# The function f is continuous on the interval [4, 15], with some of its values given in the table above. Estimate the average value of the function with a Right Rectangle Approximation, using the 4 intervals between those given points.

x 4 9 11 14 15
f(x) -6 -11 -18 -21 -25

-11.545
-14
-16.273
-18.75

1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x

1. 👍
2. 👎
3. ℹ️
4. 🚩
2. The x values given are the right-left edges of the rectangles. You want right-side values, so figure f(x) at

x = 9,11,14,15

Multiply each height by its interval's width, and the approximation is just

5*11+2*18+2*21+1*25 = 158

So, the average height is the area divided by the width: 158/11 = 14.36 (minus, of course) or -14

As for y=x^3, note that the graphs intersect at x=0,1.

So, you want 4 intervals of width 0.25, and all you do is find

f(x) at x = 1/8, 3/8, 5/8, 7/8

multiply each value by 0.25 and add 'em up.

It's just a bunch of rectangles, not something scary.

1. 👍
2. 👎
3. ℹ️
4. 🚩
3. You have to find the average value of the function.

average value = [1/(b-a)] [the integral of f(x) dx]

The integral will be solved using the right approximation.
a and b are the domain, so a = 4 and b = 15

average value = [1/(15 - 4)] [(-11)(5) + (-18)(2) + (-21)(3) + (-25)(1)]
average value = 1/11 [-179]
average value = -179/11 = -16.273

1. 👍
2. 👎
3. ℹ️
4. 🚩