Answer:
182.8125
Step-by-step explanation:
Given:
y = x^3
from [2,5] using 6 subdivisions
deltax = (5 - 2)/6 = 3/6 = 0.5
hence the subdivisions are:
[2, 2.5]; [2.5, 3]; [3, 3.5]; [3.5, 4]; [4, 3.5]; [4.5, 5]
hence the right endpoints are:
x1 = 2.5; x2 = 3; x3 = 3.5; x4 =4; x5 = 4.5; x6 = 5
now the area is given by:
A = deltax*[2.5^3 + 3^3 + 3.5^3 + 4^3+ 4.5^3 + 5^3]
A = 0.5*365.625
A = 182.8125
Area using Right Endpoint approximation is 182.8125
The area of the region is an illustration of definite integrals.
The approximation of the area of the region R is 182.8125
The given parameters are:
[tex]\mathbf{f(x) = x^3}[/tex]
[tex]\mathbf{Interval = [2,5]}[/tex]
[tex]\mathbf{n = 6}[/tex] ------ sub intervals
Using 6 sub intervals, we have the partitions to be:
[tex]\mathbf{Partitions = [2,2.5]\ u\ [2.5, 3]\ u\ [3,3.5]\ u\ [3.5,4]\ u\ [4,4.5]\ u\ [4.5,5]}[/tex]
List out the right endpoints
[tex]\mathbf{x= 2.5,\ 3,\ 3.5,\ 4,\ 4.5,\ 5}[/tex]
Calculate f(x) at these partitions
[tex]\mathbf{f(2.5) = 2.5^3 = 15.625}[/tex]
[tex]\mathbf{f(3) = 3^3 = 27}[/tex]
[tex]\mathbf{f(3.5) = 3.5^3 = 42.875}[/tex]
[tex]\mathbf{f(4) = 4^3 = 64}[/tex]
[tex]\mathbf{f(4.5) = 4.5^3 = 91.125}[/tex]
[tex]\mathbf{f(5) = 5^3 = 125}[/tex]
So, the approximated value of the definite integral is:
[tex]\mathbf{\int\limits^5_2 {f(x)} \, dx \approx \frac{1}{2}(\sum f(x))}[/tex]
This becomes
[tex]\mathbf{\int\limits^5_2 {f(x)} \, dx \approx \frac{1}{2}(15.625 + 27 + 42.875 + 64+91.125 + 125)}[/tex]
[tex]\mathbf{\int\limits^5_2 {f(x)} \, dx \approx \frac{1}{2} \times 365.625}[/tex]
[tex]\mathbf{\int\limits^5_2 {f(x)} \, dx \approx 182.8125}[/tex]
Hence, the approximation of the area of the region R is 182.8125
Read more about definite integrals at:
https://brainly.com/question/9897385
