Area

$Let\quad f(x)={ x }^{ 2 }+1\\ Let\quad g(x)={ x }^{ 2 }\\ \\ f'(x)=2x\\ f'(b)=2b\\ \\ Since\quad f(x)\quad has\quad a\quad tangent\quad at\quad point\quad P(b,{ b }^{ 2 }+1),\\ the\quad equation\quad of\quad the\quad tangent\quad is:\\ y-{ b }^{ 2 }-1=2b(x-b)\\ y=2bx-2{ b }^{ 2 }+{ b }^{ 2 }+1\\ \boxed { y=2bx-{ b }^{ 2 }+1 } \\ \\ Setting\quad { x }^{ 2 }=2bx-{ b }^{ 2 }+1\quad to\quad find\quad the\quad common\quad points\quad of\\ \quad g(x)\quad and\quad the\quad tangent,\\ { x }^{ 2 }-2bx+{ b }^{ 2 }-1=0\\ \\ Let\quad \alpha \quad and\quad \beta \quad be\quad the\quad roots\quad of\quad { x }^{ 2 }-2bx+{ b }^{ 2 }-1=0,\quad (\beta >\alpha )\\ Therefore:\quad \boxed { \alpha +\beta =2b\quad ,\quad \alpha \beta ={ b }^{ 2 }-1 } \\ \\ Since,\quad -\int _{ \alpha }^{ \beta }{ (x-\alpha )(x-\beta )dx= } \frac { { (\beta -\alpha ) }^{ 3 } }{ 6 } \\ We\quad just\quad need\quad to\quad find\quad \beta -\alpha .\\ \\ { (\beta -\alpha ) }^{ 2 }={ (\alpha +\beta ) }^{ 2 }-4\alpha \beta \\ Substituting\quad \alpha +\beta =2b\quad and\quad \alpha \beta ={ b }^{ 2 }-1,\\ { (\beta -\alpha ) }^{ 2 }={ 4b }^{ 2 }-4{ b }^{ 2 }+4=4\\ Therefore,\quad \beta -\alpha =2\quad (\beta -\alpha >0)\\ \\ -\int _{ \alpha }^{ \beta }{ ({ x }^{ 2 } } -2bx+{ b }^{ 2 }-1)dx=\frac { { 2 }^{ 3 } }{ 6 } =\frac { 8 }{ 6 } =\frac { 4 }{ 3 } \\ \\ \boxed { The\quad area\quad enclosed\quad is\quad \frac { 4 }{ 3 } . } \\$