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

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To find F(1) and F'(1) for F(x) = ∫_{-x}^{x} (dt/(1+t^2)), first evaluate the integral, which results in F(x) = arctan(x) - arctan(-x) = 2arctan(x). Next, calculate F(1) by substituting x = 1, yielding F(1) = 2arctan(1) = π. For F'(x), apply the fundamental theorem of calculus, differentiating F(x) to get F'(x) = 2/(1+x^2), and then evaluate at x = 1 to find F'(1) = 2/2 = 1. The discussion clarifies the steps needed to compute both values accurately.
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Ok i have another problemo here...

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

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

arctan(t) 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 F(x) = \int_{-x}^{x} \frac{dt}{1+t^2}

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

arctan(t) 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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