Take the infinite limit, then size the bowl

Curve y=e⁻ˣ/(1+e⁻ˣ), asymptotes y=1 and y=0, shaded R in positive quadrantSource-given figure redrawn with D3. Curve y=e⁻ˣ/(1+e⁻ˣ), asymptotes y=1 and y=0, shaded R in positive quadrant. Only labels supplied in the original question are included.xyOy=1y=0½Ry=e⁻ˣ/(1+e⁻ˣ)
Question diagram 1
Curve y=e⁻ˣ/(1+e⁻ˣ), asymptotes y=1 and y=0, shaded R in positive quadrantSource-given figure redrawn with D3. Curve y=e⁻ˣ/(1+e⁻ˣ), asymptotes y=1 and y=0, shaded R in positive quadrant. Only labels supplied in the original question are included.xyOy=1y=0½Ry=e⁻ˣ/(1+e⁻ˣ)

Given logistic-shaped curve with asymptotes and shaded region R in the positive quadrant.

Section AB of logarithmic curve, A(a,ln a), point B and horizontal guidesSource-given figure redrawn with D3. Section AB of logarithmic curve, A(a,ln a), point B and horizontal guides. Only labels supplied in the original question are included.xyOA (a, ln a)B
Question diagram 2
Section AB of logarithmic curve, A(a,ln a), point B and horizontal guidesSource-given figure redrawn with D3. Section AB of logarithmic curve, A(a,ln a), point B and horizontal guides. Only labels supplied in the original question are included.xyOA (a, ln a)B

Given section AB of logarithmic curve, with A labelled (a, ln a).

(a) C: y=e−x/(1+e−x)y=e^{-x}/(1+e^{-x}). R is bounded by C, positive xx-axis and positive yy-axis as shown; graph labels y=1,y=0,1/2,R,Oy=1,y=0,1/2,R,O.

(i) Obtain a formula in terms of nn for ∫0ne−x1+e−xdx\int_0^n\frac{e^{-x}}{1+e^{-x}}dx, n>0n>0. marks

(ii) Hence find R’s exact area. marks

(iii) Denoting the answer obtained in part (i) by qq, write down in terms of qq the area of R if C is scaled by factor 2 parallel to yy-axis. mark

(b) A girl designs a bowl by rotating section AB of y=ln⁡xy=\ln x completely about yy-axis. It must hold 300 cm3300\text{ cm}^3. A is (a,ln⁡a)(a,\ln a); the rim diameter is twice the base diameter. State B in terms of aa and find exact aa. marks

[2010/VJC/Prelim/I/11 (modified)]

Total marks: 10

Solution

(a)i

Integrating f′/ff'/f

Obtain a formula in terms of nn for ∫0ne−x1+e−x dx\displaystyle\int_0^n\frac{e^{-x}}{1+e^{-x}}\,dx, where n>0n>0.

Explore this idea

Hint

  1. Differentiate the denominator 1+e−x1+e^{-x}. Does it match the numerator?

  2. Use ∫f′f dx=ln⁡∣f∣\int\frac{f'}{f}\,dx=\ln|f|.

Answer · (a)i

qn=ln⁡2−ln⁡(1+e−n)q_n=\ln2-\ln(1+e^{-n}).

Explore the idea

A tail that thins out but never quite reaches the axis

The curve starts at 1/2 and decays towards 0. Up to x = 6 the area is already very close to ln 2.

Number of stripsn = 4
Reset the controls

Where does one strip’s area come from?

1 of 4

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
1.5
Upper value
0.5
Lower value
0
Height: upper − lower
0.5
Strip area
0.75

1.5 ×(0.5 − 0)≈ 0.75

Add this contribution to those of the other 3 strips to get the sum below. Displayed values are rounded.

Strip width
1.500
6.000 ÷ 4
Sum of strips
1.11126
Exact, ln 2 − ln(1+e⁻⁶)
0.69067
Error
+0.42059
60.90% too big

How the sum improves as n grows

  • Curve C, starting at 1/2 when x = 0
  • X-axis
  • Each of width 6/n
  • Exact area up to n = 6: ln 2 − ln(1 + e⁻⁶) ≈ 0.6907
Key idea

The tail beyond x=6x=6 adds only ln⁡(1+e−6)≈0.0025\ln(1+e^{-6})\approx0.0025. As n→∞n\to\infty it vanishes and the total approaches ln⁡2\ln2.

Ideas to try
  1. The exact area up to n = 6 is 0.6907. How close is that to ln 2 = 0.6931?
  2. Is it ever possible for the area to the right of x = 6 to be zero?
(a)ii

Improper integral

Hence find the exact area of RR, the region bounded by CC, the positive xx-axis and the positive yy-axis.

Hint

  1. The region has no right-hand edge, so the area is a limit as n→∞n\to\infty.

  2. What happens to e−ne^{-n}?

Answer · (a)ii

Area of R=ln⁡2R=\ln2.

(a)iii

Stretch of a graph

Denote the answer to part (i) by qq. Write down, in terms of qq, the area of RR if CC is scaled by a factor of 2 parallel to the yy-axis.

Hint

  1. A stretch parallel to the yy-axis by factor 2 doubles every height.

Answer · (a)iii

Total scaled area =2ln⁡2=2\ln2; finite interval [0,n][0,n] has area 2qn2q_n.

Printed 2q is the total only if q denotes the limiting area; the question definesq from the finite-n integral.

(b)i

Reading the geometry

A bowl is made by rotating the section ABAB of y=ln⁡xy=\ln x completely about the yy-axis, where A=(a,ln⁡a)A=(a,\ln a). The diameter of the rim is twice the diameter of the base. State the coordinates of BB in terms of aa.

Hint

  1. The base is the radius at AA, which is aa. The rim is twice as wide.

Answer · (b)i

B=(2a,ln⁡(2a))B=(2a,\ln(2a)).

(b)ii

Volume of revolution

The bowl must hold 300 cm3300\text{ cm}^3. Find the exact value of aa.

Explore this idea

Hint

  1. Rotating about the yy-axis, write xx in terms of yy: x=eyx=e^y.

  2. Integrate πx2\pi x^2 from y=ln⁡ay=\ln a to y=ln⁡2ay=\ln2a.

  3. Set the volume equal to 300300 and solve for aa.

Answer · (b)ii

a=200/πa=\sqrt{200/\pi} cm.

Explore the idea

A shallow dish from y = ln x

The curve climbs slowly, so a section from a to 2a is a wide, flat dish. This picture is drawn with a = √(200/π) ≈ 7.98.

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