Sum local area inequalities, then shift the axis

Graph y=1/(1+x²), shaded right-end rectangle from n−1 to nSource-given figure redrawn with D3. Graph y=1/(1+x²), shaded right-end rectangle from n−1 to n. Only labels supplied in the original question are included.xyOn−1n
Question diagram 1
Graph y=1/(1+x²), shaded right-end rectangle from n−1 to nSource-given figure redrawn with D3. Graph y=1/(1+x²), shaded right-end rectangle from n−1 to n. Only labels supplied in the original question are included.xyOn−1n

Given rectangle from n−1 to n beneath y=1/(1+x²).

(i) Prove ddx(tan⁡−1x)=1/(1+x2)\frac d{dx}(\tan^{-1}x)=1/(1+x^2). marks

The diagram sketches y=1/(1+x2)y=1/(1+x^2), with a shaded rectangle from n−1n-1 to nn.

(ii) Considering the rectangle and the area between graph and xx-axis on n−1≤x≤nn-1\le x\le n, n≥1n\ge1, show 1/(1+n2)<tan⁡−1n−tan⁡−1(n−1)1/(1+n^2)<\tan^{-1}n-\tan^{-1}(n-1). marks

(iii) Deduce 1/2+1/5+1/10+⋯+1/(1+n2)<tan⁡−1n1/2+1/5+1/10+\cdots+1/(1+n^2)<\tan^{-1}n. marks

R is bounded by y=1/(1+x2)y=1/(1+x^2), x=3x=3 and y=1/2y=1/2. Write the equation after translation 3 units in negative xx-direction. Hence or otherwise find its volume rotating completely about x=3x=3. marks

[2015/NYJC/Prelim/II/4 (modified)]

Total marks: 10

Solution

(i)

Differentiating an inverse function

Prove that ddx(tan⁡−1x)=11+x2\dfrac{d}{dx}\left(\tan^{-1}x\right)=\dfrac{1}{1+x^2}.

Hint

  1. Let y=tan⁡−1xy=\tan^{-1}x and rewrite as x=tan⁡yx=\tan y.

  2. Differentiate implicitly and use sec⁡2y=1+tan⁡2y\sec^2y=1+\tan^2y.

Answer · (i)

d(arctan⁡x)/dx=1/(1+x2)d(\arctan x)/dx=1/(1+x^2).

(ii)

Bounding an area

The curve y=11+x2y=\dfrac{1}{1+x^2} is sketched, with a rectangle of width 1 drawn under it from x=n−1x=n-1 to x=nx=n. By comparing this rectangle with the area under the curve on n−1≤x≤nn-1\le x\le n, where n≥1n\ge1, show that 11+n2<tan⁡−1n−tan⁡−1(n−1)\dfrac{1}{1+n^2}<\tan^{-1}n-\tan^{-1}(n-1).

Hint

  1. What is the rectangle's height? The curve is decreasing, so it takes the smallest value on the strip.

  2. Write the area under the curve as an integral and use part (i).

Answer · (ii)

1/(1+n2)<arctan⁡n−arctan⁡(n−1)1/(1+n^2)<\arctan n-\arctan(n-1).

n≥1n\ge1; unit-width rectangle lies below the decreasing curve.

(iii)

Telescoping sums

Deduce that 12+15+110+⋯+11+n2<tan⁡−1n\dfrac12+\dfrac15+\dfrac1{10}+\cdots+\dfrac{1}{1+n^2}<\tan^{-1}n.

Explore this idea

Hint

  1. Write part (ii) for n=1,2,3,…n=1,2,3,\dots and add the inequalities.

  2. Look at what happens to the arctangents on the right.

Answer · (iii)

∑j=1n1/(1+j2)<arctan⁡n\sum_{j=1}^n1/(1+j^2)<\arctan n.

Positive integer n; right-hand differences telescope.

Explore the idea

Right-hand rectangles sit under a decreasing curve

Set n = 6 with the right-edge rule to get six unit-width rectangles on [0, 6], exactly the ones in the question. Their total is 1/2 + 1/5 + 1/10 + … and the area under the curve is tan⁻¹ 6.

Number of stripsn = 6
Try an example

Where does one strip’s area come from?

1 of 6

Click a strip in the graph to inspect it. The darker rectangle is this strip. Its red marks use the right edge at x* = 1.

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*
1
Width Δx
1
Upper value
0.5
Lower value
0
Height: upper − lower
0.5
Strip area
0.5

1 ×(0.5 − 0)≈ 0.5

Add this contribution to those of the other 5 strips to get the sum below. Displayed values are rounded.

Strip width
1.000
6.000 ÷ 6
Sum of strips
0.92431
Exact, tan⁻¹ 6
1.40565
Error
−0.48134
34.24% too small

How the sum improves as n grows

  • Y = 1/(1+x²), decreasing for x > 0
  • Heights taken at the right edge: 1/(1+k²) at x = k
  • Area under the curve, tan⁻¹ 6 ≈ 1.4056
Key idea

Every right-hand rectangle fits under the curve, so the sum ∑11+k2\sum\frac1{1+k^2} is trapped below tan⁡−1n\tan^{-1}n.

Ideas to try
  1. Keep n = 6 and the right-edge rule: the strip total is the sum 1/2 + 1/5 + 1/10 + 1/17 + 1/26 + 1/37 ≈ 0.924.
  2. Switch to the left-edge rule. The total now exceeds the area. That is why (ii) must use the right-hand height.
(iv)

Translations

The region RR is bounded by y=11+x2y=\dfrac1{1+x^2}, the line x=3x=3 and the line y=12y=\frac12. Write down the equation of the curve after a translation of 3 units in the negative xx-direction.

Explore this idea

Hint

  1. Translating left by 3 replaces xx by x+3x+3.

  2. Where does the axis x=3x=3 go?

Answer · (iv)

y=1/[1+(x+3)2]y=1/[1+(x+3)^2].

Explore the idea

Move the axis of rotation onto the y-axis

The original curve peaks at (0, 1). Sliding everything 3 units left moves the line x = 3 onto the y-axis, and R moves with it.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Original y = 1/(1+x²)
  • Shifted curve y = 1/(1+(x+3)²)
  • Y = 1/2
  • Original axis of rotation x = 3
  • Shifted axis x = 0, the y-axis
Ideas to try
  1. Switch on “Original”, then “Shifted”, and compare the two peaks.
  2. Turn on “Region R” to see the region that will be spun.
(v)

Volume of revolution

Hence, or otherwise, find the volume generated when RR is rotated completely about the line x=3x=3.

Explore this idea

Hint

  1. After translating, rotate about the yy-axis. Write xx in terms of yy.

  2. Choose the correct branch of the square root for the region.

  3. Integrate πx2 dy\pi x^2\,dy over the heights of RR.

Answer · (v)

V=2π[1−3(arctan⁡3−π/4)+12ln⁡5]≈2.60V=2\pi[1-3(\arctan3-\pi/4)+\tfrac12\ln5]\approx2.60 cubic units (3 s.f.).

Source right-hand region 1≤x≤3; rotate about x=3.

Explore the idea

A solid trumpet from a thin sliver

R, reflected so the axis stays on the right: the line y = 1/2 becomes the flat top of a cone-like solid, and the curve carves its side.

Open the interactive view to explore this geometry in 3D.