Howard L. answered 04/03/13
Calculus Specialist with 10+ Years of Tutoring Experience
The integral of 8f(x)-9 has two parts of area, say A1= 8f(x) defined by variable x and A2= -9 defined by constant number -9. We take integral to get the area from -3 to -5.5 by changing the up-down bound order from -5.5 to -3 and the sign + to -. Then, proceed all the rest as,
Q1: Area from -3 to -5.5
\int_{-3}^{-5.5} ( 8f(x) - 9) dx
= -\int_{-5.5}^{-3} ( 8 f(x) - 9) dx
= -8 \int_{-5.5}^{-3} f(x) dx + 9 \int_ {-5.5}^{-3} dx
= (-8)(-2) + 9 \int_{-5.5}^{-3} dx
=-16 + 9x|_{-5.5}^{-3}
=16 + (-27-(-49.5))
=16 + 22.5 ( A1+ A2)
=38.5
Q2: Area from -8 to -0.5
Variable area A1= 8f(x) = 8*8=64;
Fixed area A2 = -9x|_{-8}^{-0.5}= -9 (-0.5-(-8))= -67.5
Total area A = A1 + A2 = -3.5
Q3: Area from -8 to -5.5
A1 = 8*9 =72; A2 = -9x|_{-8}^{-5.5}= -9 (-5.5-(-8)) = -22.5:
A = A1 + A2 = 49.5
Q4: Area from -3 to -0.5
A1 = 8*1 = 8; A2 = -9( -0.5 - ( -3))= -22.5
A = -14.5
Proof: Q2 = -Q1 + Q3 + Q4 (Q1 is negative because of up-down bound change)
-3.5 = -38.5 + 49.5 - 14.5