Chapter 7 · Integrals

Exercise 7.1 · Q3: An anti derivative of $e^{2x}$

Exercise 7.1Integration by inspection · Section 7.2

The question

Find an anti derivative (or integral) of the following function by the method of inspection.

e^{2x}

Exam answer

We know that $\frac{d}{dx}\left(e^{2x}\right) = 2e^{2x}$ ⁠.

\therefore\ e^{2x} = \frac{d}{dx}\left(\frac12e^{2x}\right)

$\therefore$ An anti derivative of $e^{2x}$ is $\dfrac12e^{2x}$ ⁠, and $\int e^{2x}\,dx = \dfrac12e^{2x} + C$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The exponential function is its own derivative: $\frac{d}{dx}e^x = e^x$ ⁠. So $e^{2x}$ is the obvious guess, but the chain rule makes its derivative $2e^{2x}$ ⁠, twice too big. Dividing by $2$ gives the anti derivative $\frac12 e^{2x}$ ⁠.

Given, to find, and the plan

Given: $f(x) = e^{2x}$ ⁠, defined for every real $x$ ⁠.
To find: One function $F$ with $F'(x) = e^{2x}$ for every real $x$ ⁠.
Plan: (1) Guess $e^{2x}$ from $\frac{d}{dx}e^x = e^x$ ⁠. (2) Differentiate the guess (chain rule). (3) Remove the extra factor $2$ ⁠. (4) Differentiate back to check. (5) Write the family with $+\,C$ ⁠.

Basics you need

1Anti derivative

$F$ is an anti derivative of $f$ if $F' = f$ on the interval. Inspection means finding such an $F$ from derivatives we already know. Basics: integration undoes differentiation

2The exponential is its own derivative

$\frac{d}{dx}(e^x) = e^x$ (Theorem 5, Section 5.4 of Chapter 5), so $\int e^x\,dx = e^x + C$ ⁠, item (xi) of the standard formulae in Section 7.2. Basics: integration undoes differentiation

3The chain rule

$\frac{d}{dx}e^{u(x)} = e^{u(x)}\,u'(x)$ ⁠. With $u = 2x$ the extra factor is $u' = 2$ ⁠. Basics: the chain rule backwards

Solution

Make a first guess

Because $\frac{d}{dx}e^x = e^x$ ⁠, the exponential reproduces itself under differentiation. So the natural guess for $e^{2x}$ is $e^{2x}$ itself.

The only basic function that differentiates to an exponential is the exponential. We keep the exponent $2x$ and adjust only a constant factor afterwards.

Item (xi) of the standard formulae, Section 7.2: $\int e^x\,dx = e^x + C$ ⁠. Basics: reading derivatives backwards
Differentiate the guess (chain rule)

Inner $u = 2x$ ⁠, $\frac{du}{dx} = 2$ ⁠; outer $e^u$ ⁠, whose derivative with respect to $u$ is $e^u$ ⁠:
$\begin{aligned} \frac{d}{dx}\left(e^{2x}\right) &= e^{u}\cdot\frac{du}{dx} \\ &= 2e^{2x}. \end{aligned}$
Twice too big.

The factor $2$ is the derivative of the exponent $2x$ ⁠. Note what does not happen: the exponent does not drop by one as in the power rule. $e^{2x}$ is an exponential (variable in the exponent), not a power of $x$ ⁠.

Chain rule, Theorem 4, Section 5.3.1; Theorem 5 (1), Section 5.4 (Chapter 5). Basics: the chain rule backwards
Remove the extra factor 2
$\begin{aligned} \frac{d}{dx}\left(\tfrac12 e^{2x}\right) &= \tfrac12\cdot 2e^{2x} \\ &= e^{2x}. \end{aligned}$
So $\frac12 e^{2x}$ is an anti derivative of $e^{2x}$ .

Multiplying by the constant $\frac12$ multiplies the derivative by $\frac12$ ⁠, which cancels the $2$ exactly.

Constant factor rule for derivatives (Section 5.3); the same move as Example 1 (i). Example 1
Check: differentiate the answer back

Let $F(x) = \frac12 e^{2x}$ ⁠. Then
$\begin{aligned} F'(x) &= \tfrac12\cdot e^{2x}\cdot 2 \\ &= e^{2x}. \;\checkmark \end{aligned}$
$F'$ equals the given function for every real $x$ ⁠, which is exactly what "anti derivative" means.

Property (I), Section 7.2.1. Basics: integration undoes differentiation
All anti derivatives: add the constant
$\int e^{2x}\,dx = \tfrac12 e^{2x} + C, \quad C \in \R.$
Any constant added to $\frac12 e^{2x}$ has derivative $0$ ⁠, so the family $\frac12 e^{2x} + C$ consists of anti derivatives, and by the Remark in Section 7.2 there are no others.

Remark in Section 7.2. Lesson: integration as anti differentiation

Result

An anti derivative of $e^{2x}$ is $\frac12 e^{2x}$ ⁠, valid for every real $x$ ⁠; in general

\int e^{2x}\,dx = \tfrac12 e^{2x} + C.

(Textbook answer: $\frac12 e^{2x}$ ⁠.)

Check it with numbers

At $x = 0.5$ with $h = 0.001$ ⁠, for $F(x) = \frac12 e^{2x}$ ⁠:

\begin{aligned} &\frac{F(0.501) - F(0.499)}{0.002} \\ &= \frac{1.361861916 - 1.356425349}{0.002} \\ &\approx 2.718284. \end{aligned}

The integrand there is $e^{1} \approx 2.718282$ ⁠. They agree to five decimal places (the tiny gap comes from the step $h$ ⁠, and shrinks as $h$ does).

Common mistakes

Treating $e^{2x}$ as a power. Writing $\frac{e^{2x+1}}{2x+1}$ uses the power rule, which is for $x^n$ with a fixed exponent. Here the exponent is the variable.
Multiplying by 2. $2e^{2x}$ is the derivative of $e^{2x}$ ⁠, not its integral.
Forgetting the factor altogether. $e^{2x}$ differentiates to $2e^{2x}$ ⁠, so it is not an anti derivative of $e^{2x}$ ⁠.

What to take away

For a constant $a \ne 0$ ⁠: $\int e^{ax}\,dx = \frac1a e^{ax} + C$ ⁠.
Exponentials keep their form under differentiation and integration; only the constant factor changes.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}e^x = e^x$	Steps 1, 2, 4	Theorem 5 (1), Section 5.4 (Chapter 5); standard formulae, item (xi), Section 7.2
Chain rule	Steps 2, 4	Theorem 4, Section 5.3.1 (Chapter 5)
$\frac{d}{dx}(kF) = kF'$	Step 3	Algebra of derivatives, Section 5.3 (product rule with a constant factor)
Differentiation and integration undo each other	Step 4	Property (I), Section 7.2.1
Anti derivatives differ by a constant	Step 5	Remark, Section 7.2 (Lesson: anti differentiation)

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Exercise 7.1 · Q3: An anti derivative of e2xe^{2x}e2x

Basics you need

Solution

Make a first guess

Differentiate the guess (chain rule)

Remove the extra factor 2

Check: differentiate the answer back

All anti derivatives: add the constant

Every fact used, and where it comes from

Exercise 7.1 · Q3: An anti derivative of $e^{2x}$