The Definite Integral Riemann Sums Sigma notation enables

The Definite Integral

Riemann Sums • Sigma notation enables us to express a large sum in compact form

Riemann Sums • LRAM, MRAM, and RRAM are examples of Riemann sums • Sn = This sum, which depends on the partition P and the choice of the numbers ck, is a Riemann sum for f on the interval [a, b]

Definite Integral as a Limit of Riemann Sums Let f be a function defined on a closed interval [a, b]. For any partition P of [a, b], let the numbers ck be chosen arbitrarily in the subintervals [xk-1, xk]. If there exists a number I such that no matter how P and the ck’s are chosen, then f is integrable on [a, b] and I is the definite integral of f over [a, b].

Definite Integral of a continuous function on [a, b] Let f be continuous on [a, b], and let [a, b] be partitioned into n subintervals of equal length Δx = (b-a)/n. Then the definite integral of f over [a, b] is given by where each ck is chosen arbitrarily in the kth subinterval.

Definite integral This is read as “the integral from a to b of f of x dee x” or sometimes as “the integral from a to b of f of x with respect to x. ”

Using Definite integral notation The function being integrated is f(x) = 3 x 2 – 2 x + 5 over the interval [-1, 3]

Definition: Area under a curve If y = f(x) is nonnegative and integrable over a closed interval [a, b], then the area under the curve of y = f(x) from a to b is the integral of f from a to b,

Nonpositive regions If the graph is nonpositive from a to b then

Area of any integrable function = (area above the x-axis) – (area below x-axis)

Integral of a Constant If f(x) = c, where c is a constant, on the interval [a, b], then

Evaluating Integrals using areas We can use integrals to calculate areas and we can use areas to calculate integrals. Using areas, evaluate the integrals: 1) 2)

Evaluating Integrals using areas Evaluate using areas: 3) 4) (a<b)

Evaluating integrals using areas Evaluate the discontinuous function: Since the function is discontinuous at x = 0, we must divide the areas into two pieces and find the sum of the areas = -1 + 2 = 1

Integrals on a Calculator You can evaluate integrals numerically using the calculator. The book denotes this by using NINT. The calculator function fn. Int is what you will use. = fn. Int(xsinx, x, -1, 2) is approx. 2. 04

Evaluate Integrals on calculator • Evaluate the following integrals numerically: 1) = approx. 3. 14 2) = approx. . 89

Rules for Definite Integrals 1) Order of Integration:

Rules for Definite Integrals 2) Zero:

Rules for Definite Integrals 3) Constant Multiple:

Rules for Definite Integrals 4) Sum and Difference:

Rules for Definite Integrals 5) Additivity:

Rules for Definite Integrals 6) Max-Min Inequality: If max f and min f are the maximum and minimum values of f on [a, b] then: min f ∙ (b – a) ≤ ≤ max f ∙ (b – a)

Rules for Definite Integrals 7) Domination: f(x) ≥ g(x) on [a, b] f(x) ≥ 0 on [a, b] ≥ 0

Using the rules for integration Suppose: Find each of the following integrals, if possible: a) b) c) d) e) f)

Using the rules for definite integrals Show that the value of is less than 3/2 The Max-Min Inequality rule says the max f. (b – a) is an upper bound. The maximum value of √(1+cosx) on [0, 1] is √ 2 so the upper bound is: √ 2(1 – 0) = √ 2 , which is less than 3/2

Average (Mean) Value If f is integrable on [a, b], its average (mean) value on [a, b] is: av(f) = Find the average value of f(x) = 4 – x 2 on [0, 3]. Does f actually take on this value at some point in the given interval?

Applying the Mean Value Av(f) = = 1/3(3) = 1 4 – x 2 = 1 when x = ± √ 3 but only √ 3 falls in the interval from [0, 3], so x = √ 3 is the place where the function assumes the average.

Mean Value Theorem for Definite Integrals If f is continuous on [a, b], then at some point c in [a, b],

The Fundamental Theorem of Calculus, Part I If f is continuous on [a, b], then the function F(x) = has a derivative at every point x in [a, b], and

Applications of The Fundamental Theorem of Calculus, Part I 1. 2. 3.

Applications of The Fundamental Theorem of Calculus, Part I Find dy/dx. y= Since this has an x on both ends of the integral, it must be separated.

Applications of The Fundamental Theorem of Calculus, Part I =

Applications of The Fundamental Theorem of Calculus, Part I = =

The Fundamental Theorem of Calculus, Part 2 If f is continuous at every point of [a, b], and if F is any antiderivative of f on [a, b], then This part of the Fundamental Theorem is also called the Integral Evaluation Theorem.

Trapezoidal Rule To approximate T= , use (y 0 + 2 y 1 + 2 y 2 + …. 2 yn-1 + yn) where [a, b] is partitioned into n subintervals of equal length h = (b-a)/n.

Using the trapezoidal rule Use the trapezoidal rule with n = 4 to estimate h = (2 -1)/4 or ¼, so T = 1/8( 1+2(25/16)+2(36/16)+2(49/16)+4) = 75/32 or about 2. 344

Simpson’ Rule To approximate S= , use (y 0 + 4 y 1 + 2 y 2 + 4 y 3…. 2 yn-2 +4 yn-1 + yn) where [a, b] is partitioned into an even number n subintervals of equal length h =(b –a)/n.

Using Simpson’s Rule Use Simpson’s rule with n = 4 to estimate h = (2 – 1)/4 = ¼, so S = 1/12 (1 + 4(25/16) + 2(36/16) + 4(49/16) + 4) = 7/3