Page 255 - Calculus Demystified

242
SOLUTION CHAPTER 8 Applications of the Integral
The length is

3 3
2 1/2

(1 +[f (x)] ) dx = (1 +[6x 1/2 2 1/2 dx
] )
0 0
3

1/2
= (1 + 36x) dx
0
3
1
= · (1 + 36x) 3/2
54
0
1
= [109 3/2 − 1 3/2 ]
54
(109) 3/2 − 1
= .
54
EXAMPLE 8.21
Let uscalculate the length of the graph of the function f(x) = (1/2)×
x
(e + e −x ) over the interval [1, ln 8].
SOLUTION
We calculate that
x

f (x) = (1/2)(e − e −x ).
Therefore the length of the curve is

ln 8
x −x 2 1/2

1 +[(1/2)(e − e )] dx
1
ln 8 e 2x 1 e −2x 1/2
= + + dx
1 4 2 4
ln 8
1 x −x
= e + e dx
2 1
1
x
= [e − e −x ln 8
]
1
2
63 e 1
= − + .
16 2 2e
You Try It: Set up, but do not evaluate, the integral for the arc length of the graph
√
of y = sin x on the interval π/4 ≤ x ≤ 3π/4.
Sometimes an arc length problem is more conveniently solved if we think of the
curve as being the graph of x = g(y). Here is an example.