Respuesta :
Answer:
A = 68 unit^2
Step-by-step explanation:
Given:-
The piece-wise function f(x) is defined over an interval as follows:
f(x) = { x^2+3x+4 , x < 3
f(x) = { x^2+3x+4 , x≥3
Domain : [ -3 , 4 ]
Find:-
Find the area of the region enclosed by f(x) and the x axis
Solution:-
- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.
- The first portion of function is valid over the interval [ -3 , 3 ]:
[tex]f(x) = x^2+3x+4[/tex]
- The area "A1" bounded by f(x) is given as:
[tex]A1 = \int\limits^a_b {f(x)} \, dx[/tex]
Where, The interval of the function { -3 , 3 ] = [ a , b ]:
[tex]A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2[/tex]
- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :
[tex]f(x) = 4x+10[/tex]
- The area "A2" bounded by f(x) is given as:
[tex]A2 = \int\limits^a_b {f(x)} \, dx[/tex]
Where, The interval of the function { 3 , 4 ] = [ a , b ]:
[tex]A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2[/tex]
- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:
A = A1 + A2
A = 42 + 26
A = 68 unit^2
