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.