Chapter 7 - Applications of Integration - 7.1 Exercises - Page 443: 63

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

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