Exercise 7.9 · Q8: $\int_1^2\left(\frac{1}{x}-\frac{1}{2x^2}\right)e^{2x}dx$

Basics you need

Solution

1. Check the integrand is continuous on $[1, 2]$

   $\frac{1}{x}$ and $\frac{1}{2x^2}$ are continuous wherever $x \ne 0$⁠, and $e^{2x}$ is continuous everywhere. On $[1, 2]$⁠, $x \ge 1$⁠, so the integrand is continuous there.

   The only trouble point of the integrand is $x = 0$⁠, which lies outside $[1, 2]$⁠. (The figure shows the curve heading down towards $-\infty$ as $x \to 0^+$⁠, far from our interval.)
   Sums and products of continuous functions are continuous (Chapter 5, Section 5.2.1). Lesson: the fundamental theorem

2. Substitute $t = 2x$ and change the limits

   Then $dt = 2\,dx$⁠, so $dx = \frac{1}{2}dt$⁠, and $x = \frac{t}{2}$ gives

   $$\frac{1}{x} = \frac{2}{t}, \qquad \frac{1}{2x^2} = \frac{1}{2\cdot\frac{t^2}{4}} = \frac{2}{t^2}.$$

   New limits. When $x = 1$⁠, $t = 2$⁠. When $x = 2$⁠, $t = 4$⁠.

   $$\begin{aligned}\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2}\right)e^{2x}dx &= \int_2^4\left(\frac{2}{t} - \frac{2}{t^2}\right)e^t\cdot\frac{dt}{2} \\ &= \int_2^4 e^t\left(\frac{1}{t} - \frac{1}{t^2}\right)dt.\end{aligned}$$

   Why it is valid on the whole interval. $t = 2x$ is increasing with derivative 2 and carries $[1, 2]$ onto $[2, 4]$⁠, where $t \ne 0$⁠, so the new integrand is continuous.

   The exponent $2x$ makes the pattern hard to see. Calling it $t$ simplifies the exponential to $e^t$⁠, and the $\frac{1}{2}$ from $dx$ cancels the 2s that appear in $\frac{2}{t}$ and $\frac{2}{t^2}$⁠.
   $\frac{d}{dx}(2x) = 2$⁠. Basics: substitution

3. Recognise $e^t\left[f(t) + f'(t)\right]$

   Take $f(t) = \frac{1}{t}$⁠. Then $f'(t) = -\frac{1}{t^2}$⁠, so

   $$\frac{1}{t} - \frac{1}{t^2} = f(t) + f'(t).$$

   By Section 7.6.1,

   $$\int e^t\left(\frac{1}{t} - \frac{1}{t^2}\right)dt = \frac{e^t}{t} + C.$$
   Behind the rule is integration by parts: $\int e^t\frac{1}{t}\,dt = \frac{1}{t}e^t - \int\left(-\frac{1}{t^2}\right)e^t\,dt$ (first function $\frac{1}{t}$⁠, second $e^t$⁠). The second integral cancels the $-\frac{1}{t^2}e^t$ part exactly.
   Section 7.6.1: $\int e^x[f(x) + f'(x)]dx = e^x f(x) + C$⁠. Lesson: integration by parts

4. Evaluate and simplify

   $$\begin{aligned}\left[\frac{e^t}{t}\right]_2^4 &= \frac{e^4}{4} - \frac{e^2}{2} \\ &= \frac{e^4 - 2e^2}{4} = \frac{e^2(e^2 - 2)}{4}.\end{aligned}$$

   In decimals: $\frac{e^4}{4} \approx 13.6495375$ and $\frac{e^2}{2} \approx 3.6945280$⁠, so the value is about $9.9550095$⁠.

   Common denominator 4, then take $e^2$ out of $e^4 - 2e^2$ to match the book's factored form.
   Theorem 2: F(upper) minus F(lower); $e^4 = e^2\cdot e^2$⁠. Lesson: definite integrals

5. Same result by staying in $x$

   With $t = 2x$⁠, $\frac{e^t}{t}$ becomes

   $$F(x) = \frac{e^{2x}}{2x}.$$

   Check: differentiate (quotient rule):

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

   the integrand. Then $F(2) - F(1) = \frac{e^4}{4} - \frac{e^2}{2}$⁠, as before.

   $\frac{4x}{4x^2} = \frac{1}{x}$ and $\frac{2}{4x^2} = \frac{1}{2x^2}$⁠. The check is valid for every $x \ne 0$⁠, in particular on $[1, 2]$⁠.
   Quotient rule and $\frac{d}{dx}e^{2x} = 2e^{2x}$ (Chapter 5). Basics: the chain rule backwards

6. The region

   On $[1, 2]$ the factor $\frac{1}{x} - \frac{1}{2x^2} = \frac{2x - 1}{2x^2}$ is positive, so the value is an area. The curve rises from 3.69 to 20.47, and the area 9.955 lies between $1\times 3.69$ and $1\times 20.47$⁠.
   For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

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