Exercise 7.8 · Q20: $\int_0^1\left(x e^x + \sin\frac{\pi x}{4}\right)dx$

Basics you need

Solution

1. Check continuity on $[0, 1]$

   $x$⁠, $e^x$ and $\sin\frac{\pi x}{4}$ are continuous for every real $x$⁠; so are their product and sum. The integrand is continuous on $[0, 1]$⁠.

   Theorem 2 requires continuity on the closed interval. There are no denominators, roots or logs here.
   Sums, products and compositions of continuous functions are continuous (Chapter 5). Lesson: the fundamental theorem

2. $\int x\,e^x\,dx$ by parts

   Take the first function $x$ (algebraic) and the second function $e^x$ (exponential). Then $\frac{d}{dx}x = 1$ and $\int e^x\,dx = e^x$⁠, so

   $$\begin{aligned}\int x\,e^x\,dx &= x\cdot e^x - \int 1\cdot e^x\,dx \\ &= x\,e^x - e^x + C.\end{aligned}$$
   The rule of thumb (Remark (iii), Section 7.6, often remembered as ILATE: inverse trig, log, algebraic, trig, exponential) takes the power of $x$ as the first function, because differentiating it makes it simpler (1), while $e^x$ is easy to integrate. Taking $e^x$ first would give $\frac{x^2}{2}e^x - \int\frac{x^2}{2}e^x\,dx$⁠, which is harder. This is Example 19 of the book.
   Integration by parts, Section 7.6. Lesson: integration by parts

3. $\int\sin\frac{\pi x}{4}\,dx$ by substitution

   Put $u = \frac{\pi x}{4}$⁠, so $du = \frac{\pi}{4}dx$ and $dx = \frac{4}{\pi}du$⁠:

   $$\begin{aligned}\int\sin\frac{\pi x}{4}\,dx &= \frac{4}{\pi}\int\sin u\,du = -\frac{4}{\pi}\cos u \\ &= -\frac{4}{\pi}\cos\frac{\pi x}{4} + C.\end{aligned}$$

   Together, $F(x) = x\,e^x - e^x - \frac{4}{\pi}\cos\frac{\pi x}{4}$⁠.

   $\int\sin u\,du = -\cos u$⁠. Dividing by the inner constant $\frac{\pi}{4}$ means multiplying by $\frac{4}{\pi}$⁠.
   Standard formula (iii), Section 7.2: $\int\sin x\,dx = -\cos x + C$⁠. Lesson: integration by substitution

4. Check: differentiate $F$ back

   $$\begin{aligned}F'(x) &= (e^x + x e^x) - e^x + \frac{4}{\pi}\cdot\frac{\pi}{4}\sin\frac{\pi x}{4} \\ &= x\,e^x + \sin\frac{\pi x}{4}.\ \checkmark\end{aligned}$$
   Product rule on $x e^x$ gives $e^x + x e^x$⁠; the $-e^x$ cancels the extra $e^x$⁠. The chain rule on $\cos\frac{\pi x}{4}$ gives $-\frac{\pi}{4}\sin\frac{\pi x}{4}$⁠, and the two minus signs make a plus.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

5. Evaluate $F(1) - F(0)$

   First part: $\big[x e^x - e^x\big]_0^1 = (e - e) - (0 - 1) = 1$⁠.

   Second part:

   $$\begin{aligned}\Big[-\tfrac{4}{\pi}\cos\tfrac{\pi x}{4}\Big]_0^1 &= -\frac{4}{\pi}\cos\frac{\pi}{4} + \frac{4}{\pi}\cos 0 \\ &= -\frac{4}{\pi}\cdot\frac{\sqrt2}{2} + \frac{4}{\pi} \\ &= \frac{4}{\pi} - \frac{2\sqrt2}{\pi}.\end{aligned}$$

   Adding:

   $$I = 1 + \frac{4}{\pi} - \frac{2\sqrt2}{\pi} \approx 1.3729232.$$
   At the lower limit $x e^x - e^x = 0 - 1 = -1$⁠, not 0, so the first part is $0 - (-1) = 1$⁠. In decimals: $\frac{4}{\pi} \approx 1.2732395$ and $\frac{2\sqrt2}{\pi} \approx 0.9003163$⁠, so the second part is about 0.3729232.
   $\cos\frac{\pi}{4} = \frac{\sqrt2}{2}$ and $\frac{4}{\pi}\cdot\frac{\sqrt2}{2} = \frac{2\sqrt2}{\pi}$⁠. Basics: trigonometric values

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