Question

Using integration, find the area bounded by the tangent to the curve 4y=$x^2$ at the point (2,1) and the lines whose equations are x = 2y and x = 3y - 3.

Answer 1

The given curve is 4y = $x^2\Rightarrow 4\frac{dy}{dx}=2x\therefore \frac{dy}{dx}=\frac{x}{2}\Rightarrow {\left[\frac{dy}{dx}\right]}_{at(2,1)}=\frac{2}{2}=1=m_T$
$\text{Equation of tangent at (2,1) is : y - 1 = 1(x-2)}\Rightarrow \text{y = x - 1......(i)}$
$\text{Also given lines are x = 2y.....(ii) and x = 3y - 3 ....(iii).}$
$\text{Required area =}$${\int }_2^3\left[\right(x-1)-\frac{x}{2}]dx+{\int }_3^6[\frac{x+3}{3}-\frac{x}{2}]dx\Rightarrow ={[\frac{x^2}{2}-x-\frac{x^2}{4}]}_2^3+{[\frac{x^2+6x}{6}-\frac{x^2}{4}]}_3^6\Rightarrow =[\frac{9}{2}-3-\frac{9}{4}]-[2-2-1]+[\frac{36+36}{6}-9]-[\frac{9+18}{6}-\frac{9}{4}]\Rightarrow =\frac{9}{4}-2+3-\frac{9}{4}=1\text{sq.unit}$