Area problem

Since the function, $1-x^2≤0$ on the the interval $[1,\infty]$ we know that $\left| \int_1^b 1-x^2dx \right| = -\int_1^b 1-x^2dx$ $\large \int _0^1 1-x^2dx=\left| \int _1^b 1-x^2dx \right| =-\int _1^b 1-x^2dx$ $\large -\left[ x-\frac {1}{3} {x}^{3}\right]_1^b =\left[ x-\frac { 1 }{ 3 }x^3\right]_0^1$ $\large -\left( b-\frac{1}{3}b^3- \frac{2}{3} \right)=\frac{2}{3}$ $\large -b+\frac { 1 }{ 3 } { b }^{ 3 }=0\\ \\ \large -b\left( 1-\frac { 1 }{ 3 } { b }^{ 2 } \right) =0$ $\large -b\frac { 1 }{ 3 } \left( 3-{ b }^{ 2 } \right) =0$ $\large { 3-b }^{ 2 }=\left( \sqrt { 3 } -b \right) \left( \sqrt { 3 } +b \right)=0$ since b must exist on the interval $\left[1,\infty\right]$ then $\large \quad b≥1$ $\large \therefore b=\sqrt { 3 } \approx 1.732$