If A is the area in the first quadrant enclosed by the curve C : 2 x 2 - y + 1 = 0 , the tangent to C at the…

Given,

$y=2x^2+1$

Now tangent to above equation at $\left(1,3\right)$ is given by,

$y-3=4\left(x-1\right)$

$\Rightarrow y=4x-1$

Now plotting the diagram we get,

Required area of bounded region will be,

$A={\int }_0^1\left(2x^2+1\right)dx-\text{area of} △QOT -\text{area of} △PQR +\text{area of} △QRS$

$\Rightarrow A=\left(\frac{2}{3}+1\right)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{40}$

$\Rightarrow 60A=16$