Chapter 6 | Applications of Integration
673
Solution We have f ′( x ) =3 x 1/2 , so ⎡
⎦ 2 =9 x . Then, the arc length is Arc Length = ∫ a b 1+ ⎡ ⎣ f ′( x ) ⎤
⎣ f ′( x ) ⎤
⎦ 2 dx
1
= ∫
0 1+9 xdx . Substitute u =1+9 x . Then, du =9 dx . When x =0, then u =1, and when x =1, then u =10. Thus, Arc Length = ∫ 0 1 1+9 xdx = 1 9 ∫ 1+9 x 9 dx = 1 9 ∫ 10 udu
0 1
1
u 3/2 |
1 10
⎡ ⎣ 10 10−1 ⎤
2 3
= 1 9 ·
= 2 27
⎦ ≈ 2.268 units.
6.18 Let f ( x ) = (4/3) 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.
Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration (http://cnx.org/content/m53654/latest/) . In some cases, we may have to use a computer or calculator to approximate the value of the integral. Example 6.19 Using a Computer or Calculator to Determine the Arc Length of a Function of x
Let f ( x ) = x 2 . Calculate the arc length of the graph of f ( x ) over the interval [1, 3].
Solution We have f ′( x ) =2 x , so ⎡
⎣ f ′( x ) ⎤ ⎦ 2 =4 x 2 . Then the arc length is given by Arc Length = ∫ a b 1+ ⎡ ⎣ f ′( x ) ⎤ ⎦ 2 dx = ∫ 1 3
1+4 x 2 dx .
Using a computer to approximate the value of this integral, we get ∫ 1 3 1+4 x 2 dx ≈8.26815.
6.19 Let f ( x ) = sin x . Calculate the arc length of the graph of f ( x ) over the interval [0, π ]. Use a computer or calculator to approximate the value of the integral.
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