Question

find the area of the region of the curve y2 =9x enclosed by the line y=x+2​

Answer 1

$\textbf{Given:}$

$\text{Curve is y^2=9x }$

$\text{Line is y=x+2 }$

$\textbf{To find:}$

$\text{Area of the region enclosed by the curve and the line}$

$\textbf{Solution:}$

$\textbf{To find the point of intersection:}$

$y^2=9x$..........(1)

$y=x+2$..........(2)

$\text{Using (2) in (1), we get}$

$(x+2)^2=9x$

$x^2+4+4x-9x=0$

$x^2-5x+4=0$

$(x-1)(x-4)=0$

$\implies\,x=1,4$

$\text{when x=1 , y=3 }$

$\text{when x=4 , y=6 }$

$\therefore\text{Points of intersection are (1,3)  and (4,6) }$

$\textbf{Required area}$

$=\int\limits_{a}^{b}[f(x)-g(x)]\,dx$

$=\int\limits_{1}^{4}[3\sqrt{x}-(x+2)]\,dx$

$=[3(\dfrac{x^\frac{3}{2}}{\frac{3}{2}})-\dfrac{x^2}{2}-2x]^{4}_{1}$

$=[2x^\frac{3}{2}-\dfrac{x^2}{2}-2x]\limits_{1}^{4}$

$=[2(4)^\frac{3}{2}-\dfrac{4^2}{2}-2(4)]-[2(1)^\frac{3}{2}-\dfrac{1^2}{2}-2(1)]$

$=[2(4)(2)-8-8]-[2-\dfrac{1}{2}-2]$

$=[16-8-8]-[2-\dfrac{1}{2}-2]$

$=0-[0-\dfrac{1}{2}]$

$=\dfrac{1}{2}\,\text{square units}$

$\text{}$