Compare a left-endpoint sum with the integral

Lower left-endpoint rectangles under y=2ˣ from 0 to 1, symbolic subdivision labelsSource-given figure redrawn with D3. Lower left-endpoint rectangles under y=2ˣ from 0 to 1, symbolic subdivision labels. Only labels supplied in the original question are included.xyOy=2ˣ1/n2/n3/n(n−3)/n(n−2)/n(n−1)/n1
Question diagram 1
Lower left-endpoint rectangles under y=2ˣ from 0 to 1, symbolic subdivision labelsSource-given figure redrawn with D3. Lower left-endpoint rectangles under y=2ˣ from 0 to 1, symbolic subdivision labels. Only labels supplied in the original question are included.xyOy=2ˣ1/n2/n3/n(n−3)/n(n−2)/n(n−1)/n1

Given lower rectangles beneath y=2ˣ, with all printed symbolic subdivision labels.

The graph y=2xy=2^x, 0≤x≤10\le x\le1, is shown. Rectangles, each width 1/n1/n, where nn is an integer, are drawn under the curve. Show their total area is 1/[n(21/n−1)]1/[n(2^{1/n}-1)]. marks

By considering the area bounded by y=2xy=2^x, x=1x=1 and the axes, briefly explain why ln⁡2<n(21/n−1)\ln2<n(2^{1/n}-1). marks

[2008/NJC/Prelim/I/3]

Total marks: 5

Solution

(a)

Geometric series

The graph of y=2xy=2^x for 0≤x≤10\le x\le1 is shown. Rectangles, each of width 1n\frac1n, where nn is an integer, are drawn under the curve. Show that the total area of the rectangles is 1n(21/n−1)\dfrac{1}{n\left(2^{1/n}-1\right)}.

Explore this idea

Hint

  1. Where does each rectangle get its height? The left-hand end of its strip.

  2. List the heights: 20,21/n,22/n,…2^0,2^{1/n},2^{2/n},\dots. What kind of sequence is this?

  3. Sum with the geometric series formula.

Answer · (a)

Sn=1/[n(21/n−1)]S_n=1/[n(2^{1/n}-1)].

Explore the idea

Left-endpoint rectangles sit under the curve

With the left-edge rule, each strip takes its height from its left corner. The total is exactly S_n. Slide n up and watch it creep towards the true area 1/ln 2.

Number of stripsn = 5
Try an example

Where does one strip’s area come from?

1 of 5

Click a strip in the graph to inspect it. The darker rectangle is this strip. Its red marks use the left edge at x* = 0.

upperlowerwidth Δx
Zoomed-in rectangle: width and height are scaled independently. The sample sets its upper and lower values across the whole strip.
Sample x*
0
Width Δx
0.2
Upper value
1
Lower value
0
Height: upper − lower
1
Strip area
0.2

0.2 ×(1 − 0)≈ 0.2

Add this contribution to those of the other 4 strips to get the sum below. Displayed values are rounded.

Strip width
0.200
1.000 ÷ 5
Sum of strips
1.34500
Exact, 1/ln 2
1.44270
Error
−0.09769
6.77% too small

How the sum improves as n grows

  • Y = 2ˣ
  • X-axis
  • Each has width 1/n and height 2^(k/n) from its left corner
  • Exact area 1/ln 2 ≈ 1.4427
Key idea

The left-edge sum is always below the curve and always below 1ln⁡2\frac1{\ln2}. That observation is the whole proof of part (b).

Ideas to try
  1. Keep the left-edge rule on. Is every purple bar below the curve?
  2. Read the error at n = 5. Then at n = 40. Does the sum ever overshoot?
  3. Compare with the right-edge rule: what changes?
(b)

Bounding by an integral

By considering the area bounded by y=2xy=2^x, x=1x=1 and the axes, briefly explain why ln⁡2<n(21/n−1)\ln2<n\left(2^{1/n}-1\right).

Hint

  1. Compare the sum of rectangles with the exact area under the curve.

  2. Evaluate the exact area: ∫012x dx\int_0^12^x\,dx.

  3. Rearrange the inequality.

Answer · (b)

ln⁡2<n(21/n−1)\ln2<n(2^{1/n}-1).

For positive integer n: rectangles lie strictly below the curve, so Sn<1/ln2.

Takeaway

An upper bound for ln⁡2\ln 2

As nn grows, n(21/n−1)n(2^{1/n}-1) decreases towards ln⁡2\ln2 from above. Try n=10n=10: 10(20.1−1)≈0.717710\left(2^{0.1}-1\right)\approx0.7177, only a little above ln⁡2≈0.6931\ln2\approx0.6931. This is a standard way to approximate a logarithm with only powers.