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INSTRUCTORS Raffi Hovasapian John Zhu
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Definition of Integrals

  • An integral process is just the opposite of the derivative process

Definition of Integrals

g(x) is the derivative of f(x), if f(x) = 3x + 1, find g(x)
  • f′(x) = g(x)
  • f′(x) = [d/dx] (3x + 1)
  • f′(x) = 3
g(x) = 3
g(x) is the derivative of f(x), if f(x) = 3x + c where c is a constant, find g(x)
  • f′(x) = g(x)
  • f′(x) = [d/dx] (3x + c)
  • f′(x) = 3
  • The constant could be any value as long as it's not a function of x. If that's the case, the derivative of c will always be zero.
g(x) = 3
g(x) is the derivative of f(x), if g(x) = 3, find f(x)
  • Now we have the derivative and we have to work backwards.
  • g(x) = f′(x)
  • In the previous problem we know that f(x) = 3x + c
  • We can double check to make sure it works out
  • f′(x) = [d/dx] (3x + c) = 3 = g(x)
  • f(x) is the antiderivative of g(x)
f(x) = 3x + c
g(x) is the derivative of f(x), if f(x) = sinx + 99999, find g(x)
  • f′(x) = g(x)
  • f′(x) = [d/dx] (sinx + 99999)
  • f′(x) = cosx
g(x) = cosx
Find the antiderivative, f(x), of g(x) = cosx
  • If f(x) is the antiderivative of g(x), that means that f′(x) = g(x)
  • f′(x) = [d/dx] f(x) = g(x) = cosx
  • [d/dx] f(x) = cosx
  • What f(x) would fit in this equality?
  • f(x) = sinx would work, but it's not a complete solution. There are infinitely many!
  • f(x) = sinx + 1, sinx + 7, sinx − 200 and so on
  • The inequality is satisfied with any constant value
  • General solution
  • f(x) = sin+ c
f(x) = sinx + c
Find the antiderivative, f(x), of g(x) = 2x
  • [d/dx] f(x) = 2x
  • f(x) = x2 is a possible solution. Again, this solution would work if any constant is tacked on
f(x) = x2 + c
Find the antiderivative, f(x), of g(x) = ex
  • [d/dx] f(x) = g(x)
  • [d/dx] f(x) = ex
  • f(x) = ex
General solution f(x) = ex + c
Find the antiderivative, f(x), of g(x) = [1/x] + 1
  • [d/dx] f(x) = [1/x] + 1
  • f(x) = lnx + x + c
  • Checking our work
  • f′(x) = [d/dx] (lnx + x + c)
  • f′(x) = lnx + 1 = g(x)
f(x) = lnx + x + c
Given f(x) = [1/3] x3 + [1/2] x2 + c and g(x) = x2 + x, confirm that f(x) is the antiderivative of g(x)
  • f′(x) = [d/dx] ([1/3] x3 + [1/2] x2 + c)
  • f′(x) = [3/3] x2 + [2/2] x
  • f′(x) = x2 + x = g(x)
f′(x) = g(x)
Given f(x) = tanx + [1/2] e2x + c and g(x) = sec2 x + e2x, confirm that f(x) is the antiderivative of g(x)
  • f′(x) = [d/dx] (tanx + [1/2] e2x + c)
  • f′(x) = sec2 x + [2/2] e2x
  • f′(x) = sec2 x + e2x = g(x)
f′(x) = g(x)

*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

Definition of Integrals

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
  • Definition 0:09
    • Definition
    • Example