Questions on reduction formulae

Question 1

Use integration by parts to evaluate ∫ xe^x dx between 0 and 1.

Accepted Answer

Take u = x and dv = e^x dx, so the integral is [xe^x] − ∫e^x dx = [xe^x − e^x] from 0 to 1 = (e − e) − (0 − 1) = 1. M1 for parts with u = x, A1 for the integrated expression, A1 for 1. Differentiating the power rather than the exponential is the choice that makes progress, since it lowers the index by one.

Question 2

I_n = ∫ sinⁿx dx between 0 and π/2 satisfies nI_n = (n − 1)I_n−2. Given I₀ = π/2 and I₁ = 1, find I₄ and I₅.

Accepted Answer

I₄ = (3/4)I₂ = (3/4)(1/2)(π/2) = 3π/16, about 0.5890. I₅ = (4/5)I₃ = (4/5)(2/3)(1) = 8/15, about 0.5333. M1 for applying the formula twice to reach I₄, A1 for 3π/16, M1 for the odd chain, A1 for 8/15. Even n runs down to I₀ and keeps a π; odd n runs down to I₁ and does not.

Question 3

Write down the formula for integration by parts, and state which factor should be differentiated when a reduction formula for ∫xⁿe^x dx is being derived.

Accepted Answer

∫u (dv/dx) dx = uv − ∫v (du/dx) dx, taken between the same limits when the integral is definite. B1 for the formula. Take u = xⁿ, so that the power is the factor being differentiated: each application lowers the index by one and leaves an integral of the same shape with n − 1 in place of n, which is what a reduction formula needs. B1 for differentiating the power. Differentiating the exponential instead raises the index of the power and leaves a harder integral than the one started with.

Question 4

I_n = ∫ (ln x)ⁿ dx between 1 and e. Show that I_n = e − nI_n−1 for n ≥ 1.

Accepted Answer

There is only one factor in sight, so integrate by parts with u = (ln x)ⁿ and dv = 1 dx, giving v = x. Then du/dx = n(ln x)ⁿ⁻¹ × (1/x), and the x in v cancels that 1/x: I_n = [x(ln x)ⁿ] − ∫x × n(ln x)ⁿ⁻¹/x dx = [x(ln x)ⁿ] − n∫(ln x)ⁿ⁻¹ dx. M1 for parts with v = x.
The boundary term is e × 1ⁿ = e at x = e, since ln e = 1, and 1 × 0ⁿ = 0 at x = 1, since ln 1 = 0 and n ≥ 1. A1 for the boundary term e.
What remains is n times the same integral with the index lowered by one, so I_n = e − nI_n−1. A1 for the printed formula.
The base value is I₀ = ∫1 dx = e − 1, and the formula then gives I₁ = e − (e − 1) = 1, which one round of parts on ∫ln x dx confirms. The substitution x = e^t turns this integral into ∫tⁿe^t dt from 0 to 1, which is why the recurrence is the same one that integral satisfies.

Question 5

I_n = ∫ sinⁿx dx between 0 and π/2. Show that nI_n = (n − 1)I_n−2 for n ≥ 2.

Accepted Answer

Write the integrand as sinⁿ⁻¹x × sin x and integrate by parts, differentiating the first factor. The boundary term is [−sinⁿ⁻¹x cos x], which vanishes at both limits, since cos(π/2) = 0 and sin 0 = 0. What is left is (n − 1)∫sinⁿ⁻²x cos²x dx. Replacing cos²x by 1 − sin²x splits it into (n − 1)I_n−2 − (n − 1)I_n, so I_n + (n − 1)I_n = (n − 1)I_n−2, which is the result. M1 for splitting off sin x, M1 for integrating by parts, A1 for the boundary term vanishing, M1 for replacing cos²x, A1 for the printed result.

Question 6

I_n = ∫ xⁿe^x dx between 0 and 1. Show that I_n = e − nI_n−1, and hence find I₃.

Accepted Answer

Integrating by parts with u = xⁿ gives [xⁿe^x] − n∫xⁿ⁻¹e^x dx. The boundary term is e at x = 1 and 0 at x = 0, so I_n = e − nI_n−1. Starting from I₀ = e − 1, the chain runs I₁ = e − (e − 1) = 1, I₂ = e − 2, and I₃ = e − 3(e − 2) = 6 − 2e, about 0.5634. M1 for parts with u = xⁿ, A1 for the boundary term, A1 for the printed formula, M1 for I₀ = e − 1, dM1 for running the chain, A1 for 6 − 2e.

Question 7

I_n = ∫ tanⁿx dx between 0 and π/4. Show that I_n = 1/(n − 1) − I_n−2 for n ≥ 2, and find I₄.

Accepted Answer

Split off tan²x = sec²x − 1, giving ∫tanⁿ⁻²x sec²x dx − I_n−2. The first integral is [tanⁿ⁻¹x/(n − 1)], which is 1/(n − 1) at π/4 and 0 at 0. So I_n = 1/(n − 1) − I_n−2. With I₀ = π/4 this gives I₂ = 1 − π/4, and I₄ = 1/3 − (1 − π/4) = π/4 − 2/3, about 0.1187. M1 for tan²x = sec²x − 1, A1 for the split, M1 for integrating the first part, A1 for the printed result, M1 for I₀ = π/4 and the chain, A1 for π/4 − 2/3.

Question 8

Explain why a reduction formula on its own is not an answer, and how you choose the starting value.

Accepted Answer

The formula only relates one integral to another, so it generates a chain rather than a number. You need one integral evaluated directly to anchor it. The index drops by a fixed step, so follow the chain down. A step of 2 lands on I₀ for even n and I₁ for odd n, and a step of 1 always lands on I₀. B1 for the formula only linking one integral to another, B1 for needing one evaluated directly, B1 for choosing the starting value by the step size.

Question 9

I_n = ∫ xⁿcos x dx between 0 and π/2. Show that I_n = (π/2)ⁿ − n(n − 1)I_n−2 for n ≥ 2, and write down I₀ and I₁.

Accepted Answer

Integrate by parts with u = xⁿ and dv = cos x dx, so v = sin x: I_n = [xⁿsin x] − n∫xⁿ⁻¹sin x dx. The boundary term is (π/2)ⁿ × 1 at the top and 0 at the bottom. M1 for the first application of parts, A1 for the boundary term (π/2)ⁿ.
Now integrate the remaining integral by parts again, with u = xⁿ⁻¹ and v = −cos x: ∫xⁿ⁻¹sin x dx = [−xⁿ⁻¹cos x] + (n − 1)∫xⁿ⁻²cos x dx. Its boundary term vanishes at both limits, since cos(π/2) = 0 and, for n ≥ 2, xⁿ⁻¹ = 0 at x = 0. What is left is (n − 1)I_n−2. M1 for the second application of parts, A1 for the printed formula.
The base values are I₀ = ∫cos x dx = [sin x] = 1 and, by one round of parts, I₁ = [x sin x + cos x] = π/2 − 1. B1 for both. Two applications of parts are needed because a power against a trigonometric function returns to its own shape only after the sine has become a cosine again, which is why this formula drops the index by two rather than one.

Question 10

I_n = ∫ sinⁿx dx between 0 and π/2 and J_n = ∫ cosⁿx dx between the same limits. Show, using the substitution u = π/2 − x, that I_n = J_n for every integer n ≥ 0.

Accepted Answer

Put u = π/2 − x, so x = π/2 − u and dx = −du. M1 for the substitution with its differential. When x = 0, u = π/2, and when x = π/2, u = 0, so the limits swap over; the minus sign in dx swaps them back. A1 for handling the limits.
The integrand becomes sinⁿ(π/2 − u) = cosⁿu, so I_n = ∫ cosⁿu du between 0 and π/2, which is J_n with the variable renamed. A1 for the integrand, A1 for the conclusion. A definite integral does not depend on the letter used, so renaming u as x finishes it. The result says the two families obey the same reduction formula with the same base values, which is why I₀ = J₀ = π/2 and I₁ = J₁ = 1.

Question 11

I_n = ∫ xⁿe^−2x dx from 0 to infinity. Show that I_n = (n/2)I_n−1 for n ≥ 1, and hence find I₃.

Accepted Answer

By parts with u = xⁿ and v = −½e^−2x: I_n = [−½xⁿe^−2x] + (n/2)∫xⁿ⁻¹e^−2x dx. M1 for parts with u = xⁿ. The boundary term is 0 at the origin and tends to 0 as x grows, since the exponential decays faster than any power grows, so I_n = (n/2)I_n−1. A1 for the boundary term vanishing at both ends, A1 for the printed formula.
The base value is I₀ = ∫e^−2x dx = [−½e^−2x] = ½. Running the chain, I₁ = (1/2)(1/2) = 1/4, I₂ = 1 × 1/4 = 1/4, and I₃ = (3/2)(1/4) = 3/8. B1 for I₀ with the chain to 3/8. In general I_n = n!/2ⁿ⁺¹, and 3!/2⁴ = 6/16 = 3/8 agrees. The ½ from integrating e^−2x is what turns the usual factor n into n/2, so it has to be carried through both the boundary term and the remaining integral.

Question 12

I_n = ∫ xⁿe^−x dx from 0 to infinity. Show that I_n = nI_n−1, and hence evaluate I₅.

Accepted Answer

By parts with u = xⁿ, the boundary term [−xⁿe^−x] is 0 at the origin and tends to 0 at infinity, since the exponential beats any power. What remains is n∫xⁿ⁻¹e^−x dx = nI_n−1. Since I₀ = 1, repeated use gives I_n = n!, so I₅ = 120. M1 for parts with u = xⁿ, M1 for the boundary term at infinity, A1 for it vanishing, A1 for the printed formula, B1 for I₀ = 1, M1 for the chain, A1 for 120.

Question 13

I_n = ∫ secⁿx dx between 0 and π/4. Show that (n − 1)I_n = (√2)ⁿ⁻² + (n − 2)I_n−2 for n ≥ 2, and hence find I₄.

Accepted Answer

Split the integrand as secⁿ⁻²x × sec²x and integrate by parts with u = secⁿ⁻²x and dv = sec²x dx, so v = tan x. M1 for the split, M1 for parts with v = tan x.
The boundary term is [secⁿ⁻²x tan x], which is (√2)ⁿ⁻² × 1 at π/4 and 0 at 0, since sec(π/4) = √2 and tan 0 = 0. A1 for the boundary term. The derivative of secⁿ⁻²x is (n − 2)secⁿ⁻²x tan x, so what remains is −(n − 2)∫secⁿ⁻²x tan²x dx. Replacing tan²x by sec²x − 1 splits that into −(n − 2)I_n + (n − 2)I_n−2. M1 for the identity tan²x = sec²x − 1.
So I_n = (√2)ⁿ⁻² − (n − 2)I_n + (n − 2)I_n−2, and collecting the I_n terms on the left gives the printed result. A1.
The base value is I₂ = ∫sec²x dx = [tan x] = 1. At n = 4: 3I₄ = (√2)² + 2I₂ = 2 + 2 = 4, so I₄ = 4/3. A1.

Question 14

I_n = ∫ sinⁿx dx between 0 and π/2. State the values of n for which the reduction formula nI_n = (n − 1)I_n−2 holds, and explain what goes wrong if it is applied at n = 1.

Accepted Answer

The derivation writes the integrand as sinⁿ⁻¹x × sin x and integrates by parts, producing the boundary term [−sinⁿ⁻¹x cos x] between 0 and π/2. M1 for identifying the boundary term. At the upper limit cos(π/2) = 0, so that end contributes nothing whatever n is. At the lower limit sin 0 = 0, and sinⁿ⁻¹0 is zero only when n − 1 > 0. So the formula holds for n ≥ 2. B1 for the range of n.
At n = 1 the boundary term is [−cos x] between 0 and π/2, which is 0 − (−1) = 1, not 0, so the derivation breaks there. A1 for the value 1. The formula itself would give 1 × I₁ = 0 × I₋₁ = 0, whereas I₁ = ∫sin x dx = 1. B1 for the contradiction.
The quantity it would reduce to does not exist either: I₋₁ = ∫cosec x dx over the same interval does not converge, because cosec x behaves like 1/x near the origin. B1 for that.

Question 15

I_n = ∫ sinⁿx dx between 0 and π/2, with I₀ = π/2 and I₁ = 1, and nI_n = (n − 1)I_n−2 for n ≥ 2. Write I_2m as a product of fractions times π/2, and use that product together with the odd chain to evaluate I₆ and I₇.

Accepted Answer

Rearranged, I_n = ((n − 1)/n)I_n−2, so running down from an even index 2m multiplies I₀ by one fraction at each step: I_2m = ((2m − 1)/2m)((2m − 3)/(2m − 2))…(1/2) × π/2, the odd numbers up to 2m − 1 over the even numbers up to 2m, all times π/2. M1 for running the recurrence down, A1 for the even chain landing on I₀, A1 for the product form.
At m = 3: I₆ = (5/6)(3/4)(1/2)(π/2) = (15/48)(π/2) = 5π/32, about 0.4909. A1.
An odd index runs down to I₁ = 1 instead, so no π appears: I₇ = (6/7)(4/5)(2/3)(1) = 48/105 = 16/35, about 0.4571. M1 for the odd chain, A1 for 16/35.
The two values sit close together because I_n decreases steadily: sinⁿx ≤ sinⁿ⁻¹x across the whole interval, so the sequence falls, and the ratio of consecutive terms tends to 1.