A curve `C` passes through (2,0) and the slope at `(x , y)` as `((x+1)^2+(y-3))/(x+1)dot` Find the equation of the curve. Find the area bounded by curve and x-axis in the fourth quadrant.

Answer

`y=x^(2)-2x,(4)/(3)` sq units

Explanation

Here, slope of tangent,
` (dy)/(dx) = ((x+1)^(2)+(y-3))/((x+1))`
` rArr (dy)/(dx)=(x+1)+((y-3))/((x+1)),`
Put `x+1=X and y-3=Y`
`rArr (dy)/(dx)=(dY)/(dX)`
`therefore (dY)/(dX)=X+(Y)/(X)`
`rArr (dY)/(dX)-(1)/(X)Y=X`
IF `=e^(int-(1)/(X)dx)=e^(-logX)=(1)/(X)`
` therefore " Solution is, " Y=*(1)/(X)= int X*(1)/(X) dx +c`
`rArr (Y)/(X)=X+c`

`y-3=(x+1)^(2)+c(x+1),` which passes through (2, 0).
`rArr -3=(3)^(2)+3c`
`rArr c= -4`
` therefore ` Required curve
`y=(x+1)^(2)-4(x+1)+3`
`rArr y=x^(2)-2x`
` therefore " Required area "=|int_(0)^(2)(x^(2)-2x)dx|=|((x^(3))/(3)-x^(2))_(0)^(2)|`
` =(8)/(3)-4=(4)/(3)` sq units