The Fundamental Theorem of Calculus Lesson 6.2

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

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)?

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

The Area Function The area of the slice is Now divide by h … then take the limit Divide both sides by h Why?

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

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)

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

Example Consider What is F(x), the antiderivative? Evaluate F(5) – F(0)

Properties Bringing out a constant factor The integral of a sum is the sum of the integrals

Properties Splitting an integral f must be continuous on interval containing a, b, and c

Example Consider Which property is being used to find F(x), the antiderivative? Evaluate F(4) – F(0)

Try Another What is Hint … combine to get a single power of x What is F(x)? What is F(3) – F(1)?

Assignment Lesson 6.2 Page 227 Exercises 1, 5, 9, … 41, 45 (every other odd) Second Day 3, 7, … 47 (every other odd)