If the derivative of F is f. We call F an “antiderivative” of f. Lesson: _____ Section 6.1 Constructing Antiderivatives Graphically & Numerically If the derivative of F is f. We call F an “antiderivative” of f. Ex. x2 is an antiderivative of 2x. Is it the only one? NO! 𝑥 2 +1, 𝑥 2 +2, 𝑥 2 +3, 𝑒𝑡𝑐. So any function 𝑥 2 +𝑐 is called an antiderivative of 2x. We say that f(x) = 2x has a “family of antiderivatives.”

Review: f f’ f  f’ Incr. Positive Decr. Negative Concave Up Incr. What does the derivative tell us about the antiderivative (original function)? f f’ Positive Negative Zero (with sign change) f  f’ f’ is the derivative of f f is the antiderivative of f’ Incr. Decr. Max/Min Concave Up Concave Down Inflection Pt. Incr. Decr. Max/Min

Ex1. Visualizing Antiderivatives using Slopes Given a graph of f’, sketch an approximate graph for f. f’ >0, then f is increasing! f’<0, then f is decreasing! f’ is increasing, then f is concave up! f’ is decreasing, then f is concave down! f(x) f’(x) 1 1

Ex1. Given a graph of f’, sketch an approximate graph for f. f(x) 2 2 f’(x) 1 1

Using the Fundamental Theorem of Calculus to find actual points on the graph of an antiderivative. 𝑎 𝑏 𝑓 ′ 𝑥 𝑑𝑥=𝑓 𝑏 −𝑓(𝑎) The integral tells us how the value has changed. If I know some initial point, I can use integrals to find the rest of the points.

Ex.4) Using the graph of f’(x) and given that f(0)=100, sketch a graph of f(x). Identify the coordinates of all critical & inflection points on f(x). Critical points Inflection points at x = 0, 20, 30 [ f’(x)=0 ] at x = 10, 25 [ max/min for f’(x)] 𝑓 𝟎 =𝟏𝟎𝟎 𝑓 𝟏𝟎 =𝑓(0)+𝐶ℎ𝑎𝑛𝑔𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛= 0 𝑎𝑛𝑑 10 100+ 0 10 𝑓 ′ 𝑥 𝑑𝑥 =100+ 1 2 10 20 =𝟐𝟎𝟎 𝑓 𝟐𝟎 =𝑓(10)+ 10 20 𝑓 ′ 𝑥 𝑑𝑥=200+100=𝟑𝟎𝟎 𝑓 𝟐𝟓 =𝑓(20)+ 20 25 𝑓 ′ 𝑥 𝑑𝑥=300− 1 2 (5)(10)=𝟐𝟕𝟓 𝑓 𝟑𝟎 =𝑓(25)+ 25 30 𝑓 ′ 𝑥 𝑑𝑥=275− 1 2 (5)(10)=𝟐𝟓𝟎

Ex.4) Using the graph of f’(x) and given that f(0)=100, sketch a graph of f(x). Identify the coordinates of all critical & inflection points on f(x). Critical points Inflection points (0,100), (20,300), (30,250) (10,200), (25,275) 200 300 100 10 20 30 x f(x)

𝐹 𝑏 −𝐹 0 = 0 𝑏 𝐹 ′ 𝑡 𝑑𝑡 𝐹 𝑏 =𝐹(0)+ 0 𝑏 𝐹 ′ 𝑡 𝑑𝑡 𝐹 𝑏 =2+ 0 𝑏 𝑡𝑐𝑜𝑠𝑡 𝑑𝑡 Ex.5) Suppose 𝑭 ′ 𝒕 =𝒕𝒄𝒐𝒔𝒕 𝐚𝐧𝐝 𝑭 𝟎 =𝟐. Find 𝑭 𝒃 𝐚𝐭 𝐭𝐡𝐞 𝐩𝐨𝐢𝐧𝐭𝐬 𝒃=𝟎,𝟎.𝟏,𝟎.𝟐,…, 𝟏 The FTC tells us that… 𝐹 𝑏 −𝐹 0 = 0 𝑏 𝐹 ′ 𝑡 𝑑𝑡 𝐹 𝑏 =𝐹(0)+ 0 𝑏 𝐹 ′ 𝑡 𝑑𝑡 𝐹 𝑏 =2+ 0 𝑏 𝑡𝑐𝑜𝑠𝑡 𝑑𝑡 𝟐+𝐟𝐧𝐈𝐧𝐭(𝒙𝒄𝒐𝒔𝒙, 𝒙,𝟎,𝟎.𝟏) 𝑭 𝟎 =𝟐 𝑭′ 𝒕 =𝒕𝒄𝒐𝒔𝒕 b 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 F(b) 2 2.005 2.020 2.044 2.077 2.117 2.164 2.261 2.271 2.327 2.282

What does this represent? Not sure? Graph it with your calculator and see if you can figure it out. 𝐹 𝑥 =3+ 0 𝑥 2𝑡 𝑑𝑡

Handy Reference

Handy Reference