Post

Estimating Average Value of a Function

Created by @nathanedwards
 at November 1st 2023, 2:58:24 pm.
AP_Calculus_AB-Questions
#calculus
#average-value
#riemann-sum

Question:

Let f(x) be a continuous function on the closed interval [0, 4]. The table below shows the values of f(x) for selected x-values in the interval [0, 4].

x f(x)
0 1
1 3
2 4
3 2
4 5

a) Estimate the average value of f(x) on the interval [0, 4] using the midpoint Riemann sum with 4 subintervals.

b) Determine an exact value for the average value of f(x) on the interval [0, 4] using calculus.

Answer:

a) To estimate the average value of f(x) on the interval [0, 4] using the midpoint Riemann sum with 4 subintervals, we divide the interval into 4 equal subintervals of width Δx = (4-0) / 4 = 1.

The midpoint Riemann sum is given by:

R4=Δx∗[f(x1∗)+f(x2∗)+f(x3∗)+f(x4∗)]R_4 = Δx * [f(x_1^*) + f(x_2^*) + f(x_3^*) + f(x_4^*)]

where x_i^* represents the midpoint of each subinterval.

The midpoint values of the subintervals are:

x_1^* = 0 + Δx/2 = 0 + 1/2 = 1/2, x_2^* = 1 + Δx/2 = 1 + 1/2 = 3/2, x_3^* = 2 + Δx/2 = 2 + 1/2 = 5/2, x_4^* = 3 + Δx/2 = 3 + 1/2 = 7/2.

Plugging in these values into the formula, we have:

R4=1∗[f(1/2)+f(3/2)+f(5/2)+f(7/2)]R_4 = 1 * [f(1/2) + f(3/2) + f(5/2) + f(7/2)]

Using the table, we find:

f(1/2) = 3, f(3/2) = 4, f(5/2) = 2, f(7/2) = 5.

Therefore,

R4=1∗[3+4+2+5]=14R_4 = 1 * [3 + 4 + 2 + 5] = 14

Hence, the estimate for the average value of f(x) on the interval [0, 4] using the midpoint Riemann sum with 4 subintervals is 14.

b) To determine the exact value for the average value of f(x) on the interval [0, 4] using calculus, we use the formula:

Avg=1b−a∫abf(x) dxAvg = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx

In this case, a = 0 and b = 4.

We integrate the function f(x) from 0 to 4:

∫04f(x) dx=∫011 dx+∫123 dx+∫234 dx+∫342 dx\int_{0}^{4} f(x) \, dx = \int_{0}^{1} 1 \, dx + \int_{1}^{2} 3 \, dx + \int_{2}^{3} 4 \, dx + \int_{3}^{4} 2 \, dx

Evaluating the integrals, we have:

∫011 dx=x∣01=1−0=1\int_{0}^{1} 1 \, dx = x \bigg|_{0}^{1} = 1 - 0 = 1
∫123 dx=3x∣12=3(2)−3(1)=3\int_{1}^{2} 3 \, dx = 3x \bigg|_{1}^{2} = 3(2) - 3(1) = 3
∫234 dx=4x∣23=4(3)−4(2)=4\int_{2}^{3} 4 \, dx = 4x \bigg|_{2}^{3} = 4(3) - 4(2) = 4
∫342 dx=2x∣34=2(4)−2(3)=2\int_{3}^{4} 2 \, dx = 2x \bigg|_{3}^{4} = 2(4) - 2(3) = 2

Adding up these results, we get:

∫04f(x) dx=1+3+4+2=10\int_{0}^{4} f(x) \, dx = 1 + 3 + 4 + 2 = 10

Now, using the formula for average value:

Avg=1b−a∫abf(x) dx=14−0⋅10=104=52Avg = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx = \frac{1}{4-0} \cdot 10 = \frac{10}{4} = \frac{5}{2}

Hence, the exact value for the average value of f(x) on the interval [0, 4] is 5/2.

Chat

Save To Bookmark