Exercise 7.9 · Q9 (MCQ): $\int_{1/3}^1\frac{(x-x^3)^{1/3}}{x^4}\,dx$

Basics you need

Solution

1. Check the integrand is continuous on $\left[\frac{1}{3}, 1\right]$

   On this interval $x > 0$⁠, so $x^4 \ne 0$⁠. Also $x - x^3 = x(1 - x^2) \ge 0$ for $0 \le x \le 1$⁠, so the cube root is of a non-negative number. The cube root is continuous, and so is the quotient: the integrand is continuous on $\left[\frac{1}{3}, 1\right]$⁠.

   The only danger point is $x = 0$⁠, where $x^4 = 0$⁠; the interval starts at $\frac{1}{3}$⁠, safely away from it. Knowing $x - x^3 \ge 0$ also means every value of the integrand is $\ge 0$⁠.
   Continuity of quotients and of roots (Chapter 5, Section 5.2.1). Lesson: the fundamental theorem

2. Take $x^3$ out of the cube root

   For $x > 0$⁠, $x - x^3 = x^3\left(\frac{1}{x^2} - 1\right)$⁠, so

   $$(x - x^3)^{1/3} = x\left(\frac{1}{x^2} - 1\right)^{1/3}.$$

   Dividing by $x^4$⁠:

   $$\frac{(x - x^3)^{1/3}}{x^4} = \left(\frac{1}{x^2} - 1\right)^{1/3}\cdot\frac{1}{x^3}.$$
   Inside the root, $x^3\cdot\frac{1}{x^2} = x$ and $x^3\cdot 1 = x^3$⁠, so the factorisation is correct. Taking $x^3$ out (and not $x$⁠) is what makes a perfect cube leave the root as $x$⁠, and it creates a factor $\frac{1}{x^3}$⁠, which is the derivative shape we need next.
   $(x^3 \cdot A)^{1/3} = x\,A^{1/3}$⁠. Basics: powers and roots

3. Substitute $t = \frac{1}{x^2} - 1$ and change the limits

   Then $\frac{dt}{dx} = -\frac{2}{x^3}$⁠, so

   $$\frac{dx}{x^3} = -\frac{1}{2}\,dt.$$

   New limits. When $x = \frac{1}{3}$⁠, $t = 9 - 1 = 8$⁠. When $x = 1$⁠, $t = 1 - 1 = 0$⁠.

   $$\int_{\frac{1}{3}}^{1}\frac{(x - x^3)^{1/3}}{x^4}\,dx = \int_8^0 t^{1/3}\left(-\frac{1}{2}\right)dt.$$

   Why it is valid on the whole interval. $t = x^{-2} - 1$ has the continuous derivative $-\frac{2}{x^3}$ on $\left[\frac{1}{3}, 1\right]$ (no $x = 0$ there). The derivative is negative, so $t$ decreases steadily from 8 to 0, and $t^{1/3}$ is continuous on $[0, 8]$⁠.

   $\frac{d}{dx}x^{-2} = -2x^{-3}$⁠. The leftover factor $\frac{1}{x^3}\,dx$ is exactly $-\frac{1}{2}\,dt$⁠, so everything is now in $t$⁠.
   Power rule for derivatives: $\frac{d}{dx}x^{-2} = -2x^{-3}$⁠. Basics: substitution

4. Integrate $t^{1/3}$

   $$\begin{aligned}\int_8^0 t^{1/3}\left(-\tfrac{1}{2}\right)dt &= \frac{1}{2}\int_0^8 t^{1/3}\,dt \\ &= \frac{1}{2}\cdot\frac{3}{4}\Big[t^{4/3}\Big]_0^8 \\ &= \frac{3}{8}\left(8^{4/3} - 0\right) = \frac{3}{8}\cdot 16 = 6.\end{aligned}$$
   Swapping the limits 8 and 0 changes the sign and cancels the minus (property P1). Then $8^{4/3} = \left(8^{1/3}\right)^4 = 2^4 = 16$⁠.
   $\int t^{1/3}dt = \frac{t^{4/3}}{4/3} = \frac{3}{4}t^{4/3}$⁠. Basics: the power rule

5. Same result by staying in $x$

   Returning $-\frac{1}{2}\cdot\frac{3}{4}t^{4/3}$ to $x$⁠:

   $$F(x) = -\frac{3}{8}\left(\frac{1}{x^2} - 1\right)^{4/3}.$$

   Check: differentiate. Chain rule, inner $u = x^{-2} - 1$ with $u' = -2x^{-3}$⁠:

   $$\begin{aligned}F'(x) &= -\frac{3}{8}\cdot\frac{4}{3}\left(\frac{1}{x^2} - 1\right)^{1/3}\cdot\left(-\frac{2}{x^3}\right) \\ &= \left(\frac{1}{x^2} - 1\right)^{1/3}\cdot\frac{1}{x^3},\end{aligned}$$

   which is the integrand by Step 2. Then $F(1) - F\left(\frac{1}{3}\right) = 0 - \left(-\frac{3}{8}\cdot 8^{4/3}\right) = \frac{3}{8}\cdot 16 = 6$⁠.

   $\frac{3}{8}\cdot\frac{4}{3}\cdot 2 = 1$⁠, and the two minus signs cancel.
   Chain rule (Chapter 5, Section 5.3.1). Basics: the chain rule backwards

6. The region

   At $x = \frac{1}{3}$⁠: $\left(\frac{1}{3} - \frac{1}{27}\right)^{1/3} = \left(\frac{8}{27}\right)^{1/3} = \frac{2}{3}$⁠, and dividing by $\frac{1}{81}$ gives 54. The integrand is positive inside the interval, so the integral is a positive area.
   For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

7. Examine every option

   • (A) 6. This is the value found in Steps 4 and 5. Correct.
   • (B) 0. The integrand is positive on $\left(\frac{1}{3}, 1\right)$ (Step 1), so the integral is a positive area and cannot be 0. Wrong.
   • (C) 3. Half the true value; it is what you get by mistaking $8^{4/3}$ for 8. Wrong.
   • (D) 4. Not equal to 6; a numerical integration gives 6.0000000. Wrong.
   Exactly one option equals the computed value 6. The answer key of this reprint prints "D", but option (D) is 4. The exact value $\frac{3}{8}\cdot 16 = 6$ and the numerical value 6.0000000 agree, so the printed letter does not match the integral; the correct option is (A), and the key's "D" appears to be a misprint.
   An exact value plus an independent numerical check settles an MCQ. Lesson: definite integrals

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