The area (in sq. units) bounded by the curve y = x 2 + 2 x + 1 and the tangent to it at ( 1 , 4 ) and the Y…

The equation is given as,

$y=x^2+2x+1$

Differentiate the above equation with respect to $x$,

$\frac{dy}{dx}=2x+2$

At $(1,4)$

$\frac{dy}{dx}=2\left(1\right)+2=4$

The figure below represents the area bounded.

The bounded area is,

$A={\int }_0^{1}ydx-\frac{1}{2}(OA\times AP)$

$\Rightarrow A={\int }_0^1\left(x^2+2x+1\right)dx-\frac{1}{2}\left(1\times 4\right)$

$\Rightarrow A={\left(\frac{x^3}{3}+x^2+x\right)}_0^1-2$

$\Rightarrow A=\left(\frac{1}{3}+1^2+1-0\right)-2$

$\Rightarrow A=\frac{1}{3}$ sq. units