt / (1 + t^2)), the evaluation yields F(1) = π. To find F'(1), apply the Fundamental Theorem of Calculus, which gives F'(x) = (2 / (1 + x^2)). Evaluating this at x = 1 results in F'(1) = 1. The discussion clarifies the steps for both integration and differentiation of the function.

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Ok i have another problemo here...

if [tex]F(x) = \int_{-x}^{x} \frac{dt}{1+t^2}[/tex]

Find F(1) and F'(1)...I need some assistance...the anti derivative is

[tex]arctan(t)[/tex] so then... do i set that equal to f(1) and solve for t? and then for F', take the derivative and then solve for t again? I am kind of confused... :confused:
 
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ziddy83 said:
Ok i have another problemo here...

if [tex]F(x) = \int_{-x}^{x} \frac{dt}{1+t^2}[/tex]

Find F(1) and F'(1)...I need some assistance...the anti derivative is

[tex]arctan(t)[/tex] so then... do i set that equal to f(1) and solve for t? and then for F', take the derivative and then solve for t again? I am kind of confused... :confused:

You can do the integral to get F(x) for all x in the domain of F. Then take the derivative with respect to x to get F'(x). Evaluate both at x = 1. You can also find F'(x) using the fundamental theorem of calculus. For the latter approach, you might want the break the integral into two intervals at any constant a such that -x<a<x. a = 0 would be a convenient choice, but any constant value will do.
 
ok i got it ..thank you.
 

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