A reduction formula is a recurrence in disguise: $J_n = f(n) \cdot J_{n-1} + g(n)$, or sometimes $J_n$ in terms of $J_{n-2}$. Today you will derive recurrences for genuinely new integral families ($x^n e^{-x}$, $x^n \ln x$, $(\ln x)^n$), evaluate specific $J_n$ values by descent, and recognise the deep connection between recurrences and the reduction formulae of lesson 11.
+5 XP warm-up For $J_n = \int_0^1 x^n e^{-x}\,dx$, evaluate $J_0$ directly.
For $J_n = \int_0^1 x^n e^{-x}\,dx$, evaluate $J_0$ directly. Before checking apply parts once with $u = x^n$, $dv = e^{-x}dx$. Identify which boundary term survives (and which one vanishes). Sketch your derivation below.
+5 XP to read Recurrence problems reward two habits.
Recurrence problems reward two habits. Identify the right parts split: usually the polynomial-style factor ($x^n$ or $(\ln x)^n$) becomes $u$ so that differentiating drops the index. Then track the boundary term carefully unlike the $[0,\pi/2]$ Wallis case, the boundary may contribute a non-zero constant that appears in the recurrence as $g(n)$.
The differentiate-down read: (1) put the polynomial-style factor on $u$, (2) compute the boundary term, it may not vanish, (3) rearrange the result into the standard form $J_n = f(n)J_{n-1} + g(n)$ or $J_n = f(n)J_{n-2} + g(n)$.
Example: $J_n = \int_0^1 x^n e^{-x}dx \Rightarrow J_n = -e^{-1} + nJ_{n-1}$
Take $u = x^n$, $dv = e^{-x}dx$, so $du = nx^{n-1}dx$ and $v = -e^{-x}$.
Take $u = x^n$, $dv = e^{-x}dx$, so $du = nx^{n-1}dx$ and $v = -e^{-x}$. Then $$J_n = \bigl[-x^n e^{-x}\bigr]_0^1 + n\int_0^1 x^{n-1}e^{-x}\,dx.$$
Evaluate the boundary: at $x = 1$, $-x^n e^{-x} = -e^{-1}$; at $x = 0$, $-x^n e^{-x} = 0$ (provided $n \geq 1$). So $\bigl[-x^n e^{-x}\bigr]_0^1 = -e^{-1}$.
Therefore $\boxed{J_n = nJ_{n-1} - e^{-1}}$ for $n \geq 1$, with $J_0 = \int_0^1 e^{-x}dx = 1 - e^{-1}$.
Worked through the hook. $J_1 = 1 \cdot J_0 - e^{-1} = (1 - e^{-1}) - e^{-1} = 1 - 2e^{-1}$. Then $J_2 = 2J_1 - e^{-1} = 2(1 - 2e^{-1}) - e^{-1} = 2 - 5e^{-1}$. Each step uses the same one-line recurrence.
Take $u = x^n$, $dv = e^{-x}dx$; the polynomial goes on $u$ to differentiate down · Boundary $\bigl[-x^n e^{-x}\bigr]_0^1 = -e^{-1}$ (only the $x = 1$ end contributes for $n \geq 1$) · Recurrence: $J_n = nJ_{n-1} - e^{-1}$ for $n \geq 1$, with $J_0 = 1 - e^{-1}$ · Non-homogeneous recurrence, the forcing term comes from the surviving boundary value
Pause, copy the recurrence $J_n = nJ_{n-1}-e^{-1}$, its derivation ($u = x^n$, $dv = e^{-x}dx$, boundary $= -e^{-1}$), and the base $J_0 = 1-e^{-1}$ into your book.
Quick check: For $J_n = \int_0^1 x^n e^{-x}dx$ satisfying $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$, what is $J_2$?
We just saw the model recurrence $J_n = nJ_{n-1} - e^{-1}$ for $J_n = \int_0^1 x^n e^{-x}\,dx$, where the boundary term $-e^{-1}$ (from the $x=1$ end) creates a non-homogeneous forcing term.
We just saw the model recurrence $J_n = nJ_{n-1} - e^{-1}$ for $J_n = \int_0^1 x^n e^{-x}\,dx$, where the boundary term $-e^{-1}$ (from the $x=1$ end) creates a non-homogeneous forcing term. That raises a question: how does this non-homogeneous recurrence fit the broader framework of reduction formulas? This card answers it → homogeneous recurrences arise when the boundary term vanishes; single-step index drop is typical for $x^n \times$ exponential; two-step drop occurs when a Pythagorean identity is needed.
Every reduction formula from lesson 11 is a recurrence; every integral recurrence in this lesson is a reduction formula. The difference is rhetorical, not mathematical.
Choice between $J_{n-1}$ and $J_{n-2}$ depends on what comes out of the parts identity. Single-step recurrences (involving $J_{n-1}$) are typical when the integrand splits neatly into polynomial $\times$ elementary; two-step recurrences (involving $J_{n-2}$) are typical when a Pythagorean identity must be used to recover the same family on the right.
Reduction formula $=$ integral recurrence (same idea) · Homogeneous: boundary term vanishes; non-homogeneous: boundary term contributes $g(n)$ · Single-step ($J_{n-1}$) is typical for $x^n \times$ elementary integrands · Two-step ($J_{n-2}$) is typical when a Pythagorean identity is needed
Pause, copy the homogeneous vs non-homogeneous distinction (boundary vanishes or contributes $g(n)$), the single-step ($J_{n-1}$) vs two-step ($J_{n-2}$) rule, and the Pythagorean-identity trigger for two-step into your book.
Did you get this? True or false: the recurrence $I_n = \tfrac{n-1}{n}I_{n-2}$ for the Wallis integral is homogeneous (i.e., has no forcing term $g(n)$).
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Let $J_n = \int_0^1 x^n e^{-x}\,dx$ for $n \geq 1$. Show that $J_n = nJ_{n-1} - e^{-1}$.
Use the recurrence $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$ to evaluate $J_3 = \int_0^1 x^3 e^{-x}\,dx$.
Let $L_n = \int_0^1 x^n (1-x)^n\,dx$. Show $L_n = \tfrac{n}{n+1}\cdot L_{n-1}$ is wrong by direct check, then derive the correct relation linking $L_n$ to $L_{n-1}$.
Fill the gap: For $J_n = \int_0^1 x^n e^{-x}dx$, applying integration by parts with $u = $ gives the recurrence $J_n = nJ_{n-1} + g$, where the forcing term is $g = $ .
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: in the recurrence $J_n = nJ_{n-1} - e^{-1}$ for $J_n = \int_0^1 x^n e^{-x}dx$, the term $-e^{-1}$ arises from the boundary $[-x^n e^{-x}]_0^1$ evaluated at $x = 1$.
Activities · practice with the ideas
Activities · practice with the ideas
Use the recurrence $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$ to evaluate $J_4$.
Let $K_n = \int_1^e (\ln x)^n\,dx$ for $n \geq 1$. Derive the recurrence $K_n = e - nK_{n-1}$ and state $K_0$.
Using the recurrence from question 2, evaluate $K_2 = \int_1^e (\ln x)^2\,dx$.
For $K_n = \int_0^1 x^n \ln x\,dx$ ($n \geq 0$), use integration by parts to show $K_n = -\tfrac{1}{(n+1)^2}$. (Hint: $u = \ln x$, $dv = x^n dx$. Justify why the boundary at $0$ vanishes.)
Connect to lesson 11: rewrite the Wallis recurrence $I_n = \tfrac{n-1}{n}I_{n-2}$ in the standard recurrence form $J_n = f(n)J_{n-k} + g(n)$, identifying $f$, $k$ and $g$.
Odd one out: Three of these recurrences have a vanishing boundary term ($g = 0$). Which one does NOT?
Earlier you computed $J_0 = 1 - e^{-1}$ and tracked which boundary term survived when applying parts to $\int_0^1 x^n e^{-x}\,dx$.
The non-vanishing boundary at $x = 1$ is precisely what makes this recurrence non-homogeneous: it contributes the $-e^{-1}$ forcing term in $J_n = nJ_{n-1} - e^{-1}$. Compare with the lesson 11 Wallis case, where the boundary term vanishes at both endpoints and the recurrence is homogeneous. Recognising which type of recurrence you are deriving, homogeneous or non-homogeneous, is the single most useful diagnostic in Module 15.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Given $J_n = nJ_{n-1} - e^{-1}$ with $J_0 = 1 - e^{-1}$, evaluate $J_2 = \int_0^1 x^2 e^{-x}\,dx$. Show each step of the chain. (2 marks)
Q2. Let $J_n = \int_0^1 x^n e^{-x}\,dx$ for $n \geq 1$. Use integration by parts to derive $J_n = nJ_{n-1} - e^{-1}$. Justify the value of each boundary contribution. (3 marks)
Q3. For $K_n = \int_1^e (\ln x)^n\,dx$ with $n \geq 1$, show that $K_n = e - nK_{n-1}$. Hence evaluate $K_3$. (3 marks)
Activity answers:
1. $J_1 = 1 - 2e^{-1}$, $J_2 = 2 - 5e^{-1}$, $J_3 = 6 - 16e^{-1}$, $J_4 = 4J_3 - e^{-1} = 24 - 65e^{-1}$.
2. $K_0 = e - 1$. With $u = (\ln x)^n$, $dv = dx$: $K_n = [x(\ln x)^n]_1^e - n\int_1^e (\ln x)^{n-1}dx = e \cdot 1 - 0 - nK_{n-1} = e - nK_{n-1}$.
3. $K_1 = e - 1\cdot(e-1) = 1$; $K_2 = e - 2\cdot 1 = e - 2$.
4. With $u = \ln x$, $dv = x^n dx$: $K_n = \bigl[\tfrac{x^{n+1}\ln x}{n+1}\bigr]_0^1 - \tfrac{1}{n+1}\int_0^1 x^n\,dx = 0 - \tfrac{1}{(n+1)^2} = -\tfrac{1}{(n+1)^2}$. The boundary at $x = 0$ vanishes because $x^{n+1}\ln x \to 0$ as $x \to 0^+$.
5. $f(n) = \tfrac{n-1}{n}$, $k = 2$, $g(n) = 0$, homogeneous, two-step.
Q1 (2 marks): $J_1 = 1 - 2e^{-1}$ [1]; $J_2 = 2J_1 - e^{-1} = 2 - 5e^{-1}$ [1].
Q2 (3 marks): Parts with $u = x^n$, $dv = e^{-x}dx$ [1]; identify boundary $[-x^n e^{-x}]_0^1$, at $x = 1$ this is $-e^{-1}$, at $x = 0$ this is $0$ for $n \geq 1$ [1]; the remaining integral is $nJ_{n-1}$, giving $J_n = nJ_{n-1} - e^{-1}$ [1].
Q3 (3 marks): Derivation $K_n = e - nK_{n-1}$ as in activity 2 [1]; $K_0 = e-1$, $K_1 = 1$, $K_2 = e-2$ [1]; $K_3 = e - 3K_2 = e - 3(e-2) = 6 - 2e$ [1].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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