approximate the area under the graph of ( f(x)=0.6x^{3}+6x^{2}-0.6x - 6 ) over the interval ( -7,-2 ) using…

approximate the area under the graph of ( f(x)=0.6x^{3}+6x^{2}-0.6x - 6 ) over the interval ( -7,-2 ) using 5 subintervals. use the left endpoints to find the heights of the rectangles. the area is approximately ( square ) square units (type an integer or a decimal.)

approximate the area under the graph of ( f(x)=0.6x^{3}+6x^{2}-0.6x - 6 ) over the interval ( -7,-2 ) using 5 subintervals. use the left endpoints to find the heights of the rectangles. the area is approximately ( square ) square units (type an integer or a decimal.)

Answer

Explanation:

Step1: Calculate the width of each sub - interval

The formula for the width of a sub - interval (\Delta x=\frac{b - a}{n}), where (a=-7), (b = - 2), and (n = 5). (\Delta x=\frac{-2-(-7)}{5}=\frac{-2 + 7}{5}=1)

Step2: Find the left - endpoints

The left - endpoints (x_i=a+i\Delta x) for (i = 0,1,2,3,4). When (i = 0), (x_0=-7); when (i = 1), (x_1=-6); when (i = 2), (x_2=-5); when (i = 3), (x_3=-4); when (i = 4), (x_4=-3)

Step3: Calculate the function values at the left - endpoints

  • For (x=-7): (F(-7)=0.6\times(-7)^3+6\times(-7)^2-0.6\times(-7)-6) (=0.6\times(-343)+6\times49 + 4.2-6) (=-205.8 + 294+4.2-6=86.4)
  • For (x=-6): (F(-6)=0.6\times(-6)^3+6\times(-6)^2-0.6\times(-6)-6) (=0.6\times(-216)+6\times36+3.6 - 6) (=-129.6+216+3.6 - 6=84)
  • For (x=-5): (F(-5)=0.6\times(-5)^3+6\times(-5)^2-0.6\times(-5)-6) (=0.6\times(-125)+6\times25 + 3-6) (=-75+150+3 - 6=72)
  • For (x=-4): (F(-4)=0.6\times(-4)^3+6\times(-4)^2-0.6\times(-4)-6) (=0.6\times(-64)+6\times16+2.4-6) (=-38.4 + 96+2.4-6=54)
  • For (x=-3): (F(-3)=0.6\times(-3)^3+6\times(-3)^2-0.6\times(-3)-6) (=0.6\times(-27)+6\times9+1.8-6) (=-16.2+54+1.8-6=33.6)

Step4: Calculate the sum of the areas of the rectangles

The sum (S=\sum_{i = 0}^{4}F(x_i)\Delta x) Since (\Delta x = 1), (S=F(-7)+F(-6)+F(-5)+F(-4)+F(-3)) (S=86.4 + 84+72+54+33.6) (S=330)

Answer:

(330)