Current circulation is 3000 per month. Growth rate is C’(t)=. Find C(t). C(t) = C(0) = c = C(t) =

Slides:

Advertisements

Similar presentations

Applying the well known formula:

Advertisements

Objective:To use the Fundamental Theorem of Calculus to evaluate definite integrals of polynomial functions. To find indefinite integrals of polynomial.

APPLICATIONS OF INTEGRATION

7.1 Area of a Region Between Two Curves.

Area Between Two Curves

Area Between Two Curves 7.1. Area Formula If f and g are continuous functions on the interval [a, b], and if f(x) > g(x) for all x in [a, b], then the.

1 Fundamental Theorem of Calculus Section The Fundamental Theorem of Calculus If a function f is continuous on the closed interval [a, b] and F.

Area Between Two Curves

Definite Integrals Finding areas using the Fundamental Theorem of Calculus.

Chapter 5 Key Concept: The Definite Integral

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall 5.4 Fundamental Theorem of Calculus.

4.6 Numerical Integration -Trapezoidal Rule -Simpson’s Rule

APPLICATIONS OF INTEGRATION 6. A= Area between f and g Summary In general If.

Chapter 5 .3 Riemann Sums and Definite Integrals

7 Applications of Integration

Section 5.3 – The Definite Integral

Section 7.2a Area between curves.

6.3 Definite Integrals and the Fundamental Theorem.

Constructing the Antiderivative Solving (Simple) Differential Equations The Fundamental Theorem of Calculus (Part 2) Chapter 6: Calculus~ Hughes-Hallett.

7.4: The Fundamental Theorem of Calculus Objectives: To use the FTC to evaluate definite integrals To calculate total area under a curve using FTC and.

4-4: The Fundamental Theorems Definition: If f is continuous on [ a,b ] and F is an antiderivative of f on [ a,b ], then: The Fundamental Theorem:

Announcements Topics: -sections 7.3 (definite integrals) and 7.4 (FTC) * Read these sections and study solved examples in your textbook! Work On: -Practice.

Section 4.4 The Fundamental Theorem of Calculus Part II – The Second Fundamental Theorem.

The Fundamental Theorems of Calculus Lesson 5.4. First Fundamental Theorem of Calculus Given f is  continuous on interval [a, b]  F is any function.

Lecture 18 More on the Fundamental Theorem of Calculus.

Warm Up – NO CALCULATOR Let f(x) = x2 – 2x.

Mathematics. Session Definite Integrals –1 Session Objectives  Fundamental Theorem of Integral Calculus  Evaluation of Definite Integrals by Substitution.

 Constructing the Antiderivative  Solving (Simple) Differential Equations  The Fundamental Theorem of Calculus (Part 2) Chapter 6: Calculus~ Hughes-

CHAPTER Continuity Fundamental Theorem of Calculus In this lecture you will learn the most important relation between derivatives and areas (definite.

P roblem of the Day - Calculator Let f be the function given by f(x) = 3e 3x and let g be the function given by g(x) = 6x 3. At what value of x do the.

Antidifferentiation: The Indefinite Intergral Chapter Five.

MAT 212 Brief Calculus Section 5.4 The Definite Integral.

Integration and Area Foundation- Definite Integrals.

The Fundamental Theorem of Calculus

Application: Area under and between Curve(s) Volume Generated

Right minus left. y + 1 = 0.5 y 2 – 3 0 = 0.5 y 2 – y – 4 0 = y 2 – 2y – 8 =(y – 4)(y + 2)

5.3 – The Fundamental Theorem of Calculus

How tall am I? My head is 232 feet above sea level. My feet are 226 feet above sea level. My head is 2 feet above sea level. My feet are (-4) feet ‘above’

Warm up Problems More With Integrals It can be helpful to guess and adjust Ex.

Properties of Integrals. Mika Seppälä: Properties of Integrals Basic Properties of Integrals Through this section we assume that all functions are continuous.

6.1 Areas Between Curves In this section we learn about: Using integrals to find areas of regions that lie between the graphs of two functions. APPLICATIONS.

Applications of Integration 7 Copyright © Cengage Learning. All rights reserved.

What is the volume if the grey area is revolved about the x-axis?

Announcements Topics: -sections 7.3 (definite integrals) and 7.4 (FTC) * Read these sections and study solved examples in your textbook! Work On: -Practice.

4-3: Riemann Sums & Definite Integrals Objectives: Understand the connection between a Riemann Sum and a definite integral Learn properties of definite.

5-7: The 1 st Fundamental Theorem & Definite Integrals Objectives: Understand and apply the 1 st Fundamental Theorem ©2003 Roy L. Gover

Copyright (c) 2003 Brooks/Cole, a division of Thomson Learning, Inc. Integration 5 Antiderivatives Substitution Area Definite Integrals Applications.

Definite Integrals, The Fundamental Theorem of Calculus Parts 1 and 2 And the Mean Value Theorem for Integrals.

Announcements Topics: -sections 7.3 (definite integrals) and 7.4 (FTC) * Read these sections and study solved examples in your textbook! Work On: -Practice.

Section 4.4 The Fundamental Theorem of Calculus. We have two MAJOR ideas to examine in this section of the text. We have been hinting for awhile, sometimes.

DO NOW: v(t) = e sint cost, 0 ≤t≤2∏ (a) Determine when the particle is moving to the right, to the left, and stopped. (b) Find the particles displacement.

The Fundamental Theorem of Calculus Area and The Definite Integral OBJECTIVES  Evaluate a definite integral.  Find the area under a curve over a given.

In this chapter, we explore some of the applications of the definite integral by using it to compute areas between curves, volumes of solids, and the work.

Area Between Two Curves Calculus. Think WAY back…

Area Between Curves. Objective To find the area of a region between two curves using integration.

MTH1170 Integrals as Area.

Integration Chapter 15.

6.6 Area Between Two Curves

Area of a Region Between 2 Curves

2.4 Quadratic Models.

Ch. 6 – The Definite Integral

Ch. 6 – The Definite Integral

Section 4.3 – Area and Definite Integrals

Riemann Sums and Integrals

Area & Volume Chapter 6.1 & 6.2 February 20, 2007.

Chapter 6 The Definite Integral

The Fundamental Theorems of Calculus

2. Area Between Curves.

(Finding area using integration)

Presentation transcript:

Current circulation is 3000 per month. Growth rate is C’(t)=. Find C(t). C(t) = C(0) = c = C(t) =

Fundamental Theorem of Calculus If f is continuous on [a, b] and F is any antiderivative of f, then

How tall am I? My head is 232 feet above sea level. My feet are 226 feet above sea level. My head is 2 feet above sea level. My feet are (-4) feet ‘above’ sea level.

How many feet tall am I?

What is the area of the green rectangle?

What is the area below f(x), above g(x) and between x=a and x=b ?

Set up n rectangles of width  x And height is top – bottom or f(x) – g(x)

The area of one rectangle is height times width

By the definition of the definite integral Area =

Example 1 Find the area over y=(2x-2) 2 and under y=5 Between x=0 and x=2 Just add up all of the red rectangles As they slide from x=0 to x=2 The top function is... Y=5 And the bottom function is... Y=(2x-2) 2

Example 1 Find the area over y=(2x-2) 2 and under y=5 Between x=0 and x=2 =5x-1/2 (2x-2) 3 /3 = /3-[ (-8/3)]

[5x- 0.5 (2x-2) 3 /3] [ (8/3)] - [ (-8/3)]=

Example 2 Set the two functons equal to each other Solve for x x 2 = x 3 or 0 = x 3 - x 2 By factoring 0 = x 2 ( x – 1 ) so x 2 =0 or x–1=0 Next we add up all of the red rectangles From 0 to 1 Area =

. A.. B.. C...

Area = = 1/12

The area over y=2 and under y=x 2 +3 between x=-1 and x=1 A. [ B. [ C. [

. A. [ B. [ C. [

]

Find the enclosed area. y = x = -2 x = 2 Area = =

Find the enclosed area. Area = = -ln + ln = - ln(4) + ln(8) + ln(8) – ln(4) = ln(8/4) + ln (8/4) = 2ln(2)= ln2 2 = ln(4) = sq. ft.

Find the enclosed area. Area = ln(4) = sq. ft. = ln(8) - ln(8) = 0

Example 4 Find the area bounded by y 2 = 2x + 6 and y = x – 1. Solve for x = 0.5 y 2 – 3 and we see a parabola opening to the right with the vertex at (-3, 0). Replacing y by (x-1) gives (x-1) 2 =2x + 6 x 2 - 2x + 1 = 2x + 6 x 2 - 4x – 5 = (x - 5)(x + 1) = 0 x = 5 or x = -1

Example 4 Find the area bounded by y 2 = 2x + 6 and y = x – 1. vertex at (-3, 0). x = 5 or x = -1 Area = =

. 16/3 – 0 + 1/3 (64) –[8/3-1/2-1] 72/3 – = 24 – 6 = 18

X 2 - 2x + 1 = 2x + 6 X 2 - 4x – 5 = (x - 5)(x + 1) = 0 X = 5 or x = -1 y = x – 1 so y = 4 or y = -2.

x = y + 1. x = 0.5 y 2 – 3.