Respuesta :
Answer:
enclosed area A = 2/3 units^2
Step-by-step 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 ].
[tex]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 = [ \frac{x^3}{3} - x ]\limits^1_-_1 \\\\A = [ \frac{1^3}{3} - 1 - \frac{(-1)^3}{3} - (-1) ]\\\\A = [ \frac{1}{3} - 1 + \frac{1}{3} +1 ] = \frac{2}{3}[/tex]
- The enclosed area A = 2/3 units^2.
