Exercise 7.1 · Q4: An anti derivative of $(ax+b)^2$

Basics you need

Solution

1. Guess $(ax+b)^3$ and differentiate it

   Inner function $u = ax + b$⁠, with $\frac{du}{dx} = a$⁠; outer function $u^3$⁠, with derivative $3u^2$⁠. By the chain rule,

   $$\begin{aligned} \frac{d}{dx}(ax+b)^3 &= 3u^2\cdot\frac{du}{dx} \\ &= 3(ax+b)^2\cdot a \\ &= 3a\,(ax+b)^2. \end{aligned}$$

   The guess gives $(ax+b)^2$ multiplied by the constant $3a$⁠.

   The power rule for integrals says a square comes from a cube, so the cube of the same block $ax+b$ is the natural candidate. Differentiating it shows two extra factors: $3$ from bringing the power down, and $a$ from the inner function $ax + b$⁠.
   Power rule with the chain rule: $\frac{d}{dx}u^3 = 3u^2u'$ (Table 5.3 and Theorem 4, Section 5.3.1 of Chapter 5). Basics: the chain rule backwards

2. Divide by $3a$ (when $a \ne 0$⁠)

   If $a \ne 0$⁠, multiply the guess by the constant $\frac{1}{3a}$⁠:

   $$\frac{d}{dx}\left[\frac{(ax+b)^3}{3a}\right] = \frac{3a(ax+b)^2}{3a} = (ax+b)^2.$$

   So, for $a \ne 0$⁠, $\dfrac{(ax+b)^3}{3a}$ is an anti derivative of $(ax+b)^2$.

   A constant factor passes through differentiation, so multiplying by $\frac{1}{3a}$ divides the derivative by $3a$ and cancels both extra factors at once. We may divide by $3a$ only because $a \ne 0$⁠; that is why this case is stated separately.
   Constant factor rule for derivatives (Section 5.3). Basics: the power rule

3. Check: differentiate the answer back

   Let $F(x) = \frac{1}{3a}(ax+b)^3$ with $a \ne 0$⁠. Then

   $$\begin{aligned} F'(x) &= \frac{1}{3a}\cdot 3(ax+b)^2\cdot a \\ &= (ax+b)^2. \;\checkmark \end{aligned}$$
   The derivative equals the given function for every real $x$⁠, so $F$ is an anti derivative on the whole real line.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

4. The case $a = 0$

   If $a = 0$⁠, the function is the constant $(0\cdot x + b)^2 = b^2$⁠, and the formula of Step 2 cannot be used (it would divide by $0$⁠). A constant $k$ has the anti derivative $kx$⁠, so

   $$\frac{d}{dx}\left(b^2x\right) = b^2, \quad\text{an anti derivative is } b^2x.$$
   The question does not say $a \ne 0$⁠, and the textbook answer $\frac{1}{3a}(ax+b)^3$ silently assumes it. Covering $a = 0$ makes the answer complete: in that case $ax + b$ does not depend on $x$ at all.
   $\int k\,dx = kx + C$⁠: item (i) of the standard formulae ($\int dx = x + C$⁠) with Property (IV), Section 7.2.1. Basics: the power rule

5. Another route: expand first, then compare

   Expanding the square gives a polynomial, which we can integrate term by term with the power rule:

   $$\begin{aligned} (ax+b)^2 &= a^2x^2 + 2abx + b^2, \\ \int (ax+b)^2\,dx &= \frac{a^2x^3}{3} + abx^2 + b^2x + C. \end{aligned}$$

   This looks different from Step 2, but for $a \ne 0$ the two differ only by a constant:

   $$\begin{aligned} \frac{(ax+b)^3}{3a} &= \frac{a^3x^3 + 3a^2bx^2 + 3ab^2x + b^3}{3a} \\ &= \frac{a^2x^3}{3} + abx^2 + b^2x + \frac{b^3}{3a}. \end{aligned}$$
   Two anti derivatives of the same function on an interval can differ only by a constant. Here the constant is $\frac{b^3}{3a}$⁠, which is absorbed into $C$⁠. The expanded form has the advantage that it also covers $a = 0$⁠: it then reduces to $b^2x + C$⁠, agreeing with Step 4.
   $(p+q)^2 = p^2 + 2pq + q^2$ and $(p+q)^3 = p^3 + 3p^2q$ $+\, 3pq^2 + q^3$⁠; Property (V), Section 7.2.1, for integrating term by term. Lesson: integration as anti differentiation

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