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The Fundamental Theorem of Calculus
Lesson 6.2
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Area Under a Curve f(x) a b We know the area under the curve on the interval [a, b] is given by the formula above Consider the existence of an Area Function
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Area Under a Curve f(x) A(x) a x b The area function is A(x) … the area under the curve on the interval [a, x] a ≤ x ≤ b What is A(a)? What is A(b)?
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The Area Function If the area is increased by h units
f(x) A(x+h) x x+h a b h If the area is increased by h units New area is A(x + h) Area of the new slice is A(x + h) – A(x) It's height is some average value ŷ It's width is h
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The Area Function The area of the slice is
Now divide by h … then take the limit Divide both sides by h Why?
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The Area Function The left side of our equation is the derivative of A(x) !! The derivative of the Area function, A' is the same as the original function, f
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Fundamental Theorem of Calculus
We know the antiderivative of f(x) is F(x) + C Thus A(x) = F(x) + C Since A(a) = 0 Then for x = a, 0 = F(a) + C or C = -F(a) And A(x) = F(x) – F(a) Now if x = b A(b) = F(b) – F(a)
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Fundamental Theorem of Calculus
We have said that Thus we conclude The area under the curve is equal to the difference of the two antiderivatives Often written
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Example Consider What is F(x), the antiderivative?
Evaluate F(5) – F(0)
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Properties Bringing out a constant factor
The integral of a sum is the sum of the integrals
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Properties Splitting an integral
f must be continuous on interval containing a, b, and c
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Example Consider Which property is being used to find F(x), the antiderivative? Evaluate F(4) – F(0)
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Try Another What is Hint … combine to get a single power of x
What is F(x)? What is F(3) – F(1)?
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Assignment Lesson 6.2 Page 227
Exercises 1, 5, 9, … 41, 45 (every other odd) Second Day 3, 7, … 47 (every other odd)
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