A calculus problem by Sravan Chinta

$According\quad to\quad the\quad question,\quad slope\quad of\quad the\quad curve\quad C\quad at\quad (x,y)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { { (x+1) }^{ 2 }+(y-3) }{ (x+1) } \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad or\frac { dy }{ dx } \quad \quad =(x+1)+\frac { (y-3) }{ (x+1) } \quad \quad \quad \quad \\ \frac { dy }{ dx } -(\frac { 1 }{ (x+1) } )y\quad \quad =\quad (x+1)-\frac { 3 }{ (x+1) } \quad \quad \\ \quad which\quad is\quad linear\quad differential\quad eqaution\\ I.F.={ e }^{ -\int { \frac { dy }{ (x+1) } } }\quad ={ e }^{ -\log { (x+1) } }\quad =\frac { 1 }{ (x+1) } \quad \\ Thus,solution\quad is\quad \frac { y }{ (x+1) } \quad =\quad \int { (1-\frac { 3 }{ { (x+1) }^{ 2 } } ) } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad or\quad \quad \quad \frac { y }{ (x+1) } \quad =\quad x+\frac { 3 }{ (x+1) } +C\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad or\quad \quad y\quad \quad =\quad x(x+1)+3+C(x+1)\rightarrow \quad (1)\\ As\quad the\quad curve\quad passes\quad through(2,0)\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad 0=2.3+3+C.3\quad \Rightarrow \quad \quad C\quad \quad =\quad -3\\ Thus,\quad equation\quad (1)\quad becomes\quad y=x(x+1)+3-3x-3\\ \quad \quad \quad \quad or\quad \quad y={ x }^{ 2 }-2x\quad which\quad is\quad the\quad equation\quad of\quad curve\\ This\quad can\quad be\quad written\quad as\quad \quad { (x+1) }^{ 2 }\quad =\quad (y+1)\\ [upward\quad parabola\quad with\quad vertex\quad at\quad (1,-1)\quad meeting\quad x-axis\\ at\quad (0,0)\quad and\quad (2,0)]\\ Area\quad bounded\quad by\quad the\quad curve\quad and\quad x-axis\quad in\quad fourth\quad quadrant\\ is\quad \quad \quad \quad \quad \quad \quad \quad \quad A\quad =\quad \left| \int _{ 0 }^{ 2 }{ ydx } \right| \quad =\left| \int _{ 0 }^{ 2 }{ ({ x }^{ 2 }-2x) } \right| ={ \left| \frac { { x }^{ 3 } }{ 3 } -{ x }^{ 2 } \right| }_{ 0 }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\left| \frac { 8 }{ 3 } -4 \right| \quad =\frac { 4 }{ 3 } =1.333\quad$

We know our slope equals $y' = \frac{(x+1)^2 + (y-3)}{x+1}.$ , which can be rewritten as:

$\frac{y'(x+1) - (y-3)}{(x+1)^2} = 1 \Rightarrow \frac{d}{dx} (\frac{y-3}{x+1}) = 1 \Rightarrow \frac{y-3}{x+1} = x + C$ ;

and substituting the boundary condition $y(2) = 0$ yields $C = -3$ and $y(x) = x^2 - 2x.$ The required area is then computed as:

$A = |\displaystyle{\int_{0}^{2} 2x - x^2\,\mathrm{d}x}| = \boxed{\frac{4}{3}}.$