The area (in sq. units) bounded by the curve  $y=x^2+2x+1$ and the  tangent to it at (1,4) and the y– axis is

$\begin{array}{r}\mathrm{y}={\mathrm{x}}^2+2\mathrm{x}+1\text{ and the point }=(1,4)\\ \Rightarrow \frac{dy}{dx}=2\mathrm{x}+2\\ \text{ slope }\left(\mathrm{m}\right)={\left(\frac{dy}{dx}\right)}_{(1,4)}=4\end{array}$
Equation of tangent is $y-4=4(x-1)\Rightarrow y=4x$

Required area

$\begin{array}{r}={\int }_0^1 ydx-\frac{1}{2}\times OA\times AP\\ ={\int }_0^1 \left(x^2+2x+1\right)dx-\frac{1}{2}\left(1\right)\left(4\right)\\ ={\left(\frac{x^3}{3}+x^2+x\right)}_0^1-2\\ =\frac{1}{3}\end{array}$