QCE Specialist Maths: Integration techniques; applications of integral calculus Q151

Q151 · Practice questionTechnology-activeComplex unfamiliar8 marks

QUESTION 151 (8 marks)

The region

R

is bounded by

y=\ln x

, the

x

-axis and

x=e

, as shown. A vertical line

x=c

, where

1<c<e

, divides

R

into two regions of equal area.

The region under y=ln x from x=1 to x=e, above the x axis, is shaded.

Determine the exact area of R.

[2 marks]

Determine c, correct to three decimal places.

[2 marks]

The whole region R is rotated about the x-axis. Determine the percentage of the resulting solid's volume produced by the part of R with

1\leq x\leq c

, correct to one decimal place. Use the radii of the circular cross-sections to explain, without relying only on rounded calculations, why this percentage is less than 50%.

[4 marks]

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WORKED SOLUTION

Practice marking scheme

8 marks

ANSWER

(a)

1

square unit. (b)

c=2.156

. (c)

37.8\%

, which is strictly less than half.

Worked solution

(a) Integration by parts gives

\int\ln x\,dx=x\ln x-x+C.

Thus

A=[x\ln x-x]_1^e=1

square unit.

(b) Half the area lies to the left of

c

, so

c\ln c-c+1=\frac12,\quad\text{or}\quad c\ln c-c=-\frac12.

The left area is strictly increasing for

1<c<e

, so the solution is unique. A numerical solution gives

c=2.1555352035\ldots

, hence

c=2.156

\ln x

. A second use of integration by parts gives

\int(\ln x)^2\,dx=x(\ln x)^2-2\int\ln x\,dx

=x\bigl[(\ln x)^2-2\ln x+2\bigr]+C.

Writing

F(x)=x[(\ln x)^2-2\ln x+2]

, the whole volume is

\pi(e-2)

and the left volume is

\pi[F(c)-2]

. Using the unrounded value of

c

, the percentage is

100\frac{F(c)-2}{e-2}=37.8007168\ldots\%=37.8\%.

For a strict comparison, put

h=\ln c>0

. On

1<x<c

0<\ln x<h

, so

(\ln x)^2<h\ln x

. On

c<x<e

(\ln x)^2>h\ln x

. Since each planar region has area

1/2

\int_1^c(\ln x)^2\,dx<\frac h2<\int_c^e(\ln x)^2\,dx.

The left volume is therefore strictly smaller. Equal planar areas weight radius linearly; volumes weight radius quadratically.

For the strict comparison, regard volume as radius-weighted planar area: the left radius is always below ln c and the right radius above it, apart from their shared boundary. Technology may evaluate integrals, but supporting integral setup is needed.

c rounds to 2.156; percentage rounds to 37.8%. Do not use rounded c for the final ratio.

Part a: Use integration by parts to obtain x ln x - x.

[1 mark]

Part a: Evaluate the area as exactly 1 square unit.

[1 mark]

Part b: Set up the half-area equation.

[1 mark]

Part b: Solve on (1,e) and report c=2.156.

[1 mark]

Part c: Use squared radii to set up both relevant volume integrals or their ratio.

[1 mark]

Part c: Obtain the correct antiderivative of (ln x) squared, with supporting work.

[1 mark]

Part c: Use an unrounded divider to obtain 37.8%.

[1 mark]

Part c: Justify the strict inequality using the smaller/larger radii and equal planar areas; equivalent integral inequality accepted.

[1 mark]

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Integration techniques; applications of integral calculus — Question 151

QUESTION 151 (8 marks)

Practice marking scheme