Exercise 7.10 · Q7: $\int_0^1x(1-x)^n\,dx$

Basics you need

Solution

1. The integrand is continuous on $[0, 1]$

   For $0 \le x \le 1$⁠, $1 - x \ge 0$⁠, so $(1 - x)^n$ is defined and continuous for every $n \ge 0$ (when $n$ is a whole number, $x(1 - x)^n$ is simply a polynomial). Label

   $$I = \int_0^1 x(1 - x)^n\,dx \qquad (1)$$
   P4 and Theorem 2 are used for functions continuous on the closed interval. If $n$ were negative, $(1 - x)^n$ would blow up at $x = 1$⁠; for $-1 < n < 0$ the integral still converges (as an improper integral) and the final formula still holds, but that is beyond what the question intends.
   A power $u^n$ with $n \ge 0$ is continuous for $u \ge 0$⁠. Basics: powers of x

2. Apply P4 with $a = 1$

   Here $f(x) = x(1 - x)^n$⁠. Replace every $x$ by $1 - x$⁠:

   $$\begin{aligned}f(1 - x) &= (1 - x)\big(1 - (1 - x)\big)^n \\ &= (1 - x)\,x^n .\end{aligned}$$

   So, by P4,

   $$I = \int_0^1 (1 - x)\,x^n\,dx \qquad (2)$$
   P4 holds for every continuous function on $[0, a]$ (Step 1). It is useful here because $1 - (1 - x) = x$⁠: the awkward power moves from $(1 - x)$ onto $x$⁠, where the power rule can deal with it directly.
   P4, Section 7.10. The inner bracket: $1 - (1 - x) = 1 - 1 + x = x$⁠. Lesson: properties of definite integrals

3. Expand into powers of $x$

   $$(1 - x)\,x^n = x^n - x \cdot x^n = x^n - x^{n+1} .$$

   So

   $$I = \int_0^1\left(x^n - x^{n+1}\right)dx .$$
   The power rule handles one power of $x$ at a time, so we write the integrand as a difference of two single powers.
   $x \cdot x^n = x^{1 + n}$ (add the indices). Basics: powers of x

4. Integrate with the power rule

   $$\begin{aligned}I &= \left[\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}\right]_0^1 \\ &= \left(\frac{1}{n+1} - \frac{1}{n+2}\right) - (0 - 0) .\end{aligned}$$
   The power rule needs the index not equal to $-1$⁠: here the indices are $n$ and $n + 1$⁠, and with $n \ge 0$ neither is $-1$⁠. At the upper limit, $1$ to any power is 1. At the lower limit, $0^{n+1} = 0$ and $0^{n+2} = 0$ because $n + 1 > 0$⁠.
   $\int x^m\,dx = \frac{x^{m+1}}{m+1} + C$⁠, $m \ne -1$ (standard formula (i), Section 7.2); Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

5. Simplify

   $$\begin{aligned}I &= \frac{(n + 2) - (n + 1)}{(n + 1)(n + 2)} \\ &= \frac{1}{(n + 1)(n + 2)} .\end{aligned}$$
   Put both fractions over the common denominator $(n + 1)(n + 2)$⁠; the numerators differ by exactly 1. The answer is positive, as it should be: the integrand is $\ge 0$ on $[0, 1]$⁠.
   $\frac{1}{p} - \frac{1}{q} = \frac{q - p}{pq}$ with $p = n + 1$⁠, $q = n + 2$⁠. Basics: splitting fractions

6. Test small values of $n$⁠, and the picture

   $n = 0$⁠: $\int_0^1 x\,dx = \frac12$⁠, and the formula gives $\frac{1}{1 \cdot 2} = \frac12$⁠.

   $n = 1$⁠: $\int_0^1 (x - x^2)\,dx = \frac12 - \frac13 = \frac16$⁠, and the formula gives $\frac{1}{2 \cdot 3} = \frac16$⁠.

   $n = 3$ the long way: $x(1 - x)^3 = x - 3x^2 + 3x^3 - x^4$⁠, so

   $$\int_0^1 x(1 - x)^3\,dx = \tfrac12 - 1 + \tfrac34 - \tfrac15 = \tfrac{1}{20},$$

   and the formula gives $\frac{1}{4 \cdot 5} = \frac{1}{20} = 0.05$⁠.

   Small cases computed without P4 must agree with the general formula; they do. The $n = 3$ case shows the work P4 saves: the binomial expansion of $(1 - x)^n$ has $n + 1$ terms, while form (2) always has just two.
   P4 as a picture: replacing $x$ by $a - x$ reflects the graph in $x = \frac{a}{2}$⁠. Basics: symmetry and mirror images

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