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INSTRUCTORS Raffi Hovasapian John Zhu
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Mean Value Theorem

  • f(c): average value between given interval
    • Check continuity
    • Apply formula directly to find f(c)

Mean Value Theorem

Find the average value, f(c), of of the function f(x) = sinx on the interval [0, π] using the Mean Value Theorem for Integrals
  • f(c) = [1/(π− 0)] ∫0π sinx  dx
  • f(c) = [1/(π)] (−cosx) |0π
  • f(c) = [1/(π)] (− cosπ+ cos0)
  • f(c) = [1/(π)] (1 + 1)
f(c) = [2/(π)]
Find the average value, f(c) of of the function f(x) = sinx on the interval [0, 2π] using the Mean Value Theorem for Integrals
  • f(c) = [1/(2π− 0)] ∫0 sinx  dx
  • f(c) = [1/(2π)] (−cosx) |0
  • f(c) = [1/(2π)] (− cos2π+ cos0)
  • f(c) = [1/(2π)] (−1 + 1)
f(c) = 0
Find the average value,f(c), of of the function f(x) = x2 + 7 on the interval [0, 3] using the Mean Value Theorem for Integrals
  • f(c) = [1/(3 − 0)] ∫03 x2 + 7  dx
  • f(c) = [1/3] ([(x3)/3] + 7x) |03
  • f(c) = [1/3] ([(33)/3] + 21 − (0 + 0))
  • f(c) = [1/3] 30
f(c) = 10
Find the average value,f(c), of of the function f(x) = x3 + 2x + 1 on the interval [0, 4] using the Mean Value Theorem for Integrals
  • f(c) = [1/(4 − 0)] ∫04 x3 + 2x + 1  dx
  • f(c) = [1/4] ([(x4)/4] + x2 + x) |04
  • f(c) = [1/4] ([(44)/4] + 42 + 4 − 0 − 0 − 0)
  • f(c) = 42 + 4 + 1
f(c) = 21
Find the average value,f(c), of of the function f(x) = [1/x] on the interval [−1, 1] using the Mean Value Theorem for Integrals
  • f(x) has a discontinuity at x = 0 and that is in the interval
Mean Value Theorem for Integrals cannot be applied
Find the average value,f(c), of of the function f(x) = [1/x] on the interval [1, e] using the Mean Value Theorem for Integrals
  • f(x) is continuous for x > 0
  • f(c) = [1/(e − 1)] ∫1e [1/x] dx
  • f(c) = [1/(e − 1)] lnx |1e
  • f(c) = [1/(e − 1)] (lne − ln1)
  • f(c) = [1/(e − 1)] (1)
f(c) = [1/(e − 1)]
Find the average value,f(c), of of the function f(x) = cosx + 4 on the interval [0, 2π] using the Mean Value Theorem for Integrals
  • f(c) = [1/(2π− 0)] ∫0 cosx + 4  dx
  • f(c) = [1/(2π)] (sinx + 4x) |0
  • f(c) = [1/(2π)] (sin2π+ 8 π− sin0 − 0)
  • f(c) = [1/(2π)] (8 π)
f(c) = 4
Find the average value,f(c), of of the function f(x) = [1/(√{4 − x2})] on the interval [0, 1] using the Mean Value Theorem for Integrals
  • f(x) is undefined for x ≤ −2 and x ≥ 2
  • The function is defined on the interval [0, 1]
  • f(c) = [1/(1 − 0)] ∫01 [1/(√{4 − x2})] dx
  • f(c) = ∫01 [1/(√{22 − x2})] dx
  • f(c) = sin−1 [x/2] |01
  • f(c) = sin−1 [1/2] − sin−1 0
  • f(c) = [(π)/6] − 0
f(c) = 0
Find the average value,f(c), of of the function f(x) = 4x on the interval [−5, 5] using the Mean Value Theorem for Integrals
  • f(c) = [1/(5 − (−5))] ∫−55 4x  dx
  • f(c) = [1/10] 2x2 |−55
  • f(c) = [1/10] (2(5)2 − 2(−5)2)
  • f(c) = [1/10] (0)
f(c) = 0
Find the average value,f(c), of of the function f(x) = tan[x/4] on the interval [0, π] using the Mean Value Theorem for Integrals
  • f(x) has discontinuities whenever cos[x/4] = 0
  • So near zero, f(x) is discontinuous at −2π and 2π. Our interval does not include a discontinuity
  • f(c) = [1/(π− 0)] ∫0π tan[x/4] dx
  • u = [x/4]
  • du = [1/4] dx
  • f(c) = [1/(π)] 4 ∫0π tanu  du
  • f(c) = [4/(π)] ∫0π [sinu/cosu] du
  • v = cosu
  • dv = −sinu  du
  • f(c) = [4/(π)] (−1) ∫0π [1/v] dv
  • f(c) = − [4/(π)] lnv |0π
  • f(c) = − [4/(π)] lncosu |0π
  • f(c) = − [4/(π)] lncos[x/4] |0π
  • f(c) = − [4/(π)] (lncos[(π)/4] − lncos0)
  • f(c) = − [4/(π)] (ln[1/(√2)] − ln1)
  • f(c) = − [4/(π)] (ln[1/(√2)] − 0)
  • f(c) = − [4/(π)] (ln1 − ln√2)
  • f(c) = − [4/(π)] (0 − ln2[1/2])
  • f(c) = − [4/(π)] (− [1/2] ln2)
f(c) = [2 ln2/(π)]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Mean Value Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Mean Value Theorem of Integration 0:06
    • Example 1
  • Example 2 2:29
  • Example 3 3:48
  • Example 4 6:02