Find the area enclosed by the parabola 4y = 3x^2 and the line 2y = 3x + 15?

Answer 1

First, We need to find the points of intersection of the line and the parabola. We do this by solving the following equation:

3x + 15=3/2 x²

You will find two solutions to this equation. Let x1 be one of the solutions and x2 be the other. You then need to plug in each of these solutions in either equation(line or parabola) to find the y coordinates, y1 and y2, of the points of interceptions. Thus the points of interceptions will be A(x1, y1) and B(x2,y2).

Let A1 = Area bounded by the straight line, the X-axis, and the ordinates 

y=y1, y=y2

and A2= Area bounded by the parabola, the X-axis, and the ordinates 

y=y1, y=y2

We can calculate A1 and A2 using integrals :

A1=∫(3/2 x + 15/2) dx ----the boundaries are x1 and x2

A2=∫(3/4 x²) dx ----the boundaries are x1 and x2

The area enclosed will then be A1-A2.

I let you do these computations.