Exercise 7.9 · Q10 (MCQ): $f(x)=\int_0^x t\sin t\,dt$⁠, find $f'(x)$

Basics you need

Solution

1. Check the integrand is continuous

   $t\sin t$ is a product of the continuous functions $t$ and $\sin t$⁠, so it is continuous for every real $t$⁠. Hence $f(x)$ is defined for every real $x$⁠.

   Theorem 1 requires a continuous integrand on the interval from the fixed limit to $x$⁠. Since $t\sin t$ is continuous everywhere, any interval will do.
   Products of continuous functions are continuous (Chapter 5, Section 5.2.1). Lesson: the fundamental theorem

2. Apply the first fundamental theorem

   Here $f$ is the area function $A(x) = \int_0^x g(t)\,dt$ with $g(t) = t\sin t$⁠. By Theorem 1,

   $$f'(x) = g(x) = x\sin x.$$
   Theorem 1 as printed uses $x$ in an interval $[a, b]$ with $a = 0$⁠, so it covers $x \ge 0$ directly. Step 3 computes $f$ explicitly and shows that $f'(x) = x\sin x$ holds for every real $x$⁠, negative ones included.
   Theorem 1, Section 7.8.2: the derivative of an area function is the integrand at the upper limit. Lesson: the fundamental theorem

3. Confirm by computing $f(x)$

   Integrate by parts with $u = t$ (algebraic, first by ILATE) and $dv = \sin t\,dt$⁠, so $du = dt$⁠, $v = -\cos t$⁠:

   $$\begin{aligned}\int t\sin t\,dt &= -t\cos t + \int\cos t\,dt \\ &= -t\cos t + \sin t + C.\end{aligned}$$

   By Theorem 2,

   $$\begin{aligned}f(x) &= \Big[\sin t - t\cos t\Big]_0^x \\ &= \sin x - x\cos x - (0 - 0) \\ &= \sin x - x\cos x.\end{aligned}$$

   Differentiate (product rule on $x\cos x$⁠):

   $$\begin{aligned}f'(x) &= \cos x - \left(\cos x - x\sin x\right) \\ &= x\sin x.\end{aligned}$$
   For $x > 0$ this is Theorem 2 on $[0, x]$⁠. For $x < 0$⁠, $\int_0^x$ means $-\int_x^0$ (property P1, Section 7.10), and Theorem 2 on $[x, 0]$ gives $-\big(F(0) - F(x)\big)$⁠, which is $F(x) - F(0)$⁠, the same formula; for $x = 0$ both sides are 0. So $f(x) = \sin x - x\cos x$ and $f'(x) = x\sin x$ for every real $x$⁠, which confirms Step 2 everywhere. Notice the $\cos x$ terms cancel; keeping track of that cancellation is exactly where options (A) and (D) go wrong.
   $\frac{d}{dx}(x\cos x) = \cos x - x\sin x$⁠. Lesson: integration by parts

4. See it on the graph

   Moving $x$ a little to the right adds a thin strip of width $h$ and height about $x\sin x$⁠; so $f(x + h) - f(x) \approx h\cdot x\sin x$⁠, and $f'(x) = x\sin x$⁠. At $x = 4$ the curve is below the axis, so $f$ is decreasing there even though $f(4)$ is still positive.
   Signed area: strips below the axis count as negative. Basics: area as a sum of thin strips

5. Examine every option

   Test at $x = \pi$⁠, where $f'(\pi) = \pi\sin\pi = 0$ (and $\sin\pi = 0$⁠, $\cos\pi = -1$⁠):

   • (A) $\cos\pi + \pi\sin\pi = -1 \ne 0$⁠. Wrong. (It keeps the $\cos x$ that cancels in Step 3.)
   • (B) $x\sin x$ is exactly $f'(x)$ for every $x$ (Steps 2 and 3). Correct.
   • (C) $\pi\cos\pi = -\pi \ne 0$⁠. Wrong. (It puts $\cos$ where $\sin$ belongs.)
   • (D) $\sin\pi + \pi\cos\pi = -\pi \ne 0$⁠. Wrong. (This is the derivative of $x\sin x$ itself, a step too far.)
   One value of $x$ where an option differs from $f'(x)$ rules it out. $x = \pi$ eliminates (A), (C) and (D) at once.
   Values: $\sin\pi = 0$⁠, $\cos\pi = -1$⁠. Basics: trigonometry

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