Finding the Area of a given Enclosed Region

Question

How would I find the area of the region enclosed by f(x) = 2x² and y = 12-2x, and also how would I sketch the graphs?

Answer 1

First, find the points of intersection by setting f(x) equal to y:

2x² = 12 - 2x

x = -3, 2

To find the area between the curves, we integrate the difference of the functions from x=−3 to x=2:

After evaluating this integral by separating it out and using the Power Rule, you should get 125/3 square units as your answer.

To sketch the graphs:

For f(x):

1. The vertex is at (0,0).
2. It opens upwards.
3. Points to plot: (−3,18), (0,0), and (2,8).

For y:

1. The y-intercept is 12.
2. The slope is -2.
3. Points to plot: (−3,18) and (2,8).

Hope this was helpful.

Answer 2

The graph of y=2x² is a parabola with minimum point at the origin. The graph of the 2nd function is a straight line with negative slope (NW to SE).

You need to find the points of intersection...solve the 2 equations simultaneously.

Then since the straight line is the upper bound of the area you need to integrate from the smaller x value to the larger (obtained by solving the equations as instructed above).

The integrand is the difference between the equation of the straight line and 2x².

Answer 3

2x^2= 12-2x

divide by 2

x^2 =6 -x

x^2 +x -6 =0

factor

(x+3)(x-2)

x= =-3 and 2 are the limits of evaluating the integral

the integral is of 12-2x -2x^2

raise exponents by 1, and divide by same

= 12x -x^2 -2x^3/3

evaluate from -3 to 2

= -(12(-3) -(-3)^2 - 2(-3)^3/3 ) +(12(2) - (2^2) - 2(2^3)/3)

= -36 -9 + 18 - (24 -4 -16/3)

= 125/3

= 41 2/3 units^2

the graph is of a straight line above a parabolic curve intersecting at (-3,18) and (2,8)

parabola has vertex = minimum = the origin

the line has an x intercept 6 and y intercept 12

with a rough sketch, you could by geometry and area = bh/2 = 5(18)2 = 45 = rough estimate of area= very close to 41 2/3