Question

Find the area enclosed between the parabola y squared equals to 4 x and the line y = 2 mx

Answer 1

Answer:

1 / 3m³

Step-by-step explanation:

The parabola and the line meet when

y² / 4 = x = y / 2m

<=> m y² - 2y = 0

<=> y ( m y - 2 ) = 0

So the points of intersection are at y = 0 and y = 2 / m.

For a "small piece" of area, fix y and take the horizontal rectangle between the parabola and the line.  Its width is then

y / 2m - y² / 4

and its height we take as dy.

The area is then

$\int_0^{\frac2m}\left(\frac{y}{2m} - \frac{y^2}{4}\right)\,dy\\= \left[\frac{y^2}{4m} - \frac{y^3}{12}\right]_0^{\frac2m}\\= \frac{1}{4m}\frac{4}{m^2} - \frac{1}{12}\frac{8}{m^3}\\= \frac{1}{m^3} -\frac{2}{3m^3}\\= \frac{1}{3m^3}$