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Calculus AB Area Under A Curve

Section 6: Applications of Integrals: Lecture 2 | 18:34 min

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Lecture Comments (4)

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Post by mohammad salem on March 29, 2013

example 5 is wrong, the width should be pi/4.

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Post by GUOJI LIU on January 29, 2013

he did worry.

1 answer

Last reply by: Arshin Jain
Sun Feb 9, 2014 5:16 AM

Post by James Xie on July 23, 2012

for example 5, shouldn't the answer have pi/4 instead of pi/2? The x-value of each rectangle equals pi/4 not pi/2.

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Area Under A Curve

Approximating area under curve using rectangles

Setup appropriate intervals and evaluate f(x_i)
Sum all of f(x_i)

To avoid confusion, think physical, geometric area, rather than calculus

Area Under A Curve

Find the area under the curve f(x) = x on interval [0,5] using 5 rectangles and the right-endpoint

f(1) = 1
f(2) = 2
f(3) = 3
f(4) = 4
f(5) = 5
Area under the curve is approximately equal to [(5 − 0)/5](1 + 2 + 3 + 4 + 5)

Area is approximately equal to 15

Find the area under the curve f(x) = x on interval [0,5] using 5 rectangles and the left-endpoint

f(0) = 0
f(1) = 1
f(2) = 2
f(3) = 3
f(4) = 4
Area under the curve is approximately equal to [(5 − 0)/5](0 + 1 + 2 + 3 + 4)

Area under the curve is approximately equal to 10

Find the area under the curve f(x) = x² on interval [−2,2] using 4 rectangles and the right-endpoint

f(−1) = 1
f(0) = 0
f(1) = 1
f(2) = 4
Area under the curve is approximately equal to [(2 − (−2))/4](1 + 0 + 1 + 4)

Area under the curve is approximately equal to 6

Find the area under the curve f(x) = 3cosx on interval [−[(π)/2],[(π)/2]] using 4 rectangles and the right-endpoint

f(−[(π)/4]) = 3 cos−[(π)/4] = [3/(√2)]
f(0) = 3 cos0 = 3
f([(π)/4]) = 3 cos[(π)/4] = [3/(√2)]
f([(π)/2]) = 3 cos[(π)/2] = 0
Area under the curve is approximately equal to [([(π)/2] − (−[(π)/2]))/4]([3/(√2)] + 3 + [3/(√2)] + 0)

A ≈ [(π)/4](3 + [6/(√2)])

Find the area under the curve f(x) = x + 3 on interval [0,1] using 4 rectangles and the right-endpoint

Find the width of each rectangle
[(1 − 0)/4] = .25
f(.25) = 3.25
f(.5) = 3.5
f(.75) = 3.75
f(1) = 4
Area under the curve is approximately equal to .25(3.25 + 3.5 + 3.75 + 4)

A ≈ 3.625

Find the area under the curve f(x) = x + 3 on interval [0,1] using 10 rectangles and the right-endpoint

Find the width of each rectangle
[(1 − 0)/10] = .1
f(.1) = 3.1
f(.2) = 3.2
f(.3) = 3.3
f(.4) = 3.4
f(.5) = 3.5
f(.6) = 3.6
f(.7) = 3.7
f(.8) = 3.8
f(.9) = 3.9
f(1) = 4
Area under the curve is approximately equal to [(1 − 0)/10](3.1 + 3.2 + 3.3 + 3.4 + 3.5 + 3.6 + 3.7 + 3.8 + 3.9 + 4)

A ≈ 3.55

Find the area under the curve f(x) = x² + 3x + 1 on interval [−1,1] using 2 rectangles and the right-endpoint

[(1 − (−1))/2] = 1
f(0) = 1
f(1) = 1 + 3 + 1 = 5
Area under the curve is approximately 1(1 + 5)

A ≈ 6

Find the area under the curve f(x) = sin2x on interval [0, [(π)/2]] using 2 rectangles and the left-endpoint

[([(π)/2] − 0)/2] = [(π)/4]
f(0) = sin0 = 0
f([(π)/4]) = [1/(√2)]
Area under the curve is approximately equal to [(π)/4](0 + [1/(√2)])

A ≈ [(π)/(4√2)]

Find the area under the curve f(x) = sinx on interval [−[(π)/2], [(π)/2]] using 4 rectangles and the left-endpoint

[([(π)/2] − (−[(π)/2]))/4] = [(π)/4]
f(−[(π)/2]) = sin[(−π)/2] = −1
f(−[(π)/4]) = sin[(−π)/4] = −[1/(√2)]
f(0) = sin0 = 0
f([(π)/4]) = sin[(π)/4] = [1/(√2)]
Area under the curve is approximately equal to [(π)/4](−1 − [1/(√2)] + 0 + [1/(√2)])

A ≈ −[(π)/4]

Find the area under the curve f(x) = sinx on interval [−[(π)/2], [(π)/2]] using 4 rectangles and the right-endpoint

[([(π)/2] − (−[(π)/2]))/4] = [(π)/4]
f(−[(π)/4]) = sin[(−π)/4] = −[1/(√2)]
f(0) = sin0 = 0
f([(π)/4]) = sin[(π)/4] = [1/(√2)]
f([(π)/2]) = sin[(π)/2] = 1
Area under the curve is approximately equal to [(π)/4](− [1/(√2)] + 0 + [1/(√2)] + 1)

A ≈ [(π)/4]

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Answer

Area Under A Curve

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Intro

Area Under Curve

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Intro 0:00
Area Under Curve 0:07

Definition of Integral
Left Endpoint
Right Endpoint
Midpoints

Example 1 2:40
Example 2 4:59
Example 3 8:48
Example 4 10:23
Example 5 12:30
Example 6 15:32

Mathematics: AP Calculus AB

Section 1: Functions
	Definitions & Properties of Functions	11:26
	Graphing	13:58
	Inverse Functions	6:47
	Polynomial Functions	5:04
	Trigonometric Functions	6:45
	Inverse Trigonometric Functions	5:58
	Trigonometric Identities	17:42
	Exponential Functions	5:53
	Logarithmic Functions	7:08
	Rational Functions	15:36
	Conic Sections	14:58
Section 2: Limits and Continuity
	Limit Definition & Properties	7:15
	Solving Limits with Algebra	8:01
	Rational Limit Rules	3:16
	One Sided Limits	9:57
	Special Trigonometric Limits	8:28
	Limits & Continuity	10:14
	Limits: Multiple Choice Practice	6:16
Section 3: Derivatives
	Derivative Definition & Properties	4:11
	Basic Rules of Differentiation	7:07
	Power Rule	7:14
	Trigonometric Rules	7:53
	Product Rule	11:11
	Quotient Rule	16:50
	Chain Rule	19:48
	Higher Order Derivatives	15:00
	Derivatives of Exponential Functions	13:14
	Derivatives of Logarithmic Functions	11:30
	Derivatives of Inverse Trigonometric Functions	16:54
	Implicit Differentiation	16:53
	Multiple Choice Practice: Derivatives	11:07
Section 4: Applications of Derivatives
	Tangent & Normal Lines	22:36
	Position Velocity & Acceleration	18:42
	Related Rates	26:22
	Minimum & Maximum	12:22
	Concavity	11:43
	Rolles Theorem	8:28
	Mean Value Theorem	9:39
	Differentials	12:25
	Applications of Derivatives: Multiple Choice Practice	13:21
	Applications of Derivatives: Free Response Practice	10:22
Section 5: Integrals
	Definition of Integrals	1:08
	Integrals of Power Rule	8:50
	Integrals Basic Rules of Integration	9:43
	Trigonometric Rules of Integrals	8:58
	Chain Rule	13:59
	Integrals of Exponential Functions	12:52
	Integrals of Natural Logarithmic Functions	13:00
	Integrals of Inverse Trigonometric Functions	8:29
	Integrals: Multiple Choice Practice	15:37
Section 6: Applications of Integrals
	Fundamental Theorem of Calculus	15:55
	Area Under A Curve	18:34
	Reimann Sums	10:35
	Trapezoid Rule	12:46
	Mean Value Theorem	11:22
	Second Fundamental Theorem of Calculus	4:44
	Area Between Curves	16:39
	Revolving Solids Washer Disk Methods	21:09
	Revolving Solids Cylindrical Shells Method	26:46
	Revolving Solids Known Cross Sections	27:41
	Differential Equations Eulers Method	17:54
	Differential Equations Slope Fields	16:30
	Application of Integrals: Multiple Choice Practice	14:19
	Application of Integrals: Free Response Practice	9:14
Section 7: Sample AP Test
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 1	17:50
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 2	17:32
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 3	22:14
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 4	19:35
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 5	25:43
	AP Calculus AB Practice Test: Section 2: Free Response Part 1	16:50
	AP Calculus AB Practice Test: Section 2: Free Response Part 2	21:36
	AP Calculus AB Practice Test: Section 2: Free Response Part 3	13:39

Related Books

	Cracking the AP Calculus AB Exam, 2015 Edition Author: Princeton Review ISBN: 804124809 Publisher: Princeton Review Year: 2014 The book includes 3 full length practice tests with detailed explanations, a review of all the key concepts, and targeted strateeies to ace th exam for your highest score.
	Cliffs AP: Calculus AB & BC, 3rd Edition Authors: Dale W Johnson, Kerry J King ISBN: 764586831 Publisher: Cliffs Notes Year: 2001 The book has a topic-by topic breakdown and lots of problem approach suggestions for both free response and multiple choice calculus questions. The book also includes two full length tests.

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Area Under A Curve

Slide Duration:

Table of Contents

Section 1: Functions

Definitions & Properties of Functions

11m 26s

Intro