Mathematics - Area Under Curves Question with Solution | TestHub

Given,

$(y-2{)}^2=x-1\text{ and }x-2y+4=0$

Now, plotting the graph of the above equation, we get,

Now, from above diagram we get,

Required area  $={\int }_0^{3}xdy-$ Area of $\Delta ABC$.
$={\int }_0^3\left((y-2{)}^2+1\right)dy-\frac{1}{2}\times 1\times 2$

$={\int }_0^3\left(y^2-4y+5\right)dy-1$

$={\left[\frac{y^3}{3}-2y^2+5y\right]}_0^3-1$

$=9-18+15-1=5$