1. Section 4.3 – Area and Definite Integrals
As the number of rectangles increased, the approximation of the area under the curve approaches a value.
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2. Section 4.3 – Area and Definite Integrals
Definition
If a continuous function, f(x), has an antiderivative, F(x), on the interval [a, b], then
lim 𝑛→∞ 𝑖=1 ∞ 𝑓 𝑥 𝑖 ∆𝑥= 𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹(𝑏)−𝐹(𝑏)
Parts of the Definite Integral

3. Section 4.3 – Area and Definite Integrals
Total Area
The total area under the curve will be positive.
The total area under the curve will be negative.
The total area under the curve will be zero.

4. Section 4.3 – Area and Definite Integrals
The Fundamental Theorem of Calculus
𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑒𝑣𝑒𝑟𝑦 𝑝𝑜𝑖𝑛𝑡 𝑖𝑛 𝑎, 𝑏 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑎𝑛𝑦
𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓 𝑜𝑛 𝑎, 𝑏 , 𝑡ℎ𝑒𝑛
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
Examples
2 6 5𝑥 𝑑𝑥 =
5 𝑥 = 6 2
5(6) 2 2
− =
90−10= 80
𝑥 𝑑𝑥 =
𝑥 𝑑𝑥 = 5 1
2 ln 𝑥
=2 ln 5
−2 ln 1 =
3.2189

5. Section 4.3 – Area and Definite Integrals
The Fundamental Theorem of Calculus
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
3 4 𝑥 2 +3𝑥−2 𝑑𝑥 =
3 4 𝑥 2 𝑑𝑥 𝑥 𝑑𝑥 − 𝑑𝑥 = 4 3
1 3 𝑥 3
𝑥 2
−2𝑥
1 3 (4) −2(4)
− −2 3 =
−16.5=

6. Section 4.3 – Area and Definite Integrals
The Fundamental Theorem of Calculus
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) . 32
𝑥 𝑑𝑥 =
𝑥 −1 5 − = 32 1
1 32 𝑥 −6 5 𝑑𝑥 =
−5 𝑥 1 5 1
−5 (32) 1 5
− − =
− 5 2 − − 5 1 =
2.5

7. Section 4.3 – Area and Definite Integrals
The Fundamental Theorem of Calculus
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
− 𝑒 6𝑥 𝑑𝑥 =
4 − 𝑒 6𝑥 𝑑𝑥 =
1 3
− 1 2
1 3
− 1 2
𝑒 6𝑥
= 2 3 𝑒 6𝑥
= 2 3 𝑒
− 2 3 𝑒 6 −1 2
= 2 3 𝑒 2
− 2 3 𝑒 −3
=4.8928

8. Section 4.3 – Area and Definite Integrals
The Fundamental Theorem of Calculus
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
1 2 𝑥 4 +6 𝑥 𝑑𝑥 = 2 1 2 1
𝑥 𝑑𝑥 =
𝑥 𝑥 𝑥
= 𝑥 𝑥 3 +9𝑥
(2) (2) 3 +9(2)
− (1) (1) 3 +9(1)
40.4−11.2=
29.2

9. Section 4.3 – Area and Definite Integrals
The Fundamental Theorem of Calculus
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
𝑥 2 −3𝑥 𝑥 2 𝑑𝑥 =
2 4 5− 3 𝑥 𝑑𝑥 = 4
5𝑥− 3ln x = 2
5(4)− 3ln (4)
− 5 2 − 3ln 2 =
−7.9206=
7.9206

10. Section 4.3 – Area and Definite Integrals
The Fundamental Theorem of Calculus
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
1 2 𝑒 2𝑥 𝑒 4𝑥 + 𝑒 −5𝑥 𝑑𝑥 =
0 1 𝑒 6𝑥 + 𝑒 −3𝑥 𝑑𝑥 = 1
1 6 𝑥 6 − 1 3 𝑒 −3𝑥
1 6 𝑒 6(1) − 1 3 𝑒 −3(1)
− 𝑒 6(0) − 1 3 𝑒 −3(0)
−(−0.1667)=