Chapter 7 - Applications of Integration - 7.1 Exercises - Page 443: 64

$A = \frac{{\pi - 3}}{4}$

$$\eqalign{ & y = \frac{2}{{1 + 4{x^2}}},{\text{ }}\left( {\frac{1}{2},1} \right) \cr & {\text{Differentiate }} \cr & y' = 2\frac{d}{{dx}}\left[ {{{\left( {1 + 4{x^2}} \right)}^{ - 1}}} \right] \cr & y' = - 2{\left( {1 + 4{x^2}} \right)^{ - 2}}\left( {8x} \right) \cr & y' = - \frac{{16x}}{{{{\left( {1 + 4{x^2}} \right)}^2}}} \cr & {\text{Find the slope at }}\left( {\frac{1}{2},1} \right) \cr & m = - \frac{{16\left( {1/2} \right)}}{{{{\left( {1 + 4{{\left( {1/2} \right)}^2}} \right)}^2}}} \cr & m = - 2 \cr & {\text{The equation of the tangent line at }}\left( {\frac{1}{2},1} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 1 = - 2\left( {x - \frac{1}{2}} \right) \cr & y - 1 = - 2x + 1 \cr & y = - 2x + 2 \cr & \cr & {\text{The intersection point of the functions is:}} \cr & \frac{2}{{1 + 4{x^2}}} = - 2x + 2 \cr & {\text{Solving we obtain }}x = 0 \cr & {\text{The points are }}\left( {\frac{1}{2},1} \right){\text{ and }}\left( {0,f\left( 0 \right)} \right) \cr & \cr & \frac{2}{{1 + 4{x^2}}} \geqslant - 2x + 2{\text{ for the interval }}\left[ {0,\frac{1}{2}} \right] \cr & {\text{The area is given by }}\left( {{\text{See graph below}}} \right) \cr & A = \int_0^{1/2} {\left( {\frac{2}{{1 + 4{x^2}}} - \left[ { - 2x + 2} \right]} \right)} dx \cr & A = \int_0^{1/2} {\left( {\frac{2}{{1 + 4{x^2}}} + 2x - 2} \right)} dx \cr & {\text{Integrate and evaluate}} \cr & A = \left[ {\arctan 2x + {x^2} - 2x} \right]_0^{1/2} \cr & A = \left[ {\arctan 2\left( {\frac{1}{2}} \right) + {{\left( {\frac{1}{2}} \right)}^2} - 2\left( {\frac{1}{2}} \right)} \right] - \left[ 0 \right] \cr & A = \frac{\pi }{4} + \frac{1}{4} - 1 \cr & A = \frac{{\pi - 3}}{4} \cr} $$