Thread: Area of a Plane region helppp

i cant hit the answer key;
(this is an application of Integration)
Find the area of the region bounded by the two curves
y = x^3 + 3x^2 + 2x; y = 2x^2 + 4x

answer: 37/12 sq units

Originally Posted by ^_^Engineer_Adam^_^

i cant hit the answer key;
(this is an application of Integration)
Find the area of the region bounded by the two curves
y = x^3 + 3x^2 + 2x; y = 2x^2 + 4x

answer: 37/12 sq units

Hello,

I've attached an image of the grapghs and the enclosed area.

First calculate the intersections of the 2 graphs:

x³ + 3x² + 2x = 2x² + 4x ===> x³ + x² - 2x = 0 The LHS can be factorized:

x³ + x² - 2x = x(x + 2)(x - 1)

To calculate the enclosed area you use the difference of the 2 functions:

d(x) = x³ + x² - 2x

A = |∫[from -2 to 0](x³ + x² - 2x)dx| + |∫[from 0 to 1](x³ + x² - 2x)dx|

A = |[from -2 to 0](¼ x^4 + (1/3)x³ - x²)dx| + |[from 0 to 1](¼ x^4 + (1/3)x³ - x²)dx|

A = |0 - (-8/3)| + |(-5/12) - 0| = 37/12

EB

Originally Posted by earboth

x³ + x² - 2x = x(x + 2)(x - 1)

Define f(x) = x³ + 3x² + 2x and g(x) = 2x² + 4x

In [0,1] --> f(x) < g(x); in [-2,0] --> g(x) < f(x), then the definite integral to find is