Calculus- Early Transcendentals, 2021a

228 Integration
Notice in the previous example that while we used ten equally spaced intervals, the number “10” didn’t
play a big role in the calculations until the very end. Mathematicians love abstract ideas; let’s approximate
the area of another region using n subintervals, where we do not specify a value of n until the very end.
Example 6.10: Approximating Area Using Sums
Revisit f (x)=4x − x 2 on the interval [0,4,] yet again. Approximate the area under this curve using
the Right Hand Rule with n equally spaced subintervals.
Solution. We know Δx = 4−0
n
= 4/n. Wealsofindx i = 0 + Δx(i − 1)=4(i − 1)/n. The Right Hand Rule
uses x i+1 ,whichisx i+1 = 4i/n. We construct the Right Hand Rule Riemann sum as follows.
n
n
( ) 4i
∑ f (x i+1 )Δx = ∑ f Δx
i=1
i=1
n
[
n
= ∑ 4 4i ( ) ]
4i
2
i=1
n − Δx
n
n
( ) 16Δx
n
( ) 16Δx
= ∑ i −
i=1
n
∑
i=1
n 2 i 2
( ) 16Δx n∑ ( ) 16Δx n∑
=
i −
n
i=1
n 2 i 2
i=1
( )
( ) 16Δx n(n + 1) 16Δx n(n + 1)(2n + 1)
= · −
n 2 n 2 (recall Δx = 4/n)
6
32(n + 1) 32(n + 1)(2n + 1)
= −
n
3n 2
(now simplify)
= 32 (1 − 1 )
3 n 2
The result is an amazing, easy to use formula. To approximate the area with ten equally spaced
subintervals and the Right Hand Rule, set n = 10 and compute
(
32
1 − 1 )
3 10 2 = 10.56.
Recall how earlier we approximated the area with 4 subintervals; with n = 4, the formula gives 10, our
answer as before.
It is now easy to approximate the area with 1,000,000 subintervals! Hand-held calculators will round
off the answer a bit prematurely giving an answer of 10.66666667. (The actual answer is 10.666666666656.)
We now take an important leap. Up to this point, our mathematics has been limited to geometry and
algebra (finding areas and manipulating expressions). Now we apply calculus. Foranyfinite n, we know
that the corresponding Right Hand Rule Riemann sum is:
32
3
(1 − 1 n 2 )
.