INSTRUCTORS Raffi Hovasapian John Zhu

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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.