1. Fundamental Theorem of Calculus
Lesson 5-3a
Fundamental Theorem of Calculus

2. Ice Breaker
A stone was dropped off a cliff and hit the ground with a speed of 120 ft/s. What is the height of the cliff?
A(t) = -16 ft/s²
V(t) = A(t) = dt = t + V0 ∫
Since the stone was dropped: 0 = V0
V(th) = 120 = -16th th = 7.5
S(t) = V(t) = (-16t) dt = t² + S0 ∫
S(7.5) = -8(7.5)² + S0
S0 = 450

3. Objectives
Understand both forms of the Fundamental Theorem of Calculus
Understand the relation between integration and differentiation

4. Vocabulary
Definite Integral – is a number, not a function

5. Differentiation vs Definite Integration∆x ∆x ∆x ∆x ∆y
Secant Line ∆y
Tangent Line ∆y ∆y
Area of
rectangle
Area under the curve a b a b a b a b
∆y f(b) – f(a)
slope = =
∆x b – a
area = ∆y ∆x
= f(b) (b-a) ∆y
slope ≈ ∆x
area ≈ ∆y ∆x b ∆y
lim = mt ∆x
∆x→0 ∫
= f’(a)
lim ∆y ∆x = area = f(x) dx
∆x→0 a

6. Fundamental Theorem of Calculus, Part 2
If f is continuous on [a,b], then
Where F is any anti-derivative of f, that is a function such that F’ = f(x)
Your notes has a proof of this if you are interested.
This is just our area problem combined with the anti-derivatives we learned from 4-10. ∫ b a
f(x) dx = F(b) – F(a)

7. Three Important Observations
If you can find an anti-derivative, you can evaluate the definite integral without using Riemann Sums
Convenient Notation:
It is not necessary to include the integration constant, because it just cancels out
f(x) dx = F(b) – F(a) ∫ a b

8. ∫ ∫ ∫ Practice Problems 1) (x² + 3) dx ⅓x³ + 3x |2
⅓x³ + 3x |
[⅓2³ + 3(2)] – [⅓1³ + 3(1)]
8/3 + 6 – 1/3 – 3 = 5 ⅓
x=1
x=2
2) (3/2)(x) dx ∫ 1 4
(3/2)(2/3)x3/2 |
[43/2] – [13/2]
8 – 1 = 7
x=1
x=4
3) sec²(x) dx ∫
π/4
tan (x) |
[tan(π/4)] – [tan(0)]
1 – 0 = 1
x=0
x=π/4

9. ∫ ∫ Practice Problems 4) |2x – 3| dx = -(2x – 3) dx + (2x – 3) dx =y x 5 3 -1 2
4) |2x – 3| dx = ∫
-(2x – 3) dx + (2x – 3) dx = ∫
3/2 2
-(x² - 3x) | (x² - 3x) |
x=0
x=3/2
x=2
-[(3/2)² - 3(3/2)] – -[0² - 3(0)] + [(2)² - 3(2)] – [(3/2)² - 3(3/2)]
-[9/4 - 9/2]– -[0² - 3(0)] + [4 - 6] – [9/4 - 9/2]
9/ – (-9/4) = 2 ¼ + 1/4 = 2.5

10. Example 5
Find the area of the region bounded by y = 2x2 – 3x + 2, the x-axis and the lines x = 0 and x = 2. y x
(2x² – 3x + 2) dx = ∫ 2
(⅔x³ – (3/2)x² + 2x) | =
x =0
x =2
(⅔(2)³ – (3/2)2² + 2(2)) - (⅔(0)³ – (3/2)0² + 2(0)) =
(16/3 – 6 + 4) (0) = 10/3

11. Summary & Homework Summary: Homework: Definite Integrals are a number
Evaluated at endpoints of integration
Homework:
Day One: pg : 19, 22, 27, 28,
Day Two: pg : 3, 7, 9, 61 (see appendix E)

12. Ice Breaker
Police investigating an accident measured skid marks of 200 ft before the car came to a stop. From the manufacturer they found that the brakes had a constant deceleration of 16 ft/s². How fast was the car traveling when the brakes were first applied?
A(t) = -16 ft/s² S0 = 0
V(t) = A(t) = dt = t + V0
V(t) = 0 = -16t + V ts = V0
16ts = 80 ft/s = V0
S(t) = V(t) = (-16t + V0) dt = t² + V0 + S0
S(ts) = 200 = -8ts² + (16ts)ts
200 = 8t² t = 5