Given, $y=x^{2}+2 x+1$
Differentiating w.r.t. $x$, we get
$\left.\therefore \frac{d y}{d x}\right|_{P(1,4)}=2+2=4$
$\therefore $ Equation of tangent at $P(1,4)$
$ y-4 =4(x-1) $
$y-4 =4 x-4 $
$y =4 x$
Required area $=\int\limits_{0}^{1} y d x-\frac{1}{2} \times O A \times A P$
$=\int\limits_{0}^{1}\left(x^{2}+2 x+1\right) d x-\frac{1}{2} \times 1 \times 4$
$=\left(\frac{x^{3}}{3}+x^{2}+x\right)_{0}^{1}-2=\left(\frac{1}{3}+1+1-0\right)-2$
$=\frac{7}{3}-2=\frac{1}{3}$
