Question

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.)

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:

\(330\)