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5.3 – The Definite Integral and the Fundamental Theorem of Calculus

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1 5.3 – The Definite Integral and the Fundamental Theorem of Calculus
Math 140 5.3 – The Definite Integral and the Fundamental Theorem of Calculus

2 Q: How can we find the area under any curve
Q: How can we find the area under any curve? For example, let’s try to find the area under 𝑓 π‘₯ = π‘₯ 3 βˆ’9 π‘₯ 2 +25π‘₯βˆ’19

3 What is the area of this?

4 1 rectangle

5 2 rectangles

6 4 rectangles

7 50 rectangles

8 If the left sides of our rectangles are at π‘₯ 1 , π‘₯ 2 , …, π‘₯ 𝑛 and each rectangle has width Ξ”π‘₯, then the areas of the rectangles are 𝑓 π‘₯ 1 Ξ”π‘₯, 𝑓 π‘₯ 2 Ξ”π‘₯, …, 𝑓 π‘₯ 𝑛 Ξ”π‘₯. So, the sum of the areas of the rectangles (called a Riemann sum) is: 𝑓 π‘₯ 1 Ξ”π‘₯+𝑓 π‘₯ 2 Ξ”π‘₯+…+𝑓 π‘₯ 𝑛 Ξ”π‘₯ = 𝑓 π‘₯ 1 +𝑓 π‘₯ 2 +…+𝑓 π‘₯ 𝑛 Ξ”π‘₯

9 So, the sum of the areas of the rectangles (called a Riemann sum) is: 𝑓 π‘₯ 1 Ξ”π‘₯+𝑓 π‘₯ 2 Ξ”π‘₯+…+𝑓 π‘₯ 𝑛 Ξ”π‘₯ = 𝑓 π‘₯ 1 +𝑓 π‘₯ 2 +…+𝑓 π‘₯ 𝑛 Ξ”π‘₯

10 So, the sum of the areas of the rectangles (called a Riemann sum) is: 𝑓 π‘₯ 1 Ξ”π‘₯+𝑓 π‘₯ 2 Ξ”π‘₯+…+𝑓 π‘₯ 𝑛 Ξ”π‘₯ = 𝑓 π‘₯ 1 +𝑓 π‘₯ 2 +…+𝑓 π‘₯ 𝑛 Ξ”π‘₯

11 As the number of rectangles (𝑛) increases, the above sum gets closer to the true area. That is, Area under curve= π’π’Šπ’Ž 𝐧→+∞ 𝒇 𝒙 𝟏 +𝒇 𝒙 𝟐 +…+𝒇 𝒙 𝒏 πœŸπ’™

12 π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = lim nβ†’+∞ 𝑓 π‘₯ 1 +𝑓 π‘₯ 2 +…+𝑓 π‘₯ 𝑛 Ξ”π‘₯ The above integral is called a definite integral, and π‘Ž and 𝑏 are called the lower and upper limits of integration.

13 The Fundamental Theorem of Calculus
If 𝑓(π‘₯) is continuous on the interval π‘Žβ‰€π‘₯≀𝑏, then 𝒂 𝒃 𝒇 𝒙 𝒅𝒙 =𝑭 𝒃 βˆ’π‘­(𝒂) where 𝐹(π‘₯) is any antiderivative of 𝑓(π‘₯) on π‘Žβ‰€π‘₯≀𝑏.

14 Ex 1. Evaluate the given definite integrals using the fundamental theorem of calculus. βˆ’1 1 2 π‘₯ 2 𝑑π‘₯

15 Ex 1. Evaluate the given definite integrals using the fundamental theorem of calculus. βˆ’1 1 2 π‘₯ 2 𝑑π‘₯

16 Ex 1 (cont). Evaluate the given definite integrals using the fundamental theorem of calculus. 0 1 ( 𝑒 2π‘₯ βˆ’ π‘₯ ) 𝑑π‘₯

17 Ex 1 (cont). Evaluate the given definite integrals using the fundamental theorem of calculus. 0 1 ( 𝑒 2π‘₯ βˆ’ π‘₯ ) 𝑑π‘₯

18 Note: Definite integrals can evaluate to negative numbers
Note: Definite integrals can evaluate to negative numbers. For example, 1 2 (1βˆ’π‘₯) 𝑑π‘₯ =βˆ’ 1 2

19 Just like with indefinite integrals, terms can be integrated separately, and constants can be pulled out. π‘Ž 𝑏 𝑓 π‘₯ ±𝑔(π‘₯) 𝑑π‘₯ = π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ Β± π‘Ž 𝑏 𝑔(π‘₯) 𝑑π‘₯ π‘Ž 𝑏 π‘˜π‘“ π‘₯ 𝑑π‘₯ =π‘˜ π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ Also, π‘Ž π‘Ž 𝑓 π‘₯ 𝑑π‘₯ =0 𝑏 π‘Ž 𝑓 π‘₯ 𝑑π‘₯ =βˆ’ π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯

20 π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = π‘Ž 𝑐 𝑓 π‘₯ 𝑑π‘₯ + 𝑐 𝑏 𝑓 π‘₯ 𝑑π‘₯ (here, 𝑐 is any real number)

21 π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = π‘Ž 𝑐 𝑓 π‘₯ 𝑑π‘₯ + 𝑐 𝑏 𝑓 π‘₯ 𝑑π‘₯ (here, 𝑐 is any real number)

22 π‘Ž 𝑏 𝑓 π‘₯ 𝑑π‘₯ = π‘Ž 𝑐 𝑓 π‘₯ 𝑑π‘₯ + 𝑐 𝑏 𝑓 π‘₯ 𝑑π‘₯ (here, 𝑐 is any real number)

23 Ex 2. Let 𝑓(π‘₯) and 𝑔(π‘₯) be functions that are continuous on the interval βˆ’3≀π‘₯≀5 and that satisfy βˆ’3 5 𝑓 π‘₯ 𝑑π‘₯ =2 βˆ’3 5 𝑔 π‘₯ 𝑑π‘₯ =7 1 5 𝑓 π‘₯ 𝑑π‘₯ =βˆ’8 Use this information along with the rules definite integrals to evaluate the following. βˆ’3 5 2𝑓 π‘₯ βˆ’3𝑔 π‘₯ 𝑑π‘₯ βˆ’3 1 𝑓 π‘₯ 𝑑π‘₯

24 Ex 3. Evaluate: 0 1 8π‘₯ π‘₯ 𝑑π‘₯

25 Ex 3. Evaluate: 0 1 8π‘₯ π‘₯ 𝑑π‘₯

26 Net Change If we have 𝑄′(π‘₯), then we can calculate the net change in 𝑄(π‘₯) as π‘₯ goes from π‘Ž to 𝑏 with an integral: 𝑄 𝑏 βˆ’π‘„ π‘Ž = π‘Ž 𝑏 𝑄 β€² π‘₯ 𝑑π‘₯

27 Net Change Ex 4. Suppose marginal cost is 𝐢 β€² π‘₯ =3 π‘₯βˆ’4 2 dollars per unit when the level of production is π‘₯ units. By how much will the total manufacturing cost increase if the level of production is raised from 6 units to 10 units?


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