Using the Fundamental Theorem of Calculus...

**PTL** · August 5th 2009, 06:49 AM

I'm asked to state the fundamental theorem of calculus, and use it to show that
d/dx
integral from x to tan x $\frac{1}{1+t^{2}}=\frac{x^{2}}{1+x^{2}}$ .

I've been trying to use the wikipedia definition:
Fundamental theorem of calculus - Wikipedia, the free encyclopedia

So I have f(t)=1/(1+t^2)
a = x, b= tan x

But what do I use for the F(x)?

Or is one supposed to use some corrollary?

And to 'show something using the Fun. Theorem of Caculus', does one simply say: "Thus, by the Fundamental Theorem of Calculus.. qed."?

**red_dog** · August 5th 2009, 07:04 AM

$F(x)=\arctan x$

**PTL** · August 5th 2009, 07:57 AM

So that gives arctan(tan(x)) - arctan(x)?

**Random Variable** · August 5th 2009, 08:09 AM

You don't need to know F(x) explicitly.

$\frac{d}{dx} \int^{\tan x}_{x} \frac{1}{1+t^{2}} \ dt$

let $F(x) = \int^{x}_{0} \frac{1}{1+t^{2}} \ dt$

then $\int^{\tan x}_{x} \frac{1}{1+t^{2}} \ dt = F(\tan x) - F(x)$

$\frac{d}{dx} \int^{\tan x}_{x} \frac{1}{1+t^{2}} \ dt = F'( \tan x) \frac{d}{dx} (\tan x) - F'(x)$

$= \frac {\sec^{2} x}{1+ \tan^{2}x} - \frac{1}{1+x^{2}}$

$= 1 - \frac{1}{1+x^{2}} = \frac{x^{2}}{1+x^{2}}$

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