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The Fundamental Theorems of Calculus. Lesson 5.4. Don't let this happen to you!. a b. Average Value Theorem. Consider function f(x) on an interval Area = Consider the existence of a value for x = c such that a < x < b and. c. a b.
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The Fundamental Theorems of Calculus Lesson 5.4
a b Average Value Theorem • Consider function f(x) on an interval • Area = • Consider the existence of a value for x = c such that a < x < b and c
a b Average Value of Function • Also The average value is c
Area Under a Curve f(x) • 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 a b
Area Under a Curve f(x) • 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)? A(x) x a b 0
h The Area Function f(x) • 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 A(x+h) x x+h a b
Why? The Area Function • The area of the slice is • Now divide by h … then take the limit Divide both sides by h
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
First Fundamental Theorem of Calculus • Given f is • continuous on interval [a, b] • F is any function that satisfies F’(x) = f(x) • Then
First Fundamental Theorem of Calculus • The definite integralcan be computed by • finding an antiderivative F on interval [a,b] • evaluating at limits a and b and subtracting • Try
Area Under a Curve • Consider • Area =
Area Under a Curve • Find the area under the following function on the interval [1, 4]
Second Fundamental Theorem of Calculus • Often useful to think of the following form • We can consider this to be a function in terms of x View Movie on next slide
Second Fundamental Theorem of Calculus • This is a function View Demo
Second Fundamental Theorem of Calculus • Suppose we aregiven G(x) • What is G’(x)?
Since this is a constant … Second Fundamental Theorem of Calculus • Note that • Then • What about ?
Second Fundamental Theorem of Calculus • Try this
Second Fundamental Theorem of Calculus • An application from Differential Equations • Consider • Now integrate both sides • And
Assignment • Lesson 5.4 • Page 327 • Exercises 1 – 49 odd