Question

What is the area of the region between the graphs of f(x) = x^2 + 2xf(x) = x2 +2x, and g(x) = 2x+1g(x) = 2x+1g(x) = 2x + 1?

Answer 1

enclosed area A = 2/3 units^2

Explanation:

Given:-

- The two functions are given:-

f ( x ) = x^2 + 2x

g ( x ) = 2x + 1

Find:-

What is the area of the region between the graphs of f(x) and g(x)

Solution:-

- We will determine the interval or limits in which the two given function f(x) and g(x).

- We will find the interval by determining the point of intersection between the two graphs. Hence,

f ( x ) = g ( x )

x^2 + 2x = 2x + 1

x^2 = 1

x = ± 1 , y = 3 , -1

- The pair of intersection coordinates are ( -1 , -1 ) and ( 1 , 3 ). The functions are bounded by the intervals:

-1 ≤ x ≤ 1

-1 ≤ y ≤ 3

- Now sketch each function to visualize the enclosed region. ( See attachment ). We see that in the range ( -1 ≤ x ≤ 1 ) the function g ( x ) > f ( x ).

- So we will find the area along the y-direction with the interval ( -1 ≤ x ≤ 1 ) will define the limits of integration. The enclosed area is given by:

- The limits of integration are [ a , b ] = [ 1 , -1 ].

`A = \int\limits^a_b {( y_2 (x) - y_1(x) )} \, dx \\\\y_2 ( x ) > y_1 ( x )\\\\A = \int\limits^a_b {( g (x) - f(x) )} \, dx \\\\A = \int\limits^a_b {( x^2 + 2x - 2x - 1 )} \, dx \\\\ A = \int\limits^a_b {( x^2 - 1 )} \, dx \\\\A = [ (x^3)/(3) - x ]\limits^1_-_1 \\\\A = [ (1^3)/(3) - 1 - ((-1)^3)/(3) - (-1) ]\\\\A = [ (1)/(3) - 1 + (1)/(3) +1 ] = (2)/(3)`

- The enclosed area A = 2/3 units^2.