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College Calculus: Level I Area Between Curves

Section 7: Application of Integrals, part 1: Lecture 1 | 19:59 min

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Lecture Comments (6)

	0 answers Post by Eti Sinha on September 8, 2014 For additional example #5, is it possible to find the area underneath the curve by: Integral of ( 1/y ) dy, from 1/3 to 3? Using dy as opposed to dx?
	0 answers Post by Rufino Hernandez on July 5, 2014 Asides from getting the area of this triangle using this method, you can also get it using another method. The Area of a triangle is square root of (s(s-a)(s-b)(s-c)). Where s=(1/2)(a+b+c). Where a is one side of the triangle, b is another side of the triangle, and c is the third side of the triangle. Since the Triangle's lower angle is the same as the angle made by the function y=\|x\|, it can be deduced that the angle made is 90 degrees. If it is a 90 degree triangle, then we can simply use the Area of a triangle to come to the conclusion of the area; A=(1/2)bh, where b=base and h=height. Since we have the end points of the sides of the triangle, we can simply derive sides. For example, the right side has an endpoint of (2,2); y=2, x=2. Using this we can make a second triangle where the sides are 2 and 2 which can lead us to find the third side which is square root of (8) = 2 square root of(2). We now know the length of the right side of the triangle. Using this same same method, we can derive the length of the base of the triangle (bottom left). The endpoint of this side is (-2/3, -2/3). So we can make a third triangle that has these sides and derive the length of the second side of the original triangle. This second side will be (-2/3)^2+(-2/3)^2=c^2 ==> (4/9)+(4/9)= c^2 ==> 8/9=c^2 ==> c= (2 * square root of (2))/3. Now since we have the height and base, we can simply put this into the Area of a triangle; A=(1/2)(base)(height) ==> A=(1/2)((2square root of (2))/2)(2 * square root of (2) ==> 4/3. Or we can solve for the third side of the triangle and input them into the (The Area of a triangle is square root of (s(s-a)(s-b)(s-c)). Where s=(1/2)(a+b+c). Where a is one side of the triangle, b is another side of the triangle, and c is the third side of the triangle.) equation and get the same results. This method will yield the same answer as the method stated in this video.
	0 answers Post by Fadel Hanoun on December 17, 2013 What about when x = 5pi/4, where Tan = 1?
	0 answers Post by Troy Ling on February 4, 2011 k= -15/16
	1 answer Last reply by: Nopparat SAntisathaporn Wed Oct 6, 2010 2:36 AM Post by Nopparat SAntisathaporn on October 6, 2010 Hi there, I don't know how to do number 7 in this test http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2007/90286-exm-07.pdf Can you please explain how to solve for K. Thank You.

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Area Between Curves

I recommend always starting with a sketch and drawing in a “sample rectangle.”
If an interval is not given, you may need to set the two functions equal in order to determine the interval involved.
If two curves cross, then you will need to break up the integral into more than one integral.
Sometimes, geometrical considerations can help you to check your results or even provide a non-calculus way to finish a problem.

Area Between Curves

Find the area enclosed by y = 4 and y = x²

First, we find the bounds. To do so, we set the equations equal to each other
x² = 4
x = ±2
A = ∫₋₂² 4 − x² dx
A = (4x − [(x³)/3]) |₋₂²
A = 8 − [8/3] − (−8 − [(−8)/3])
A = 16 − [16/3]

A = [32/3]

Find the area under the velocity curve v(t) = 3t² + t from t = 0 to t = 2

A = ∫₀² 3t² + t − 0 dt
The subtraction by zero has no effect. It is just to illustrate that we're taking the x-axis as the lower-bound.
A = (t³ + [(t²)/2]) |₀²
A = 8 + 2 − 0 − 0
The area under a velocity curve also happens to be the displacement!
∫_t₁^t₂ v(t) dt = x(t₂) − x(t₁)

A = 10

Find the area under y = e^x from x = 0 to x = ln10

A = ∫₀^ln10 e^x dx
A = e^x |₀^ln10
A = e^ln10 − e⁰
A = 10 − 1
A = 9

A = 9

Find the area under y = e^[t/3] from t = 0 to t = 6

A = ∫₀⁶ e^[t/3] dt
u = [t/3]
du = [1/3] dt
A = 3 ∫₀⁶ e^u du
A = 3 e^[t/3] |₀⁶
A = 3 (e² − e⁰)

A = 3(e² − 1)

Find the area bounded by y = x² and y = |x|

Find the bounds of the integral
x² = |x|
x² = √{x²}
x⁴ = x²
x² = 1
x = ±1
y = |x| has different behavior on either side of zero. We can treat this as a piecewise function with the area as the sum of two different integrals
A = ∫₋₁⁰ −x − x² dx + ∫₀¹ x − x² dx
A = (−[(x²)/2] − [(x³)/3]) |₋₁⁰ + ([(x²)/2] − [(x³)/3]) |₀¹
A = (−0 − 0) − (−[((−1)²)/2] − [((−1)³)/3]) + ([1/2] − [1/3]) − 0
A = [1/2] − [1/3] + [1/2] − [1/3]
A = 1 − [2/3]

A = [1/3]

Find the area bound by x = y³ and the y-axis from y = 0 to y = 2

A = ∫₀² y³ dy
A = [(y⁴)/4] |₀²
A = [(2⁴)/4] − 0

A = 4

Find the area bound by y = sinx and the x-axis from x = 0 to x = 2π

A = ∫₀^2π sinx dx
A = − cosx |₀^2π
A = −cos2π− (− cos0)
A = −1 + 1 = 0
An area of zero doesn't quite make sense here, so using the straight integral is insufficient. What we can do is treat this as two separate integrals, one where the area is above the x-axis and one where it is below and add their effective area.
sinx = 0, x = 0, π, 2π in this interval.
A = ∫₀^π sinx dx + (−∫_π^2π sinx dx)
The negative sign is there to compensate for the "negative area"
A = −cosx |₀^π + cosx |_π^2π
A = − cosπ+ cos0 + cos2π− cosπ
A = 1 + 1 + 1 + 1

A = 4

Find the area bound by x = y² and y = x − 2

We could find the integral in terms of dx, but the we would have to split it up into at least two separate integrals. Besides, x = y² is not a function of x (there are two outputs for every x input).
x = y²
x = y + 2
Find the limits of the integral
y² = y + 2
y² − y − 2 = 0
(y + 1)(y − 2) = 0
y = 2, −1
A = ∫₋₁² (y + 2) − y² dy
A = ([(y²)/2] + 2x − [(y³)/3]) |₋₁²
A = (2 + 4 − [8/3]) − ([1/2] − 2 + [1/3])
A = 8 − [1/2] − [9/3]
The area above the

A = [9/2]

Find the area bound by y = x² + 1 and y = −x² + 3

Find the limits of the integral
x² + 1 = −x² + 3
2x² = 2
x² = 1
x = ±1
A = ∫₋₁¹ (−x² + 3) − (x² + 1) dx
A = ∫₋₁¹ −2x² + 2 dx
A = (−[(2x³)/3] + 2x) |₋₁¹
A = −[2/3] + 2 − (−[(−2)/3] − 2)
A = −[2/3] + 2 − [2/3] + 2
A = 4 − [4/3]

A = [8/3]

Find the area bound by y = x² and y = 2x

Find the limits
x² = 2x
x² − 2x = 0
x(x − 2) = 0
x = 0, 2
A = ∫₀² 2x − x² dx
A = x² − [(x³)/3] |₀²
A = 4 − [8/3] − 0

A = [4/3]

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Answer

Area Between Curves

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Area Between Two Curves

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Lecture Example 1

Lecture Example 2

Lecture Example 3

Additional Example 4

Additional Example 5

Intro 0:00
Area Between Two Curves 0:12

Graphic Description

Lecture Example 1 1:44
Lecture Example 2 5:39
Lecture Example 3 8:45
Additional Example 4
Additional Example 5

College Calculus 1 Online Course

Section 1: Overview of Functions
	Review of Functions	26:29
	Compositions of Functions	12:29
Section 2: Limits
	Average and Instantaneous Rates of Change	20:59
	Limit Investigations	22:37
	Algebraic Evaluation of Limits	28:19
	Formal Definition of a Limit	23:39
	Continuity and the Intermediate Value Theorem	19:09
Section 3: Derivatives, part 1
	Limit Definition of the Derivative	22:52
	The Power Rule	26:01
	The Product Rule	14:54
	The Quotient Rule	19:17
	Applications of Rates of Change	17:43
	Trigonometric Derivatives	26:58
	The Chain Rule	23:47
	Inverse Trigonometric Functions	27:05
	Equation of a Tangent Line	15:52
Section 4: Derivatives, part 2
	Implicit Differentiation	30:05
	Higher Derivatives	13:16
	Logarithmic and Exponential Function Derivatives	17:42
	Hyperbolic Trigonometric Function Derivatives	14:30
	Related Rates	29:05
	Linear Approximation	23:52
Section 5: Application of Derivatives
	Absolute Minima and Maxima	18:57
	Mean Value Theorem and Rolle's Theorem	20:00
	First Derivative Test, Second Derivative Test	27:11
	L'Hopital's Rule	23:09
	Curve Sketching	40:16
	Applied Optimization	25:37
	Newton's Method	25:13
Section 6: Integrals
	Approximating Areas and Distances	36:50
	Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus	22:02
	Substitution Method for Integration	23:19
Section 7: Application of Integrals, part 1
	Area Between Curves	19:59
	Volume by Method of Disks and Washers	24:22
	Volume by Method of Cylindrical Shells	30:29
	Average Value of a Function	16:31
Section 8: Extra
	Graphs of f, f', f''	23:58
	Slope Fields for Differential Equations	18:32
	Separable Differential Equations	17:04

Related Books

	Calculus by Larson, 9th Edition Authors: Ron Larson, Bruce H. Edwards ISBN: 0547167024 Publisher: Brooks Cole Year: 2009 The book is filled with examples and detailed explanations. Capstone exercises are also included for each section and synthesize the main concepts into a single example.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 1 Author: Ron Larson ISBN: 547213093 Publisher: Brooks Cole Year: 2008 This solutions manual includes worked out solutions to every odd-numbered exercise in Calculus of a Single Variable, 9th edition.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 2 Authors: Ron Larson, Bruce H. Edwards ISBN: 547213107 Publisher: Brooks Cole Year: 2009 This solutions manual includes worked out solutions to every odd-numbered exercise in Multivariable Calculus, 9th edition.

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Area Between Curves

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Table of Contents

Section 1: Overview of Functions

Review of Functions

26m 29s

Intro