Calculus- Early Transcendentals, 2021a

6.1. Displacement and Area 229
Both common sense and high–level mathematics tell us that as n gets large, the approximation gets better.
In fact, if we take the limit as n → ∞, wegettheexact area. Thatis,
32
lim
(1 − 1 )
n→∞ 3 n 2 = 32 (1 − 0)
3
= 32
3 = 10.6
This is a fantastic result. By considering n equally–spaced subintervals, we obtained a formula for an
approximation of the area that involved our variable n. As n grows large – without bound – the error
shrinks to zero and we obtain the exact area.
♣
This section started with a fundamental calculus technique: make an approximation, refine the approximation
to make it better, then use limits in the refining process to get an exact answer. That is precisely
what we just did.
Let’s practice this again.
Example 6.11: Approximating Area With a Formula, Using Sums
Find a formula that approximates the area under f (x) =x 3 on the interval [−1,5] using the Right
Hand Rule and n equally spaced subintervals, then take the limit as n → ∞ to find the exact area.
Solution. We have Δx = 5−(−1)
n
= 6/n. Wehavex i =(−1)+(i − 1)Δx; as the Right Hand Rule uses x i+1 ,
we have x i+1 =(−1)+iΔx.
The Riemann sum corresponding to the Right Hand Rule is (followed by simplifications):
n
∑ f (x i+1 )Δx =
i=1
=
=
=
n
∑
i=1
n
∑
i=1
n
∑
i=1
n
∑
i=1
= Δx 4 n ∑
i=1
f (−1 + iΔx)Δx
(−1 + iΔx) 3 Δx
(
(iΔx) 3 − 3(iΔx) 2 + 3iΔx − 1 ) Δx
(
i 3 Δx 4 − 3i 2 Δx 3 + 3iΔx 2 − Δx )
i 3 − 3Δx 3 n ∑
= Δx 4 ( n(n + 1)
2
i=1
i 2 + 3Δx 2 n ∑
i=1
i −
n
∑
i=1
Δx
) 2
3 n(n + 1)(2n + 1) 2 n(n + 1)
− 3Δx + 3Δx − nΔx
6
2
= 1296
n 4 · n2 (n + 1) 2
− 3 216 n(n + 1)(2n + 1)
· + 3 36 n(n + 1)
4 n3 6 n 2 − 6
2
= 156 + 378
n + 216
n 2
Once again, we have found a compact formula for approximating the area with n equally spaced subintervals
and the Right Hand Rule. Using ten subintervals, we have an approximation of 195.96 (these
rectangles are shown in Figure 6.11). Using n = 100 gives an approximation of 159.802.