Approximate the area under the graph of $F(x)=0.9 x^{3}+9 x^{2}-0.9 x-9$ over the interval $[-7,-2]$ using 5 subintervals. Use the left endpoints to find the heights of the rectangles.
The area is approximately square units.
(Type an integer or a decimal.)
Answer
Final Answer: The area under the graph of \(F(x)=0.9 x^{3}+9 x^{2}-0.9 x-9\) over the interval \([-7,-2]\) using 5 subintervals and the left endpoints to find the heights of the rectangles is approximately \(\boxed{495.0}\) square units.
Step 1 ：Determine the width of each subinterval. The total width of the interval is \(-2 - (-7) = 5\), so each subinterval has width \(5/5 = 1\).
Step 2 ：Calculate the left endpoints of each subinterval. These are \(-7, -6, -5, -4, -3\).
Step 3 ：Calculate the function value at each left endpoint to get the height of each rectangle. The heights are \(129.6, 126, 108, 81, 50.4\).
Step 4 ：Sum these values and multiply by the width of the subinterval to get the total area. The area is approximately \(495.0\) square units.
Step 5 ：Final Answer: The area under the graph of \(F(x)=0.9 x^{3}+9 x^{2}-0.9 x-9\) over the interval \([-7,-2]\) using 5 subintervals and the left endpoints to find the heights of the rectangles is approximately \(\boxed{495.0}\) square units.
