For f be an increasing function on given interval and g be its inverse, then ∫ a b 1 + f ' ( x ) 2 ⋅ d x = ∫ f ( a ) f ( b ) 1 + g ' ( y ) 2 ⋅ d y

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Answer 1

Question

Chapter 9.2, Problem 65E

To determine

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

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Answer 2

Question

Chapter 9.2, Problem 65E

To determine

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

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Answer 3

Question

Chapter 9.2, Problem 65E

To determine

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

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Answer 4

Question

Chapter 9.2, Problem 65E

To determine

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

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Answer 5

Chapter 9.2, Problem 65E

To determine

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

Answer 6

Chapter 9.2, Problem 65E

To determine

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

Answer 7

Chapter 9.2, Problem 65E

To determine

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 9.1 - Prob. 1PQ Ch. 9.1 - Prob. 2PQ Ch. 9.1 - Prob. 3PQ Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E

For f be an increasing function on given interval and g be its inverse, then ∫ a b 1 + f ' ( x ) 2 ⋅ d x = ∫ f ( a ) f ( b ) 1 + g ' ( y ) 2 ⋅ d y

Solution Summary: The author explains that if f be an increasing function on given interval and g be its inverse, the arc length of the curves will be equal.

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To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$

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Chapter 9 Solutions

For f be an increasing function on given interval and g be its inverse, then ∫ a b 1 + f ' ( x ) 2 ⋅ d x = ∫ f ( a ) f ( b ) 1 + g ' ( y ) 2 ⋅ d y

Solution Summary: The author explains that if f be an increasing function on given interval and g be its inverse, the arc length of the curves will be equal.

Videos

To Show: For f be an increasing function on given interval and g be its inverse, then ∫ab1+f'(x)2⋅dx=∫f(a)f(b)1+g'(y)2⋅dy

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Chapter 9 Solutions

To Show:

For $f$ be an increasing function on given interval and $g$ be its inverse, then $\int_{a}^{b} \sqrt{1 + f' {(x)}^{2}} \cdot d x = \int_{f (a)}^{f (b)} \sqrt{1 + g' {(y)}^{2}} \cdot d y$