To approximate the area under the function [tex] f(x) = x^2 + 2 [/tex] from [tex] a = 0 [/tex] to [tex] b = 10 [/tex] using a right-hand sum with 5 intervals, you first need to divide the interval [tex][a, b][/tex] into 5 equal parts. The width [tex] \Delta x [/tex] of each interval is given by:
[tex] \Delta x = \frac{{b - a}}{n} [/tex]
where [tex] n [/tex] is the number of intervals. In this case:
[tex] \Delta x = \frac{{10 - 0}}{5} = 2 [/tex]
Thus, each interval will be 2 units wide. The right-hand sum means you evaluate the function at the right endpoint of each interval. So we need to evaluate [tex] f(x) [/tex] at [tex] x = 2 [/tex], [tex] x = 4 [/tex], [tex] x = 6 [/tex], [tex] x = 8 [/tex], and [tex] x = 10 [/tex].
Now, let's calculate the sum:
[tex] R_5 = f(2) \Delta x + f(4) \Delta x + f(6) \Delta x + f(8) \Delta x + f(10) \Delta x [/tex]
Let's calculate the value of [tex] f(x) [/tex] at each of these points:
[tex] f(2) = 2^2 + 2 = 4 + 2 = 6 [/tex]
[tex] f(4) = 4^2 + 2 = 16 + 2 = 18 [/tex]
[tex] f(6) = 6^2 + 2 = 36 + 2 = 38 [/tex]
[tex] f(8) = 8^2 + 2 = 64 + 2 = 66 [/tex]
[tex] f(10) = 10^2 + 2 = 100 + 2 = 102 [/tex]
Now we multiply each by [tex] \Delta x [/tex] (which is 2):
[tex] R_5 = 6(2) + 18(2) + 38(2) + 66(2) + 102(2) [/tex]
[tex] R_5 = 12 + 36 + 76 + 132 + 204 [/tex]
And sum these values to get:
[tex] R_5 = 12 + 36 + 76 + 132 + 204 = 460 [/tex]
So the right-hand sum approximation of the area under the curve from [tex] x = 0 [/tex] to [tex] x = 10 [/tex] is 460 square units. Remember, this is an approximation of the actual area; the exact area can be found using definite integrals.
