Practise › Questions › Reduction formulae

Reduction formulae questions

Integrate by parts once and the answer contains a smaller version of the same integral. Turn that into a recurrence and the whole family follows from one base case.

7 original questions · 34 marks · the reduction formulae notes · Further Pure 2

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 3 marks · use · warm-up

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

  Worked 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. That is exactly the move a reduction formula automates.

• 4 marks · warm-up

  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.

  Worked 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.

• 5 marks

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

  Worked 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. The I_n reappearing on the right is the point of the method; collect it on the left rather than treating it as a new integral.

• 6 marks

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

  Worked 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.

• 6 marks

  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.

  Worked 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.

• 3 marks · explain

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

  Worked 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.

• 7 marks · stretch

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

  Worked 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.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

Practise reduction formulae one question at a time

The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.