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8.1 Finding the Average Value of a Function on an Interval

2 min read•february 15, 2024

Jesse

Jesse

Anusha Tekumulla

Anusha Tekumulla

Jesse

Jesse

Anusha Tekumulla

Anusha Tekumulla

Attend a live cram event

Review all units live with expert teachers & students

    Welcome back to AP Calculus with Fiveable! This topic focuses on finding the average value of a continuous function using definite integrals.

    🔢 Average Value of a Function

    The average value of a function will allow us to solve problems that involve the accumulation of change over an interval, which will later be used to understand more difficult topics of integration.

    For questions that require the average value of a function, we are never given a finite number of data points. Therefore, we must use integration to determine what the average value is.

    This idea is fairly simple once you memorize a key piece of information: if f is continuous on [a,b],[a,b], then the average value of f on [a,b[a,b] is the following.

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

    Untitled

    Image Courtesy of ExamSolutions

    🔍Average Value of a Function Steps

    Here are some steps to help break down the formula!

    1. Set up the integral so that you integrate f(x) from a to b with respect to x, which will calculate the area under the curve between these two limits.
    2. Then place the fraction in front of the integral, which is simply the reciprocal of the difference between a and b.
    3. Evaluating this expression allows you to get the average y-value of this function between [a,b][a,b]

    ✏️ Average Value of a Function Walkthrough

    If the formula still seems a little difficult to understand due to its notation, practice questions are the best way to better understand its use!

    Consider the function f(x)=2x2−3x+5f(x) = 2x^2-3x+5 on the interval [1,4]. Find the average value of this function on the interval.

    In this case, a = 1 and b = 4. So we begin by subbing the 1 and 4 into both the denominator of the fraction in front of the integral and the limits of the integral.

    Average Value=14−1∫14(2x2−3x+5) dx\text{Average Value} = \frac{1}{4-1} \int_{1}^{4} (2x^2 - 3x + 5) \, dx

    Next, take the integral of f(x).

    =13[23x3−32x2+5x]14= \frac{1}{3} \left[ \frac{2}{3}x^3 - \frac{3}{2}x^2 + 5x \right]_{1}^{4}

    Finally, we can sub in the limits and evaluate.

    13[(23(4)3−32(4)2+5(4))−(23(1)3−32(1)2+5(1))]=232 \frac{1}{3} \left[ \left( \frac{2}{3}(4)^3 - \frac{3}{2}(4)^2 + 5(4) \right) - \left( \frac{2}{3}(1)^3 - \frac{3}{2}(1)^2 + 5(1) \right) \right] =\boxed{\frac{23}{2}}

    Time for you to practice some questions yourself! ⬇️

    📝 Average Value of a Function Practice Problems

    Give each of these problems a try before you move onto the solutions!

    1. What is the average value of 5x2+45x^2+4 on the interval 0≤x≤60\le x\le 6?
    2. What is the average value of x3−x2x^3-x^2 on the interval 2≤x≤52\le x\le 5?
    3. What is the average value of sin(x)+cos(x)sin(x)+cos(x) on the interval 0≤x≤π0\le x\le \pi?

    Average Value of a Function Question Solutions

    Question 1 Solution

    Average Value=16−0∫06(5x2+4) dx\text{Average Value} = \frac{1}{6-0} \int_{0}^{6} (5x^2+4) \, dx
    =16[53x3+4x]06= \frac{1}{6} \left[ \frac{5}{3}x^3 +4x \right]_{0}^{6}
    16[(53(6)3+4(6))−(53(0)3+4(0))]=64 \frac{1}{6} \left[ \left( \frac{5}{3}(6)^3 + 4(6) \right) - \left( \frac{5}{3}(0)^3 + 4(0) \right) \right] =\boxed{64}

    Question 2 Solution

    Average Value=15−2∫25(x3−x2) dx\text{Average Value} = \frac{1}{5-2} \int_{2}^{5} (x^3-x^2) \, dx
    =13[14x4−13x3]25= \frac{1}{3} \left[ \frac{1}{4}x^4 - \frac{1}{3} x^3 \right]_{2}^{5}
    13[(14(5)4−13(5)3)−(14(2)4−13(2)3)]=1514 \frac{1}{3} \left[ \left( \frac{1}{4}(5)^4 - \frac{1}{3}(5)^3 \right) - \left( \frac{1}{4}(2)^4 - \frac{1}{3}(2)^3 \right) \right] =\boxed{\frac{151}{4}}

    Question 3 Solution

    Average Value=1π−0∫0πsin(x)+cos(x) dx\text{Average Value} = \frac{1}{\pi-0} \int_{0}^{\pi} sin(x)+cos(x) \, dx
    =1π[−cos(x)+sin(x)]0π= \frac{1}{\pi} \left[ -cos(x)+sin(x)\right]_{0}^{\pi}
    1π[(−cos(π)+sin(π))−(−cos(0)+sin(0))]=2π \frac{1}{\pi} \left[ \left( -cos(\pi) + sin(\pi) \right) - \left( -cos(0) + sin(0) \right) \right] =\boxed{\frac{2}{\pi}}

    ⭐ Closing

    Great job! This topic often shows up as part (a) of FRQs, so keep this in mind for the AP.

    8.1 Finding the Average Value of a Function on an Interval

    2 min read•february 15, 2024

    Jesse

    Jesse

    Anusha Tekumulla

    Anusha Tekumulla

    Jesse

    Jesse

    Anusha Tekumulla

    Anusha Tekumulla

    Attend a live cram event

    Review all units live with expert teachers & students

      Welcome back to AP Calculus with Fiveable! This topic focuses on finding the average value of a continuous function using definite integrals.

      🔢 Average Value of a Function

      The average value of a function will allow us to solve problems that involve the accumulation of change over an interval, which will later be used to understand more difficult topics of integration.

      For questions that require the average value of a function, we are never given a finite number of data points. Therefore, we must use integration to determine what the average value is.

      This idea is fairly simple once you memorize a key piece of information: if f is continuous on [a,b],[a,b], then the average value of f on [a,b[a,b] is the following.

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

      Untitled

      Image Courtesy of ExamSolutions

      🔍Average Value of a Function Steps

      Here are some steps to help break down the formula!

      1. Set up the integral so that you integrate f(x) from a to b with respect to x, which will calculate the area under the curve between these two limits.
      2. Then place the fraction in front of the integral, which is simply the reciprocal of the difference between a and b.
      3. Evaluating this expression allows you to get the average y-value of this function between [a,b][a,b]

      ✏️ Average Value of a Function Walkthrough

      If the formula still seems a little difficult to understand due to its notation, practice questions are the best way to better understand its use!

      Consider the function f(x)=2x2−3x+5f(x) = 2x^2-3x+5 on the interval [1,4]. Find the average value of this function on the interval.

      In this case, a = 1 and b = 4. So we begin by subbing the 1 and 4 into both the denominator of the fraction in front of the integral and the limits of the integral.

      Average Value=14−1∫14(2x2−3x+5) dx\text{Average Value} = \frac{1}{4-1} \int_{1}^{4} (2x^2 - 3x + 5) \, dx

      Next, take the integral of f(x).

      =13[23x3−32x2+5x]14= \frac{1}{3} \left[ \frac{2}{3}x^3 - \frac{3}{2}x^2 + 5x \right]_{1}^{4}

      Finally, we can sub in the limits and evaluate.

      13[(23(4)3−32(4)2+5(4))−(23(1)3−32(1)2+5(1))]=232 \frac{1}{3} \left[ \left( \frac{2}{3}(4)^3 - \frac{3}{2}(4)^2 + 5(4) \right) - \left( \frac{2}{3}(1)^3 - \frac{3}{2}(1)^2 + 5(1) \right) \right] =\boxed{\frac{23}{2}}

      Time for you to practice some questions yourself! ⬇️

      📝 Average Value of a Function Practice Problems

      Give each of these problems a try before you move onto the solutions!

      1. What is the average value of 5x2+45x^2+4 on the interval 0≤x≤60\le x\le 6?
      2. What is the average value of x3−x2x^3-x^2 on the interval 2≤x≤52\le x\le 5?
      3. What is the average value of sin(x)+cos(x)sin(x)+cos(x) on the interval 0≤x≤π0\le x\le \pi?

      Average Value of a Function Question Solutions

      Question 1 Solution

      Average Value=16−0∫06(5x2+4) dx\text{Average Value} = \frac{1}{6-0} \int_{0}^{6} (5x^2+4) \, dx
      =16[53x3+4x]06= \frac{1}{6} \left[ \frac{5}{3}x^3 +4x \right]_{0}^{6}
      16[(53(6)3+4(6))−(53(0)3+4(0))]=64 \frac{1}{6} \left[ \left( \frac{5}{3}(6)^3 + 4(6) \right) - \left( \frac{5}{3}(0)^3 + 4(0) \right) \right] =\boxed{64}

      Question 2 Solution

      Average Value=15−2∫25(x3−x2) dx\text{Average Value} = \frac{1}{5-2} \int_{2}^{5} (x^3-x^2) \, dx
      =13[14x4−13x3]25= \frac{1}{3} \left[ \frac{1}{4}x^4 - \frac{1}{3} x^3 \right]_{2}^{5}
      13[(14(5)4−13(5)3)−(14(2)4−13(2)3)]=1514 \frac{1}{3} \left[ \left( \frac{1}{4}(5)^4 - \frac{1}{3}(5)^3 \right) - \left( \frac{1}{4}(2)^4 - \frac{1}{3}(2)^3 \right) \right] =\boxed{\frac{151}{4}}

      Question 3 Solution

      Average Value=1π−0∫0πsin(x)+cos(x) dx\text{Average Value} = \frac{1}{\pi-0} \int_{0}^{\pi} sin(x)+cos(x) \, dx
      =1π[−cos(x)+sin(x)]0π= \frac{1}{\pi} \left[ -cos(x)+sin(x)\right]_{0}^{\pi}
      1π[(−cos(π)+sin(π))−(−cos(0)+sin(0))]=2π \frac{1}{\pi} \left[ \left( -cos(\pi) + sin(\pi) \right) - \left( -cos(0) + sin(0) \right) \right] =\boxed{\frac{2}{\pi}}

      ⭐ Closing

      Great job! This topic often shows up as part (a) of FRQs, so keep this in mind for the AP.

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      About Us
      About FiveableBlogCareersCode of ConductTerms of UsePrivacy PolicyCCPA Privacy Policy
      Resources
      Cram ModeAP Score CalculatorsStudy GuidesPractice QuizzesGlossaryCram EventsMerch ShopCrisis Text LineHelp Center

      © 2024 Fiveable Inc. All rights reserved.

      AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

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