4-4: Fun Theorems, 2 Objectives: Assignment: 1. To find the average value • P. 291-292: 47-50, 54-60, of a function 63 2. To understand and use the Second Fundamental • P. 293: 75-91 odd Theorem of Calculus Warm Up Find the value of π guaranteed by the MVT for integrals for π(π₯) = 1 + π₯ 2 on the interval −1,2 . Mean Value Theorem (Integrals) If π is a continuous function on π, π , then there exists a number π in π, π such that π π(π₯) ππ₯ = π π π − π π Mean Value Rectangle There exists a rectangle whose base is the width of the interval, whose height is a function value and whose area is equal to the area under the curve. Objective 1 You will be able to find the average value of a function Average Value If π is integrable on π, π , then the average value of π on π, π is 1 π−π π π π₯ ππ₯ π Average Value If π is integrable on π, π , then the average value of π on π, π is 1 π−π π π π₯ ππ₯ π Average Value Partition π, π such that βπ₯ = value in the πth subinterval. π−π , π where π₯π∗ is any π π₯1∗ + π π₯2∗ + π π₯3∗ + β― + π π₯π∗ Average = π Almost a Riemann sum 1 = π π π π−π π−π π₯π∗ π=1 1 = π−π π π π=1 π₯π∗ π−π π Take the limit as π → ∞ = 1 π−π 1 = π−π π π π₯π∗ βπ₯ π=1 π π π₯ ππ₯ π Exercise 1 Find the average value of π π₯ = 3π₯ 2 − 2π₯ on 1,4 . Exercise 2 If the function π π₯ = π₯ 3 has an average value of 9 on the closed interval 0, π , then what is the value of π? Exercise 3 Show that the average velocity of a car over a time interval π‘1 , π‘2 is equal to the average of value of its velocity function. The average value of a function π is equivalent to the slope of the secant line of the antiderivative of π. Objective 2 You will be able to understand and use the Second Fundamental Theorem of Calculus Definite Versus Indefinite Remember that a definite integral yields a number; however… Definite Integral Indefinite Integral Number Family of functions Net area above/below π Antiderivative of π′ Definite Versus Indefinite Remember that a definite integral yields a number; however… Definite Integral Number Net area above/below π There is a way to define a function based on a definite integral Exercise 4 Evaluate πΉ π₯ = your answer. π₯ 0 π‘ + 1 ππ‘ and interpret π π‘ =π‘+1 πΉ(π₯) is an accumulation function: πΉ 2 =4 Area over 0,2 = 4 πΉ 4 = 12 Area over 0,4 = 12 πΉ(π₯) Number Versus Function Remember that a definite integral yields a number; however… Definite Integral as a Number: π π π₯ ππ₯ Variable Definite Integral as a Function: π₯ πΉ π₯ = π π‘ ππ‘ π π Constant Constants Function of π₯ Function of π‘ Exercise 5 Evaluate πΉ π₯ = and π/2. π₯ cos π‘ ππ‘ 0 at π₯ = 0, π/6, π/4, π/3, Now, what do you get if you take the derivative of πΉ(π₯)? 2nd Fundamental Theorem of Calculus If π is continuous on and open interval containing π, then for every π₯ in the interval Insert Proof Here π ππ₯ π₯ π π‘ ππ‘ = π π₯ π Exercise 6 Evaluate π ππ₯ π₯ 0 π‘ 2 + 1 ππ‘ Exercise 7 Find the derivative of πΉ π₯ = π₯3 cos π‘ ππ‘. π/2 4-4: Fun Theorems, 2 Objectives: Assignment: 1. To find the average value • P. 291-292: 47-50, 54-60, of a function 63 2. To understand and use the Second Fundamental • P. 293: 75-91 odd Theorem of Calculus