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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of Calculus AB
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Second Fundamental Theorem of Calculus

    • Check continuity
    • Apply formula directly
  • Solvable without second fundamental theorem, however, much lengthier

Second Fundamental Theorem of Calculus

F(x) = ∫0x 3e4t dt. Find [dF/dx]
  • F(x) = ∫0x 3e4t dt
  • F(x) = [(3e4t)/4] |0x
  • F(x) = [(3e4x)/4] − [(3e4(0))/4]
  • F(x) = [(3e4x)/4] − [3/4]
  • [dF/dx] = [d/dx] ([(3e4x)/4] − [3/4])
  • [dF/dx] = 4([(3e4x)/4]) − 0
df = 3e4x
Find [d/dx] ∫0x 3e4t dt using the Second Fundamental Theorem of Calculus
  • 3e4t is continuous
[d/dx] ∫0x 3e4t dt = 3e4x
[d/dx] ∫13x t2 + 2t  dt
  • [d/dx] ∫13x t2 + 2t  dt = [d/dx] ([(t3)/3] + t2) |13x
  • [d/dx] ∫13x t2 + 2t  dt = [d/dx] ([((3x)3)/3] + (3x)2 − [(13)/3] − (12))
  • [d/dx] ∫13x t2 + 2t  dt = [d/dx] (9x3 + 9x2) − [d/dx] [10/3]
  • [d/dx] ∫13x t2 + 2t  dt = 27x2 + 18x
[d/dx] ∫13x t2 + 2t  dt = 27x2 + 18x)
Find [d/dx] ∫13x t2 + 2t  dt using the Second Fundamental Theorem of Calculus
  • Polynomial, so continuous everywhere.
  • [d/dx] ∫13x t2 + 2t  dt = (3x)2 [d/dx] (3x) + 2(3x) [d/dx] (3x)
  • [d/dx] ∫13x t2 + 2t  dt = 9x2(3) + 6x(3)
[d/dx] ∫13x t2 + 2t  dt = 27x2 + 18x
Find [d/dx] ∫−20x tant  dt using the second fundamental theorem of calculus.
  • tanx = [sinx/cosx] Discontinuous at x = ... −[(π)/2], [(π)/2], ...
Second Fundamental Theorem cannot be applied.
Find [d/dx] ∫1x t et dt using the Second Fundamental Theorem of Calculus
  • t et is continuous
[d/dx] ∫1x t et dt = x ex
Find [d/dx] ∫13x2 cost  dt using the Second Fundamental Theorem of Calculus
  • cost is continuous
  • [d/dx] ∫13x2 cost  dt = cos(3x2) [d/dx] (3x2)
[d/dx] ∫13x2 cost  dt = 6x cos(3x2)
Find [d/dx] ∫5x ln(t2 + 1)  dx
  • lnx is undefined for x ≤ 0
  • t2 + 1 > 0, so this will always be defined. Continuous for all reals.
[d/dx] ∫5x ln(t2 + 1)  dx = ln(x2 + 1)
Find [d/dx] ∫1ex cost dt using the Second Fundamental Theorem of Calculus
  • cost is continuous
  • [d/dx] ∫1ex cost dt = cos(ex) [d/dx] ex
[d/dx] ∫1ex cost dt = ex cosex
Find [d/dx] ∫12x − 3 3t3 dt using the Second Fundamental Theorem of Calculus
  • Polynomial, so continuous everywhere
  • [d/dx] ∫12x − 3 3t3 dt = 3(2x − 3)3 [d/dx] (2x − 3)
[d/dx] ∫12x − 3 3t3 dt = 6(2x − 3)3

*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

Second Fundamental Theorem of Calculus

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
  • Second Fundamental Theorem of Calculus 0:07
    • Definition
  • Example 1 1:08
  • Example 2 2:07
  • Example 3 2:48
  • Example 4 3:23