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Chapter 6 | Applications of Integration
Figure 6.38 A representative line segment approximates the curve over the interval [ x i −1 , x i ].
By the Pythagorean theorem, the length of the line segment is (Δ x ) 2 + ⎛ ⎝ Δ y i ⎞
⎠ 2 . We can also write this as
Δ x 1+ ⎛ ⎝ ⎛
⎞ ⎠ /(Δ x ) ⎞ ⎠ 2 . Now, by the Mean Value Theorem, there is a point x
i * ∈ [ x i −1 , x i ] such that
⎝ Δ y i
⎞ ⎠ /(Δ x ). Then the length of the line segment is given by Δ x 1+ ⎡
⎣ f ′( x i * ) ⎤
⎛ ⎝ Δ y i
⎦ 2 . Adding up the lengths of all
f ′( x i * ) =
the line segments, we get
Arc Length ≈ ∑ i =1 n
1+ ⎡
⎣ f ′( x i * ) ⎤
⎦ 2 Δ x .
This is a Riemann sum. Taking the limit as n →∞, we have Arc Length = lim n →∞ ∑ i =1 n 1+ ⎡ ⎣ f ′( x i * ) ⎤
b
⎦ 2 Δ x = ∫
⎦ 2 dx .
1+ ⎡
⎣ f ′( x ) ⎤
a
We summarize these findings in the following theorem.
Theorem 6.4: Arc Length for y = f ( x ) Let f ( x ) be a smooth function over the interval ⎡ ⎣ a , b ⎤
⎦ . Then the arc length of the portion of the graph of f ( x ) from
the point ⎛
⎝ a , f ( a ) ⎞
⎛ ⎝ b , f ( b ) ⎞
⎠ to the point
⎠ is given by
(6.7)
Arc Length = ∫ a b
⎦ 2 dx .
1+ ⎡
⎣ f ′( x ) ⎤
Note that we are integrating an expression involving f ′( x ), so we need to be sure f ′( x ) is integrable. This is why we require f ( x ) to be smooth. The following example shows how to apply the theorem.
Example 6.18 Calculating the Arc Length of a Function of x
Let f ( x ) =2 x 3/2 . Calculate the arc length of the graph of f ( x ) over the interval [0, 1]. Round the answer to three decimal places.
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