Chapter 5: Applications of Integration
Section 5.4: Arc Length
Example 5.4.4
Obtain the length of the curve defined by y=sinx2,0≤x≤2.
Solution
Mathematical Solution
Application of the appropriate member of Table 5.4.1 leads to
∫021+ⅆⅆ x sinx2ⅆx = ∫021+4 x2cos2x2ⅆx ≐3.620249143
Maple can provide no antiderivative for the integrand 1+4 x2cos2x2. (There probably is no antiderivative amongst the elementary functions.) The integral is evaluated numerically by techniques akin to evaluating a Riemann sum, but only more efficient. Some of these techniques will be studied in Chapter 6.
Maple Solution
Figure 5.4.4(a) shows the tutor applied to finding the arc length of sinx2 on the interval 0,2.
In the graph, the function is in red; the integrand, in blue; and the arc-length function, in green.
Although an arc-length function exists, its rule can't be expressed analytically without an antiderivative for the arc-length integral. However, the tutor numerically evaluates the integral at enough points for a graph of this function to be drawn.
Figure 5.4.4(a) Arc Length tutor
The ArcLength command
Tools≻Load Package: Student Calculus 1
Loading Student:-Calculus1
Apply the ArcLength command with the output option set to integral.
Press the Enter key to obtain the unevaluated arc-length integral.
Context Panel: Approximate≻10 (digits)
ArcLengthsinx2,0..2,output=integral
∫021+4x2cosx22ⅆx
→at 10 digits
3.620249143
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