Please answer both questions. Approximate the area under the graph of F(x) = 0.8

Question

Please answer both questions.

Approximate the area under the graph of F(x) = 0.8x^3 + 8x^2 - 0.8x - 8 over the interval [- 8, - 3] using 5 subintervals. Use the left endpoints to find the height of the rectangles. The area is approximately (Type an integer or a decimal.) Approximate the area under the graph of f(x) = 0.05x64 - 2.25x^2 + 86 over the interval [2, 10] by dividing the interval into 4 subintervals. Use the left endpoint of each subinterval. The area under the graph of f(x) = 0.05x^4 - 2.25x^2 + 86 over the interval [2, 10] is approximately (Simplify your answer. Type an integer or a decimal.)

Explanation / Answer

let the 5 left end point of the interval be:
x=-8,-7,-6,-5,-4
Length of each interval =1

F(x) = 0.8*x^3 + 8*x^2 -0.8*x -8
F(-8) = 0.8*(-8)^3 + 8*(-8)^2 -0.8*(-8) -8
=100.8
F(-7) = 0.8*(-7)^3 + 8*(-7)^2 -0.8*(-7) -8
=115.2
F(-6) = 0.8*(-6)^3 + 8*(-6)^2 -0.8*(-6) -8
= 544.96
F(-5) = 0.8*(-5)^3 + 8*(-5)^2 -0.8*(-5) -8
=96
F(-4) = 0.8*(-4)^3 + 8*(-4)^2 -0.8*(-4) -8
=72

So,
area = 100.8*1 + 115.2*1 + 544.96*1 + 96*1 + 72*1
= 928.96
Answer: 928.96

I am allowed to answer only 1 question at a time

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---