Page 254 - Calculus Demystified

Applications of the Integral
CHAPTER 8
usual planar distance formula: 241

2 2 1/2
j = [x j − x j−1 ] +[f(x j ) − f(x j−1 )] .
We denote the quantity x j − x j−1 by x and apply the definition of the derivative
to obtain

f(x j ) − f(x j−1 )

≈ f (x j ).
x
Now we may rewrite the formula for j as
2
2 1/2

j ≈ ([8x] +[f (x j )8x] )
2 1/2

= (1 +[f (x j )] ) 8x.
Summing up the lengths j (Fig. 8.30) gives an approximate length for the curve:
k k
2 1/2

length of curve ≈ j = (1 +[f (x j )] ) 8x.
j=1 j=1

Fig. 8.30

But this last is a Riemann sum for the integral

b
2 1/2

= (1 +[f (x)] ) dx. (.)
a
As the mesh of the partition becomes finer, the approximating sum is ever closer to
what we think of as the length of the curve, and it also converges to this integral.
Thus the integral represents the length of the curve.

EXAMPLE 8.20
Let uscalculate the arc length of the graph of f(x) = 4x 3/2 over the interval
[0, 3].