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