QCE Specialist Maths: Integration techniques Q252

Integration techniques — Question 252

Original QCE Vault practice · 13 marks

Q252 · Practice questionTechnology-activeVery complex unfamiliar13 marks

QUESTION 252 (13 marks)

For nonnegative integers

n

, define

I_n=\int_0^{\pi/2}\cos^{2n}x\,dx

. A solid ceramic bead is formed by rotating the region

0\le y\le\cos^2x

0\le x\le\pi/2

, about the

x

-axis. Its density is uniform. All angles are in radians.

The region beneath y=cos squared x from x=0 to pi/2, hatched above the x-axis, with a generic dashed vertical cutting line x=a. The line is schematic and does not reveal the answer.

Use integration by parts to establish

I_n=\frac{2n-1}{2n}I_{n-1}

for

n\ge1

[4 marks]

Prove by mathematical induction that

I_n=\frac{\pi}{2}\frac{(2n)!}{2^{2n}(n!)^2}

for all

n\ge0

[3 marks]

Determine the exact volume of the bead.

[2 marks]

A plane

x=a

divides the bead into two pieces of equal mass. Form an exact equation for

a

, determine

a

to four decimal places, and justify uniqueness. Do not replace equal mass with equal axial length.

[4 marks]

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WORKED SOLUTION

Practice marking scheme

13 marks

ANSWER

(a) The stated reduction formula. (b) The stated formula holds for every nonnegative integer. (c)

V=3\pi^2/16

cubic units. (d)

\frac{3a}{8}+\frac{\sin2a}{4}+\frac{\sin4a}{32}=\frac{3\pi}{32}

;

a=0.3142

Worked solution

(a) Take

u=\cos^{2n-1}x

and

dv=\cos x\,dx

. The boundary term

[\cos^{2n-1}x\sin x]_0^{\pi/2}

is zero. Therefore

I_n=(2n-1)\int_0^{\pi/2}\sin^2x\cos^{2n-2}x\,dx=(2n-1)(I_{n-1}-I_n)

. Rearranging gives

2nI_n=(2n-1)I_{n-1}

(b) For

n=0

I_0=\pi/2

, as required. Assume the formula for

n=k

. The recurrence gives

I_{k+1}=\frac{2k+1}{2k+2}\frac{\pi}{2}\frac{(2k)!}{2^{2k}(k!)^2}=\frac{\pi}{2}\frac{(2k+2)!}{2^{2k+2}((k+1)!)^2}

. Hence the induction step holds, proving the formula for all

n\ge0

\cos^2x

, so

V=\pi\int_0^{\pi/2}\cos^4x\,dx=\pi I_2

. The formula gives

I_2=3\pi/16

, hence

V=3\pi^2/16

(d) Uniform density makes equal mass equivalent to equal volume. Thus

\int_0^a\cos^4x\,dx=I_2/2=3\pi/32

. The identity

\cos^4x=(3+4\cos2x+\cos4x)/8

gives the stated equation. Numerical solution yields

a=0.3142176412\ldots

. The accumulated volume is strictly increasing on

0<a<\pi/2

, since its derivative is

\cos^4a>0

, so the solution is unique.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Choose valid parts and calculate the derivative.

[1 mark]

Part a: Evaluate the boundary term as zero.

[1 mark]

Part a: Use

\sin^2x=1-\cos^2x

to relate adjacent integrals.

[1 mark]

Part a: Rearrange to the reduction formula.

[1 mark]

Part b: Check the base case and state a clear induction hypothesis.

[1 mark]

Part b: Use the recurrence to derive the factorial expression for

k+1

[1 mark]

Part b: Conclude the result for every nonnegative integer.

[1 mark]

Part c: Use the square of the radius in the volume integral.

[1 mark]

Part c: Evaluate the exact volume.

[1 mark]

Part d: Form the half-volume condition.

[1 mark]

Part d: Integrate the fourth power to obtain the exact equation.

[1 mark]

Part d: Solve to the requested accuracy.

[1 mark]

Part d: Use positive cross-sectional area to justify uniqueness.

[1 mark]

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Integration techniques — Question 252

QUESTION 252 (13 marks)

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