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5.3 β The Definite Integral and the Fundamental Theorem of Calculus
Math 140 5.3 β The Definite Integral and the Fundamental Theorem of Calculus
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Q: How can we find the area under any curve
Q: How can we find the area under any curve? For example, letβs try to find the area under π π₯ = π₯ 3 β9 π₯ 2 +25π₯β19
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What is the area of this?
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1 rectangle
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2 rectangles
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4 rectangles
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50 rectangles
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If the left sides of our rectangles are at π₯ 1 , π₯ 2 , β¦, π₯ π and each rectangle has width Ξπ₯, then the areas of the rectangles are π π₯ 1 Ξπ₯, π π₯ 2 Ξπ₯, β¦, π π₯ π Ξπ₯. So, the sum of the areas of the rectangles (called a Riemann sum) is: π π₯ 1 Ξπ₯+π π₯ 2 Ξπ₯+β¦+π π₯ π Ξπ₯ = π π₯ 1 +π π₯ 2 +β¦+π π₯ π Ξπ₯
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So, the sum of the areas of the rectangles (called a Riemann sum) is: π π₯ 1 Ξπ₯+π π₯ 2 Ξπ₯+β¦+π π₯ π Ξπ₯ = π π₯ 1 +π π₯ 2 +β¦+π π₯ π Ξπ₯
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So, the sum of the areas of the rectangles (called a Riemann sum) is: π π₯ 1 Ξπ₯+π π₯ 2 Ξπ₯+β¦+π π₯ π Ξπ₯ = π π₯ 1 +π π₯ 2 +β¦+π π₯ π Ξπ₯
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As the number of rectangles (π) increases, the above sum gets closer to the true area. That is, Area under curve= πππ π§β+β π π π +π π π +β¦+π π π ππ
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π π π π₯ ππ₯ = lim nβ+β π π₯ 1 +π π₯ 2 +β¦+π π₯ π Ξπ₯ The above integral is called a definite integral, and π and π are called the lower and upper limits of integration.
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The Fundamental Theorem of Calculus
If π(π₯) is continuous on the interval πβ€π₯β€π, then π π π π π
π =π π βπ(π) where πΉ(π₯) is any antiderivative of π(π₯) on πβ€π₯β€π.
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Ex 1. Evaluate the given definite integrals using the fundamental theorem of calculus. β1 1 2 π₯ 2 ππ₯
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Ex 1. Evaluate the given definite integrals using the fundamental theorem of calculus. β1 1 2 π₯ 2 ππ₯
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Ex 1 (cont). Evaluate the given definite integrals using the fundamental theorem of calculus. 0 1 ( π 2π₯ β π₯ ) ππ₯
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Ex 1 (cont). Evaluate the given definite integrals using the fundamental theorem of calculus. 0 1 ( π 2π₯ β π₯ ) ππ₯
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Note: Definite integrals can evaluate to negative numbers
Note: Definite integrals can evaluate to negative numbers. For example, 1 2 (1βπ₯) ππ₯ =β 1 2
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Just like with indefinite integrals, terms can be integrated separately, and constants can be pulled out. π π π π₯ Β±π(π₯) ππ₯ = π π π π₯ ππ₯ Β± π π π(π₯) ππ₯ π π ππ π₯ ππ₯ =π π π π π₯ ππ₯ Also, π π π π₯ ππ₯ =0 π π π π₯ ππ₯ =β π π π π₯ ππ₯
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π π π π₯ ππ₯ = π π π π₯ ππ₯ + π π π π₯ ππ₯ (here, π is any real number)
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π π π π₯ ππ₯ = π π π π₯ ππ₯ + π π π π₯ ππ₯ (here, π is any real number)
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π π π π₯ ππ₯ = π π π π₯ ππ₯ + π π π π₯ ππ₯ (here, π is any real number)
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Ex 2. Let π(π₯) and π(π₯) be functions that are continuous on the interval β3β€π₯β€5 and that satisfy β3 5 π π₯ ππ₯ =2 β3 5 π π₯ ππ₯ =7 1 5 π π₯ ππ₯ =β8 Use this information along with the rules definite integrals to evaluate the following. β3 5 2π π₯ β3π π₯ ππ₯ β3 1 π π₯ ππ₯
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Ex 3. Evaluate: 0 1 8π₯ π₯ ππ₯
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Ex 3. Evaluate: 0 1 8π₯ π₯ ππ₯
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Net Change If we have πβ²(π₯), then we can calculate the net change in π(π₯) as π₯ goes from π to π with an integral: π π βπ π = π π π β² π₯ ππ₯
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Net Change Ex 4. Suppose marginal cost is πΆ β² π₯ =3 π₯β4 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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