Answer:
[tex](a,b) \ne (7,-9)[/tex]
Step-by-step explanation:
Given
[tex]f(x) = ax + b[/tex]
[tex]Interval = [0,4][/tex]
[tex]Area = 16[/tex]
Required
[tex]a, b \ne[/tex]
The area is calculated as:
[tex]Area = \int\limits^a_b {f(x)} \, dx[/tex]
So, we have:
[tex]Area = \int\limits^4_0 {(ax + b)} \, dx[/tex]
Integrate
[tex]Area = (\frac{ax^2}{2} + bx)|\limits^4_0[/tex]
Expand
[tex]Area = [(\frac{a*4^2}{2} + b*4)] - [(\frac{a*0^2}{2} + b*0)][/tex]
[tex]Area = [(\frac{a*4^2}{2} + b*4)][/tex]
[tex]Area = 8a+ 4b[/tex]
This gives:
[tex]8a + 4b = 16[/tex]
Divide through by 4
[tex]2a + b = 4[/tex]
From the list of options, the possible values of a and b are:
[tex](a,b) = (-2,8) \to 2 * -2 + 8 = 4[/tex]
[tex](a,b) = (1,2) \to 2 * 1 + 2 = 4[/tex]
[tex](a,b) = (3,-2) \to 2 * 3 -2 = 4[/tex]
[tex](a,b) = (5,-6) \to 2 * 5 -6 = 4[/tex]
The odd one is:
[tex](a,b) = (7,-9) \to 2 * 7 -9 = 5[/tex]
Hence:
[tex](a,b) \ne (7,-9)[/tex]
