Presentation on theme: "1 §12.4 The Definite Integral The student will learn about the area under a curve defining the definite integral."— Presentation transcript:
1 §12.4 The Definite Integral The student will learn about the area under a curve defining the definite integral.
2 Introduction We have been studying the indefinite integral or the antiderivative of a function. We now introduce the definite integral. We now introduce the definite integral. This integral will be the area bounded by f (x), the x axis, and the vertical lines x = a and x = b with notation:
3 Area One way we can determine the area we want is by filling it with rectangles and calculating the sum of the areas of the rectangles. The width of each rectangle is Δx = 1. If we use the left end then the height of each of the four rectangles is f (1), f (2), f (3) and f(4) respectively. The sum of the rectangles is then L 4 = f (1) Δx + f (2) Δx + f (3) Δx + f (4) Δx f (x) =.5 x 2 + 2 1 2 3 4 5 Let L 4 denote that we are using 4 rectangles and we are determining the height on the left side of the rectangle L 4 = 2.5 + 4 + 6.5 + 10 = 23
4 Area We can repeat this using the right side of the rectangle to determine the height The width of each rectangle is again Δx = 1. The height of each of the four rectangles is now f (2), f (3), f (4) and f(5) respectively. The sum of the rectangles is then R 4 = 4 + 6.5 + 10 + 14.5 = 35 f (x) =.5 x 2 + 2 1 2 3 4 5 The average of these two areas would be an even better approximation. Area (23 + 35)/2 = 29.
5 More on Area The previous average of 29 is very close to the actual area of 28.666…. Our accuracy could be improved if we let the number or rectangles increase and Δx would then become smaller. The error in our process can be calculated if the function is monotone. That is if the function is only increasing or only decreasing. f (x) =.5 x 2 + 2 1 2 3 4 5
6 Error of the Area If we calculate the left and right areas as before using n rectangles and if a is the left side of the area and b the right side of the area then: Error =|Area – L n | |f (b) – f (a)| Δx = f (x) =.5 x 2 + 2 1 2 3 4 5 For our previous example this is computed as: Error 12 Theorem 1 |R n – L n |
7 Theorem 2 If f (x) is either increasing or decreasing on [a,b], then its left and right sums approach the same real number as n → ∞. The number approached as n → ∞ by the left and right sums in this theorem is the area between the graph of f and the x axis from x = a to x = b.
8 Approximating Area From the previous area example f (x) =.5 x 2 +2 with a = 1 and b = 5 and Δx = 1. R 4 = 4 + 6.5 + 10 + 14.5 = 35 f (x) =.5 x 2 + 2 1 2 3 4 5 L 4 = 2.5 + 4 + 6.5 + 10 = 23 Error 12 With error bound
9 Approximating Area continued f (x) =.5 x 2 + 2 1 2 3 4 5 If we wanted a particular accuracy, say 0.05 we could use this last formula to calculate, n, the number of rectangles needed. And solving for n yields n = 960. Meaning that 960 rectangles would be need to guarantee an error that does not exceed 0.05 !!!!
10 Definite Integral as Limit of Sums Let f be a function on interval [a, b]. Partition [a, b] into n subintervals of equal length ∆ x = (b – a)/n with endpoints a = x 0 < x 1 < x 2, … < x n – 1 < x n = b, Then This last sum is called a Riemann sum and each c k is required to be in the subinterval [x k – 1, x k ].
11 A Visual Presentation of the Above a = x 0 x n = bx 1 x 2 x n -1... c 1 c 2 c n f (c 1 ) f (c 2 ) Δx The area under the curve is approximated by the Riemann sum
12 Area Let’s revisit our original area problem and calculate the Riemann sum using the midpoints for c k. The width of each rectangle is again Δx = 1. The height of each of the four rectangles is now f (1.5), f (2.5), f (3.5) and f(4.5) respectively. The sum of the rectangles is then S 4 = 3.125 + 5.125 + 8.125 + 12.125 = 28.5 f (x) =.5 x 2 + 2 1 2 3 4 5 This is quite close to the actual area of 28.666....
13 Definite Integral Theorem 3. Let f be a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as n → ∞. Definition. Let f be a continuous function on [a, b]. The limit I of the Riemann sums for f on [a, b] is called the definite integral of f from a to b, denoted The integrand is f (x), the lower limit of integration is a, and the upper limit of integration is b.
14 Negative Values If f (x) is positive for some values of x on [a,b] and negative for others, then the definite integral symbol Represents the cumulative sum of the signed areas between the graph of f (x) and the x axis where areas above are positive and areas below negative. y = f (x) a bA B
15 Examples Calculate the definite integrals by referring to the figure with the indicated areas. Area A = 3.5 Area B = 12 y = f (x) a bA B c - 3.5 12 - 3.5 + 12 = 8.5
18 Summary. We summed rectangles under a curve using both the left and right ends and the centers and found that as the number of rectangles increased accuracy of the area under the curve increased. We found error bounds for these sums and average. We defined the definite integral as the limit of these sums and found that it represented the area between the function and the x axis.. We learned how to deal with areas under the x axis.