Average Value of Functions on Intervals - AP Calculus AB

  1. What is the integral form used to calculate the average value of f(x)f(x) on [a,b][a, b]?
    1b−a×∫abf(x) dx\frac{1}{b-a} \times \int_a^b f(x) \, dx. This is the mathematical definition using the fundamental theorem of calculus.
  2. What is the average value of f(x)=2x+3f(x) = 2x + 3 on [0,4][0, 4]?
    77. Apply the formula: 14∫04(2x+3) dx=14⋅28=7\frac{1}{4}\int_0^4 (2x+3)\,dx = \frac{1}{4} \cdot 28 = 7.
  3. Describe how average value of a function is related to its definite integral.
    Average value is the integral value divided by the interval length. Dividing by interval length converts total accumulation to average rate.
  4. What is the average value of f(x)=1xf(x) = \frac{1}{x} on [1,2][1, 2]?
    ln⁡(2)1\frac{\ln(2)}{1}. Apply the formula: 11∫121x dx=ln⁡(2)\frac{1}{1}\int_1^2 \frac{1}{x}\,dx = \ln(2).
  5. What is the significance of the interval [a,b][a, b] in average value calculations?
    It defines the domain over which the average is computed. The interval determines the bounds of integration and the divisor (b−a)(b-a).
  6. What is the average value of f(x)=x2f(x) = x^2 on the interval [0,2][0, 2]?
    43\frac{4}{3}. Apply the formula: 12∫02x2 dx=12⋅83=43\frac{1}{2}\int_0^2 x^2\,dx = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}.
  7. Find the average value of f(x)=e−xf(x) = \text{e}^{-x} on [0,1][0, 1].
    1−1e1 - \frac{1}{\text{e}}. Apply the formula: 11∫01e−x dx=1−1e\frac{1}{1}\int_0^1 e^{-x}\,dx = 1 - \frac{1}{e}.
  8. Calculate the average value of f(x)=5f(x) = 5 over the interval [2,6][2, 6].
    55. The average value of any constant function equals the constant itself.
  9. Find the average value of f(x)=4−x2f(x) = 4 - x^2 on [−2,2][-2, 2].
    83\frac{8}{3}. Apply the formula: 14∫−22(4−x2) dx=14⋅323=83\frac{1}{4}\int_{-2}^2 (4-x^2)\,dx = \frac{1}{4} \cdot \frac{32}{3} = \frac{8}{3}.
  10. Why is the average value important in mathematical modeling?
    It provides a summary measure of function behavior over an interval. It gives a single representative value characterizing overall function behavior.
  11. How does the average value formula change if [a,b][a, b] is [−b,b][-b, b]?
    The formula remains the same; the interval length is $2b$. The interval length becomes $2b$ but the formula structure stays the same.
  12. What is the average value of f(x)=x+2f(x) = x + 2 on [−1,1][-1, 1]?
    22. Apply the formula: 12∫−11(x+2) dx=12⋅4=2\frac{1}{2}\int_{-1}^1 (x+2)\,dx = \frac{1}{2} \cdot 4 = 2.
  13. Calculate the average value of f(x)=x3−x2f(x) = x^3 - x^2 on [0,2][0, 2].
    23\frac{2}{3}. Apply the formula: 12∫02(x3−x2) dx=12⋅43=23\frac{1}{2}\int_0^2 (x^3-x^2)\,dx = \frac{1}{2} \cdot \frac{4}{3} = \frac{2}{3}.
  14. Calculate the average value of f(x)=3x+2f(x) = 3x + 2 on [1,4][1, 4].
    99. Apply the formula: 13∫14(3x+2) dx=13⋅27=9\frac{1}{3}\int_1^4 (3x+2)\,dx = \frac{1}{3} \cdot 27 = 9.
  15. What is the average value of f(x)=6−2xf(x) = 6 - 2x on [0,3][0, 3]?
    33. Apply the formula: 13∫03(6−2x) dx=13⋅9=3\frac{1}{3}\int_0^3 (6-2x)\,dx = \frac{1}{3} \cdot 9 = 3.
  16. What is the average value of f(x)=cos(x)f(x) = \text{cos}(x) on [0,π][0, \text{π}]?
    00. Apply the formula: 1π∫0πcos⁡(x) dx=1π⋅0=0\frac{1}{\pi}\int_0^{\pi} \cos(x)\,dx = \frac{1}{\pi} \cdot 0 = 0.
  17. Find the average value of f(x)=1x2f(x) = \frac{1}{x^2} on [1,3][1, 3].
    13\frac{1}{3}. Apply the formula: 12∫131x2 dx=12⋅23=13\frac{1}{2}\int_1^3 \frac{1}{x^2}\,dx = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}.
  18. What is the average value of f(x)=x2+1f(x) = x^2 + 1 on [0,2][0, 2]?
    73\frac{7}{3}. Apply the formula: 12∫02(x2+1) dx=12⋅143=73\frac{1}{2}\int_0^2 (x^2+1)\,dx = \frac{1}{2} \cdot \frac{14}{3} = \frac{7}{3}.
  19. What is the average value of f(x)=x2+xf(x) = x^2 + x on [1,3][1, 3]?
    163\frac{16}{3}. Apply the formula: 12∫13(x2+x) dx=12⋅323=163\frac{1}{2}\int_1^3 (x^2+x)\,dx = \frac{1}{2} \cdot \frac{32}{3} = \frac{16}{3}.
  20. Calculate the average value of f(x)=7x−3f(x) = 7x - 3 on [2,5][2, 5].
    1919. Apply the formula: 13∫25(7x−3) dx=13⋅57=19\frac{1}{3}\int_2^5 (7x-3)\,dx = \frac{1}{3} \cdot 57 = 19.
  21. If f(x)f(x) is continuous on [a,b][a, b], what theorem justifies the average value calculation?
    The Mean Value Theorem for Integrals. This theorem guarantees existence of a point where function equals its average.
  22. What does the average value of a function represent in a physical context?
    The average value represents the mean level of the function over the interval. It's the constant value that would yield the same total area under the curve.
  23. Calculate the average value of f(x)=2x3f(x) = 2x^3 on [1,2][1, 2].
    7.57.5. Apply the formula: 11∫122x3 dx=11⋅7.5=7.5\frac{1}{1}\int_1^2 2x^3\,dx = \frac{1}{1} \cdot 7.5 = 7.5.
  24. What condition must f(x)f(x) satisfy for its average value to be calculated over [a,b][a, b]?
    f(x)f(x) must be continuous on [a,b][a, b]. Continuity ensures the integral exists and the Mean Value Theorem applies.
  25. What is the average value of f(x)=2xf(x) = 2x on [0,3][0, 3]?
    33. Apply the formula: 13∫032x dx=13⋅9=3\frac{1}{3}\int_0^3 2x\,dx = \frac{1}{3} \cdot 9 = 3.
  26. Define the definite integral in the context of average value of a function.
    The integral computes the total accumulation of the function over the interval. The integral measures the signed area under the curve over the interval.
  27. Find the average value of f(x)=1xf(x) = \frac{1}{x} on [1,4][1, 4].
    ln⁡(4)3\frac{\ln(4)}{3}. Apply the formula: 13∫141x dx=13⋅ln⁡(4)=ln⁡(4)3\frac{1}{3}\int_1^4 \frac{1}{x}\,dx = \frac{1}{3} \cdot \ln(4) = \frac{\ln(4)}{3}
  28. Find the average value of f(x)=exf(x) = \text{e}^x on [0,1][0, 1].
    e−1\text{e} - 1. Apply the formula: 11∫01ex dx=e−1\frac{1}{1}\int_0^1 e^x\,dx = e - 1.
  29. State the formula for the average value of a function on an interval [a,b][a, b].
    1b−a×∫abf(x) dx\frac{1}{b-a} \times \int_a^b f(x) \, dx. The standard formula divides the definite integral by the interval length.
  30. Identify the first step to find the average value of a function f(x)f(x) on [a,b][a, b].
    Calculate the definite integral of f(x)f(x) from aa to bb. The definite integral gives the total accumulation before dividing by interval length.

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