Post

Average Value of a Function on an Interval

Created by @nathanedwards
 at October 31st 2023, 10:07:39 pm.
AP_Calculus_AB-Questions
#power-rule
#average-value
#definite-integral

Question:

Let f(x)=3x2−2x+1f(x) = 3x^2 - 2x + 1 over the interval [1,5][1, 5]. Find the average value of f(x)f(x) on this interval.

Answer:

The average value of a function on an interval can be found using the formula:

Average Value=1b−a∫abf(x) dx \text{{Average Value}} = \frac{1}{{b-a}}\int_a^b f(x) \, dx

where aa and bb represent the lower and upper bounds of the interval, respectively.

In this case, a=1a = 1 and b=5b = 5. To find the average value, we need to compute the definite integral of f(x)f(x) on the interval [1,5][1, 5].

Average Value=15−1∫15f(x) dx \text{{Average Value}} = \frac{1}{{5-1}} \int_1^5 f(x) \, dx

Evaluating the integral:

Average Value=14∫15(3x2−2x+1) dx \text{{Average Value}} = \frac{1}{4} \int_1^5 (3x^2 - 2x + 1) \, dx
Average Value=14(∫153x2 dx−∫152x dx+∫151 dx) \text{{Average Value}} = \frac{1}{4} \left(\int_1^5 3x^2 \, dx - \int_1^5 2x \, dx + \int_1^5 1 \, dx \right)

Applying the power rule of integration, we find:

Average Value=14([x3]15−[x2]15+[x]15) \text{{Average Value}} = \frac{1}{4} \left( \left[ x^3 \right]_1^5 - \left[ x^2 \right]_1^5 + \left[ x \right]_1^5 \right)
Average Value=14((53−13)−(52−12)+(5−1)) \text{{Average Value}} = \frac{1}{4} \left( (5^3 - 1^3) - (5^2 - 1^2) + (5 - 1) \right)
Average Value=14((125−1)−(25−1)+4) \text{{Average Value}} = \frac{1}{4} \left( (125 - 1) - (25 - 1) + 4 \right)
Average Value=125−1−25+1+44 \text{{Average Value}} = \frac{125 - 1 - 25 + 1 + 4}{4}
Average Value=1044 \text{{Average Value}} = \frac{104}{4}
Average Value=26 \boxed{\text{{Average Value}} = 26}

Therefore, the average value of f(x)f(x) on the interval [1,5][1, 5] is 26.

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