`y = 2/ (1+4x^2) , (1/2, 1)` Set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the given point.

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Given, `y= 2/(1+4x^2)` and the point is `(x_0,y_0)= (1/2,1)`

First we have to find equation of the tangent line 

by using the following formula:

  `y=f'(x_0)(x-x_0)+y_0`

In our case `f'(x)=y'=(2/(1+4x^2))' = -16x/(1+4x^2)^2`

`f'(1/2)=-16(1/2)/(1+4(1/2)^2)^2=-(16(1/2))/(1+1)^2=-2`

By plugging that into the formula for equation of the tangent line we get

`y=(-2)(x-1/2)+1`

`y=-2x+1+1=2-2x`

`y=2-2x`

Now we can draw both graphs of the function and its tangent line

Now we need to calculate the area of the region, but before that we need to find point of intersection of the two graphs. We can do that by solving the following equations:

`y=2-2x`

`y=2/(1+4x^2)`

=>

`2-2x=2/(1+4x^2)`

`1-x=1/(1+4x^2)`

=> `(1+4x^2)(1-x)=1`

=> `1+4x^2-x-4x^3=1`

=> `4x^2-x-4x^3=0`

=>-`x(-4x+1+4x^2)=0`

=>`-x(2x-1)^2=0`

=>` x=0 or 2x-1=0`

=> `x=0 x=1/2`

so the area is given as

Area=

`int_0^(1/2) ((2/(1+4x^2))-(2-2x))dx`

=`int_0^(1/2) (2/(1+4x^2))dx-int_0^(1/2) (2-2x)dx`

`=[tan^(-1)(2x)]_0^(1/2) -[2x-x^2]_0^(1/2)`

=`[tan^(-1)(1) -tan^(-1)(0)]-[1-1/4]`

=`pi/4 -3/4`

= `0.0354`

The area is `0.0354.`

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