Exercise 7.10 · Q19: $\int_0^a fg\,dx=2\int_0^a f\,dx$

Basics you need

Solution

1. Name the integral

   Let

   $$I = \int_0^{a} f(x)g(x)\,dx .$$

   Since $f$ and $g$ are continuous on $[0, a]$⁠, so is their product, and $I$ is a definite (finite) number.

   The question does not say so, but integrals over $[0, a]$ only make sense in this book for functions continuous on $[0, a]$ (Remark (iv) after Theorem 2). We need $I$ finite in Step 4, where we move $I$ from one side of an equation to the other.
   Products of continuous functions are continuous (Chapter 5, Section 5.2.1).

2. Apply P4 to the product

   P4 with $h(x) = f(x)g(x)$ replaces every $x$ by $a - x$⁠:

   $$I = \int_0^{a} f(a - x)\,g(a - x)\,dx .$$
   P4 holds for every continuous function on $[0, a]$⁠; the product $fg$ is one (Step 1). We use it because both given conditions are statements about $f(a - x)$ and $g(a - x)$⁠, so we want those expressions to appear.
   P4, Section 7.10. Lesson: properties

3. Use the two conditions

   The first condition gives $f(a - x) = f(x)$⁠. The second, rearranged, gives $g(a - x) = 4 - g(x)$⁠. Substituting both:

   $$\begin{aligned}I &= \int_0^{a} f(x)\,\big(4 - g(x)\big)\,dx \\ &= 4\int_0^{a} f(x)\,dx - \int_0^{a} f(x)g(x)\,dx \\ &= 4\int_0^{a} f(x)\,dx - I .\end{aligned}$$
   Both conditions hold for every $x$ in $[0, a]$⁠, so they can be substituted inside the integral at every point. Splitting the integral of a difference and taking out the constant 4 are the linearity properties of the integral. The last integral is the original $I$ again: the integral has reproduced itself.
   $g(x) + g(a - x) = 4$ is the same as $g(a - x) = 4 - g(x)$⁠. Linearity: Section 7.2.1, Properties (III) and (IV).

4. Solve for $I$

   Add $I$ to both sides of $I = 4\int_0^{a} f(x)\,dx - I$⁠:

   $$2I = 4\int_0^{a} f(x)\,dx ,$$

   and divide by 2:

   $$\int_0^{a} f(x)g(x)\,dx = 2\int_0^{a} f(x)\,dx .$$

   This is what we had to show. The proof used only P4 and the two conditions, so it holds for every such pair $f, g$ and every $a$⁠.

   The steps are reversible algebra on finite numbers (Step 1), so the identity follows. Note that nothing was assumed about $g$ beyond the stated condition: $g$ need not be symmetric itself.
   Solving $I = c - I$ for a number $I$⁠: $I = \frac{c}{2}$⁠.

5. Check the identity on two examples

   Example 1: $a = \pi$⁠, $f(x) = \sin x$⁠, $g(x) = 2 + \cos x$⁠. Conditions: $\sin(\pi - x) = \sin x$⁠, and $g(x) + g(\pi - x)$ $= 4 + \cos x - \cos x = 4$⁠. Then

   $$\begin{aligned}&\int_0^{\pi}\sin x\,(2 + \cos x)\,dx \\ &= 2\int_0^{\pi}\sin x\,dx + \left[\frac{\sin^2 x}{2}\right]_0^{\pi} \\ &= 2 \cdot 2 + 0 = 4,\end{aligned}$$

   and $2\int_0^{\pi}\sin x\,dx = 2 \cdot 2 = 4$⁠. Both sides equal 4.

   Example 2: $a = 3$⁠, $f(x) = x^2(3 - x)^2$⁠, $g(x) = 2 + e^x - e^{3 - x}$⁠. Conditions: $f(3 - x) = (3 - x)^2x^2 = f(x)$⁠, and $g(x) + g(3 - x) = 4$ because the exponentials cancel. Numerically, $\int_0^3 f(x)g(x)\,dx = 16.2$ and $2\int_0^3 f(x)\,dx = 2 \times 8.1 = 16.2$⁠.

   A general proof is complete on its own, but a numerical check catches algebra slips. In Example 1, $\int_0^{\pi}\sin x\cos x\,dx = 0$ is exactly the part of $g$ that is antisymmetric about $\frac{\pi}{2}$ cancelling out. In Example 2, $\int_0^3 x^2(3 - x)^2\,dx = \frac{3^5}{30} = 8.1$⁠.
   $\int\sin x\cos x\,dx = \frac{\sin^2 x}{2} + C$ (substitution $t = \sin x$⁠). Basics: substitution

6. The condition on $g$ is needed

   Keep $a = \pi$ and $f(x) = \sin x$⁠, but take $g(x) = x$⁠. Now $g(x) + g(\pi - x) = \pi$⁠, not 4. Integration by parts gives

   $$\int_0^{\pi} x\sin x\,dx = \Big[-x\cos x + \sin x\Big]_0^{\pi} = \pi ,$$

   while $2\int_0^{\pi}\sin x\,dx = 4$⁠. Since $\pi \ne 4$⁠, the identity fails without the condition.

   The same proof with $g(x) + g(a - x) = k$ gives $\int_0^a fg\,dx = \frac{k}{2}\int_0^a f\,dx$⁠. Here $k = \pi$⁠: $\frac{\pi}{2} \times 2 = \pi$⁠, which matches. The number 4 in the question is just the case $k = 4$⁠.
   Integration by parts (Section 7.6) with $x$ as the first function and $\sin x$ as the second (ILATE: algebraic before trigonometric), since $x$ becomes simpler when differentiated and $\int\sin x\,dx = -\cos x$⁠. Lesson: integration by parts

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