1. If the derivative of F is f. We call F an “antiderivative” of f.
Lesson: _____ Section 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.”

2. 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

3. 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

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

6. 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.

7. 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= 𝑎𝑛𝑑 10
𝑓 ′ 𝑥 𝑑𝑥
= =𝟐𝟎𝟎
𝑓 𝟐𝟎 =𝑓(10) 𝑓 ′ 𝑥 𝑑𝑥= =𝟑𝟎𝟎
𝑓 𝟐𝟓 =𝑓(20) 𝑓 ′ 𝑥 𝑑𝑥=300− 1 2 (5)(10)=𝟐𝟕𝟓
𝑓 𝟑𝟎 =𝑓(25) 𝑓 ′ 𝑥 𝑑𝑥=275− 1 2 (5)(10)=𝟐𝟓𝟎

8. 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)

9. 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

10. 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

11. Handy Reference

12. Handy Reference