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http://www.lahc.edu/math/frankma.htm

Given a 2D region R and we would like to find its area.

R

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Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b.

R

More on Area

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b.

a=x b=x

R

More on Area

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b.

a=x b=x

R

More on Area

Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b.

a=x b=x

R

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L(x)

Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b.

a=x b=x

R

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L(x)

Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

a=x0 b=xn

Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b]

x1 x2 xi–1 xi

R

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L(x)

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

a=x0 b=xn

Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi].

*

x1 x2 xi–1 xi

R

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L(x)

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

a=x0 b=x

Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi].

x1 x2

x1*

xi–1 xi

R

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L(x)

*

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

a=x0 b=x

Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi].

x1 x2

x1* x2

*xi–1 xi

R

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L(x)

*

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

a=x0 b=x

Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi].

x1 x2

x1* x2

* x3*

xi–1 xi

xi*

R

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L(xi)

*

*

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

a=x0 b=x

Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. Let Δx = the width of each subinterval.

x1 x2

x1* x2

* x3*

xi–1 xi

xi*

R

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*

L(xi)*

Given a 2D region R and we would like to find its area. Take a ruler and measure R from one end to the other, let x be the measurements and let R span from x = a to x = b. Let L(x) = cross–sectional length at x.

a=x0 b=x

Let the sequence {a=x0, x1, x2, .. xn=b} be a regular partition of [a, b] and select an arbitrary point xi in each sub-interval [x i–1, xi]. Let Δx = the width of each subinterval. The rectangle with L(xi) as height and Δx as width approximates the area in R that is spanned from xi–1 to xi.

x1 x2

x1* x2

* x3*

xi–1 xi

xi*

R

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*

*

Δx

L(xi)*

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

R

The Riemann sum

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

R

L(x1)Δx*The Riemann sum

L(x1)*

Δx

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

R

L(x1)Δx+ * L(x2)Δx+ … *The Riemann sum

L(x1) L(x2)* *

ΔxΔx

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

L(xi)

Δx

R

L(x1)Δx+ * L(x2)Δx+ … * L(xn)Δx*The Riemann sum

L(x1) L(x2)* **

ΔxΔx

L(xn)*

Δx

xn*

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

L(xi)

Δx

R

L(x1)Δx+ *i=1

n∑ L(xi)Δx *L(x2)Δx+ … * L(xn)Δx = *The Riemann sum

of all such rectangles approximates the area of R.

L(x1) L(x2)* **

ΔxΔx

xn*

Δx

L(xn)*

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

L(xi)

Δx

In fact the mathematical definition of the area of R is the limit

R

L(x1)Δx+ *i=1

n∑ L(xi)Δx *L(x2)Δx+ … * L(xn)Δx = *The Riemann sum

of all such rectangles approximates the area of R.

n∑ L(xi)Δx as Δx 0 or n ∞,*of the Riemann sum

L(x1) L(x2)* **

ΔxΔx

xn*

Δx

L(xn)*

so

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

L(xi)

Δx

In fact the mathematical definition of the area of R is the limit

the area of R =∫x=a

bL(x) dx

R

L(x1)Δx+ *i=1

n∑ L(xi)Δx *L(x2)Δx+ … * L(xn)Δx = *The Riemann sum

of all such rectangles approximates the area of R.

n∑ L(xi)Δx as Δx 0 or n ∞,*of the Riemann sum

by the FTC

L(x1) L(x2)* **

ΔxΔx

xn*

Δx

L(xn)*

so

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a=x0 b=xx1 x2 xi–1 xi

x1* x2

* x3* xi

*

L(xi)

Δx

In fact the mathematical definition of the area of R is the limit

the area of R =∫x=a

bL(x) dx

R

L(x1)Δx+ *i=1

n∑ L(xi)Δx *L(x2)Δx+ … * L(xn)Δx = *The Riemann sum

of all such rectangles approximates the area of R.

n∑ L(xi)Δx as Δx 0 or n ∞,*of the Riemann sum

by the FTC

Theorem. The area of a 2D region is the definite integral of its cross–section (length) function.

L(x1) L(x2)* **

ΔxΔx

xn*

Δx

L(xn)*

so

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

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Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

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y = –x2 + 2x

y = x2

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area.

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y = –x2 + 2x

y = x2

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.x2 = –x2 + 2x

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.x2 = –x2 + 2x2x2 = 2x

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.x2 = –x2 + 2x2x2 = 2x2x(x – 1) = 0 so x = 0, 1.

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.x2 = –x2 + 2x2x2 = 2x2x(x – 1) = 0 so x = 0, 1.

Hence the area is ∫x=0

1

(–x2 + 2x) – x2 dx

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.x2 = –x2 + 2x2x2 = 2x2x(x – 1) = 0 so x = 0, 1.

Hence the area is ∫x=0

1

(–x2 + 2x) – x2 dx = ∫1

–2x2 + 2x dx0

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.x2 = –x2 + 2x2x2 = 2x2x(x – 1) = 0 so x = 0, 1.

Hence the area is ∫x=0

1

(–x2 + 2x) – x2 dx = ∫1

–2x2 + 2x dx

= –23

x3 + x2 |0

0

1

Example A. a. Find the area bounded by y = –x2 + 2x and y = x2

We need to find the span of the area. They are the x–coordinates of the intersections of the curves.

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y = –x2 + 2x

y = x2

Set the equations equal to find the intersection points.x2 = –x2 + 2x2x2 = 2x2x(x – 1) = 0 so x = 0, 1.

Hence the area is ∫x=0

1

(–x2 + 2x) – x2 dx = ∫1

–2x2 + 2x dx

= –23

x3 + x2 |0

0

1

= 13

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

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b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

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f(x) = 2x – x3

g(x) = –x2

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts.

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f(x) = 2x – x3

g(x) = –x2

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x).

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f(x) = 2x – x3

g(x) = –x2

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top.

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f(x) = 2x – x3

g(x) = –x2

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top. We set the equations equal to solve for the span.

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f(x) = 2x – x3

g(x) = –x2

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top. We set the equations equal to solve for the span.

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f(x) = 2x – x3

g(x) = –x2

–x2 = 2x – x3

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top. We set the equations equal to solve for the span.

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f(x) = 2x – x3

g(x) = –x2

–x2 = 2x – x3

x3 – x2 – 2x = 0

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top. We set the equations equal to solve for the span.

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f(x) = 2x – x3

g(x) = –x2

–x2 = 2x – x3

x3 – x2 – 2x = 0 x(x2 – x – 2) = 0

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top. We set the equations equal to solve for the span.

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f(x) = 2x – x3

g(x) = –x2

–x2 = 2x – x3

x3 – x2 – 2x = 0 x(x2 – x – 2) = 0

x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top. We set the equations equal to solve for the span.

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f(x) = 2x – x3

g(x) = –x2

–x2 = 2x – x3

x3 – x2 – 2x = 0 x(x2 – x – 2) = 0

x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.

Therefore the span for the 1st area is from x = –1 to x = 0

b. Find the area bounded by f(x) = 2x – x3 and g(x) = –x2

The bounded area consists of two parts. The first part is bounded above by g(x) and below by f(x). The second part is has f(x) on top. We set the equations equal to solve for the span.

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f(x) = 2x – x3

g(x) = –x2

–x2 = 2x – x3

x3 – x2 – 2x = 0 x(x2 – x – 2) = 0

x(x – 2)(x + 1) = 0 so x = –1, 0, and 2.

Therefore the span for the 1st area is from x = –1 to x = 0 and the span for the 2nd area is from x = 0 to x = 2.

∫x= –1

0

–x2 – (2x – x3) dx

More on Areaf(x) = x – x3

g(x) = –x2

Hence the total bounded area is

–1 0 2

∫x= –1

0

–x2 – (2x – x3) dx

More on Areaf(x) = x – x3

g(x) = –x2

Hence the total bounded area is

+ ∫x= 0

2

2x – x3 – (–x2) dx

–1 0 2

∫x= –1

0

–x2 – (2x – x3) dx

More on Areaf(x) = x – x3

g(x) = –x2

Hence the total bounded area is

+ ∫x= 0

2

2x – x3 – (–x2) dx

=

–1 0 2

∫–1

0x3 – x2 – 2x dx

∫x= –1

0

–x2 – (2x – x3) dx

More on Areaf(x) = x – x3

g(x) = –x2

Hence the total bounded area is

+ ∫x= 0

2

2x – x3 – (–x2) dx

=

–1 0 2

∫–1

0x3 – x2 – 2x dx + ∫0

2– x3 + x2 + 2x dx

∫x= –1

0

–x2 – (2x – x3) dx

More on Areaf(x) = x – x3

g(x) = –x2

Hence the total bounded area is

+ ∫x= 0

2

2x – x3 – (–x2) dx

=

–1 0 2

∫–1

0x3 – x2 – 2x dx + ∫0

2– x3 + x2 + 2x dx

x4

4– = x3

3– x2 |

–1

0

∫x= –1

0

–x2 – (2x – x3) dx

More on Areaf(x) = x – x3

g(x) = –x2

Hence the total bounded area is

+ ∫x= 0

2

2x – x3 – (–x2) dx

=

–1 0 2

∫–1

0x3 – x2 – 2x dx + ∫0

2– x3 + x2 + 2x dx

x4

4– = x3

3– x2 |

–1

0

+ –x4

4+ x3

3+ x2 |

0

2

∫x= –1

0

–x2 – (2x – x3) dx

More on Areaf(x) = x – x3

g(x) = –x2

Hence the total bounded area is

+ ∫x= 0

2

2x – x3 – (–x2) dx

=

–1 0 2

∫–1

0x3 – x2 – 2x dx + ∫0

2– x3 + x2 + 2x dx

x4

4– = x3

3– x2 |

–1

0

+ –x4

4+ x3

3+ x2 |

0

2

= 512 + 8

3

= 3712

A type II region R is a region that is bounded on the right by a curve x = f(y) and the left by x = g(y) from y = a to y = b.

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A type II region R is a region that is bounded on the right by a curve x = f(y) and the left by x = g(y) from y = a to y = b.

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y = a

y = bx = f(y)x = g(y)

A type II region R is a region that is bounded on the right by a curve x = f(y) and the left by x = g(y) from y = a to y = b.Its cross–sectional length is L(y) = f(y) – g(y).

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y = a

y = bx = f(y)x = g(y)

A type II region R is a region that is bounded on the right by a curve x = f(y) and the left by x = g(y) from y = a to y = b.Its cross–sectional length is L(y) = f(y) – g(y).

More on Area

y = a

y = bx = f(y)x = g(y)

L(y) = f(y) – g(y)

A type II region R is a region that is bounded on the right by a curve x = f(y) and the left by x = g(y) from y = a to y = b.Its cross–sectional length is L(y) = f(y) – g(y).

More on Area

y = a

y = bx = f(y)x = g(y)

L(y) = f(y) – g(y)

Therefore the area of R = f(y) – g(y) dy ∫y=a

b

More on AreaExample B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

More on AreaExample B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

More on Area

The shaded region is the area in question.

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 4

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 4

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4.

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.

4

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.

As a type I region, the span is from x = 0 to x = 4.

4

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.

As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function.

4

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.

As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function.

4

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.

As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function. Therefore to find the area of the region, we have to split it into two pieces, I and II as shown.

4

More on Area

The shaded region is the area in question. For the span of the region, set

√x = x – 2

Example B. Find the area that is bounded by the positive x–axis, y = x – 2 and y = √x.

y = x – 2

y = √x

x = x2 – 4x + 40 = x2 – 5x + 40 = (x – 4)(x – 1) so x = 1 and 4. However x = 4 is the only good solution.

As a type I region, the span is from x = 0 to x = 4. But the lower boundary is not a single function. Therefore to find the area of the region, we have to split it into two pieces, I and II as shown.

I II

4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

2 4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2 4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2

∫x= 2

4

√x – (x – 2) dx

4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2

∫x= 2

4

√x – (x – 2) dx

= 2x3/2

3 + |0

2

4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2

∫x= 2

4

√x – (x – 2) dx

= 2x3/2

3 + 2x3/2

3( – x2

2 + 2x ) |0

2

2

4

|

4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2

∫x= 2

4

√x – (x – 2) dx

= 2x3/2

3 + 2x3/2

3( – x2

2 + 2x ) |0

2

2

4

|

= 2(2)3/2

3 +

4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2

∫x= 2

4

√x – (x – 2) dx

= 2x3/2

3 + 2x3/2

3( – x2

2 + 2x ) |0

2

2

4

|

= 2(2)3/2

3 + 2(4)3/2

3[( – 42

2 + 8 )

– 2(2)3/2

3 – 22

2 + 4 )

( ]

4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2

∫x= 2

4

√x – (x – 2) dx

= 2x3/2

3 + 2x3/2

3( – x2

2 + 2x ) |0

2

2

4

|

= 2(2)3/2

3 + 2(4)3/2

3[( – 42

2 + 8 )

– 2(2)3/2

3 – 22

2 + 4 )

( ]

= 2(2)3/2

3 + 163 – 2(2)3/2

3 – 2

4

More on Area

y = x – 2

y = √x

The total area is (area I) + (area II) i.e.

I II

∫x= 0

2

√x dx +

2

∫x= 2

4

√x – (x – 2) dx

= 2x3/2

3 + 2x3/2

3( – x2

2 + 2x ) |0

2

2

4

|

= 2(2)3/2

3 + 2(4)3/2

3[( – 42

2 + 8 )

– 2(2)3/2

3 – 22

2 + 4 )

( ]

= 2(2)3/2

3 + 163 – 2(2)3/2

3 – 2 = 103

4

More on Area

However, we may view this as a type II region.

y = x – 2

y = √x

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2.

y = x – 2

y = √xy = 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions.

y = x – 2

y = √xy = 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2

y = x – 2

y = √xy = 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2

y = x – 2

y = √xy = 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2

y = 2y = √xso x = y2

y = x – 2 so x = y + 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2

y = 2y = √xso x = y2

y = x – 2 so x = y + 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.

y = x – 2 so x = y + 2

y = √xso x = y2

L(y) = y + 2 – y2

y = 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.

y = x – 2 so x = y + 2

y = √xso x = y2

L(y) = y + 2 – y2

Therefore the area is

y + 2 – y2 dy∫y = 0

2

y = 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.

y = x – 2 so x = y + 2

y = √xso x = y2

L(y) = y + 2 – y2

Therefore the area is

y + 2 – y2 dy∫y = 0

2

= + y2

2 2y

– y3

3 0

2

|

y = 2

More on Area

However, we may view this as a type II region. It spans from y = 0 to y = 2. Solve for x for the boundary functions. The right boundary is x = f(y) = y + 2 and the left boundary is x = g(y) = y2 so L(y) = (y + 2) – y2 = y + 2 – y2.

y = x – 2 so x = y + 2

y = √xso x = y2

L(y) = y + 2 – y2

Therefore the area is

y + 2 – y2 dy∫y = 0

2

= + y2

2 2y

– y3

3 0

2

|

= 6 – 83 = 10

3

y = 2