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The Fundamental Theorem of Calculus (Part I). The overall idea behind the Fundamental Theorem of Calculus is that differentiation. and integration. are operational opposites;. In the same way that multiplication. and division are. Graphical Proof. 2 Reminders. 1).
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The overall idea behind the Fundamental Theorem of Calculus is that differentiation and integration are operational opposites; In the same way that multiplication and division are.
2 Reminders 1) 2) A function is like a machine that takes in an input, manipulates it according to some rule, and spits out an output. Example: is a function that gives you the distance an object is from the surface of the earth at any given time, t. Think of it as “the distance away from earth function.” You input a time, t, and the function spits out a distance the object is form the earth.
Start of proof… Lets take some function, f(t), that has the following graph: f(t) The area under the curve depends on where x is (‘a’ is any constant and x is a variable). Therefore we can say that the area under the curve is a function of x. a x1 x2 x3 t Let’s call it F(x) and think of it as “the area so far” function. Area1= F(x1) = 4.0000 Area2= F(x2) = 5.9532 Area3= F(x3) = 8.0000
f(t) f(x) a x t h 2) Now lets say we move x a small distance, h, to x+h. This means that the change in our “area so far” function, F(x), is: h x+h which is approximately equal to: with base h and height f(x)). (the area of the rectangle So What we end up with is:
EXACTLY When will ?
Therefore we can say that and we know that is (the definition of the derivative of F(x) )
So we know that a) Since F(x) is “the area so far” function, we can say that b) If we take d/dx of both sides in part b, we get c) Since We can combine statements a and c
What we are saying by the statement is that the derivative of the integral of a function gives us the original function. In other words, differentiation can undo integration, and vise versa
What we are also saying by the statement is that the integral is the of the original function Example:
If f is continuous at every point on the interval [a, b], And if F is any anti-derivative of f on [a, b] then: Example 1: Solution: We saw earlier that a simple anti-derivative of x2 is x3/3
Example 2: Solution: A simple anti-derivative of (x3+1) is
Note that the total area will not simply be Find the area of the region between the curve y = 4 – x2 And the x-axis on [0, 3] Example 3: What we need to do is find The zeros of f, separate our interval into sub-intervals, integrate each sub-interval separately, and then add the absolute value of each integral Sub-interval1 [0, 2] : Sub-interval1 [2, 3] : Area of entire shaded region =
Steps for Finding Total Area Between Graph and x-axis on [a, b] 1) Partition [a, b] with the zeros of f. 2) Integrate f over each sub-interval 3) Add the absolute value of the integrals
Finding Area Using FNINT Example: Find the area of the region between the curve and the x-axis on [-3, 3] Solution: We want to find the area of the red region, but instead of separating [-3,3] into sub-intervals, we will simply find the integral of the absolute value of y. The absolute value of y simply flips all the regions below the x-axis to the top. So on your calculator type: FNINT(abs(xcos(2x)), x, -3, 3)