Chapter 7 · Integrals

Exercise 7.1 · Q16: $\int (2x - 3\cos x + e^x)\,dx$

Exercise 7.1Integration as anti-differentiation · Section 7.2

The question

Find the following integral:

\int (2x - 3\cos x + e^x)\,dx

Exam answer

Let $I = \displaystyle\int\left(2x - 3\cos x + e^x\right)dx$ ⁠.

\begin{aligned} I &= 2\int x\,dx - 3\int\cos x\,dx \\ &\quad + \int e^x\,dx \\ &= 2\cdot\frac{x^2}{2} - 3\sin x + e^x + C \end{aligned}

$\therefore\ I = x^2 - 3\sin x + e^x + C$ ⁠.

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

The one idea this problem is built on

The integrand is a sum of three functions whose antiderivatives we already know from the derivative table: $x^2$ has derivative $2x$ ⁠, $\sin x$ has derivative $\cos x$ ⁠, and $e^x$ is its own derivative. Integration is linear (sums split up, constants come out), so we integrate each term on its own and add.

Given, to find, and the plan

Given: The integrand $f(x) = 2x - 3\cos x + e^x$ ⁠, defined for every real $x$ ⁠.
To find: $\int f(x)\,dx$ ⁠.
Plan: (1) Split into three integrals, constants in front. (2) Use the standard formulae for $x^n$ ⁠, $\cos x$ and $e^x$ ⁠. (3) Differentiate the answer to check it.

Basics you need

1Integration undoes differentiation

$\int f(x)\,dx = F(x) + C$ means $F'(x) = f(x)$ ⁠. Each standard integral is a derivative formula read backwards. Basics: integration undoes differentiation

2Three standard integrals

From the list in Section 7.2: (i) $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ ⁠; (ii) $\int \cos x\,dx = \sin x + C$ ⁠; (xi) $\int e^x\,dx = e^x + C$ ⁠. Basics: the power rule Basics: trigonometric facts for integrals

3Linearity

$\int (k_1 f_1 + k_2 f_2)\,dx$ $= k_1\!\int f_1\,dx + k_2\!\int f_2\,dx$ ⁠, and the same for three or more terms (Property V, Section 7.2.1). Lesson: integration as anti-differentiation

Solution

Split the integral (Property V)
$\begin{aligned} &\int (2x - 3\cos x + e^x)\,dx \\ &\quad = 2\int x\,dx - 3\int \cos x\,dx + \int e^x\,dx. \end{aligned}$
Property V is the rule that integration respects sums and constant multiples, with $k_1 = 2$ ⁠, $k_2 = -3$ ⁠, $k_3 = 1$ ⁠. Taking the constants out leaves three integrals that are each exactly a formula from the list. The minus sign belongs to the constant $-3$ ⁠, so it stays attached to the cosine term.

Property V, Section 7.2.1. Lesson: integration as anti-differentiation
Use the standard formulae
$\begin{aligned} 2\int x\,dx &= 2\cdot\frac{x^2}{2} = x^2, \\ -3\int \cos x\,dx &= -3\sin x, \\ \int e^x\,dx &= e^x. \end{aligned}$
Adding, with a single constant of integration,
$\begin{aligned} &\int (2x - 3\cos x + e^x)\,dx \\ &\quad = x^2 - 3\sin x + e^x + C. \end{aligned}$
Formula (i) with $n = 1$ gives $\frac{x^2}{2}$ ⁠, and the 2 in front cancels the 2 below. Formula (ii) holds because $\frac{d}{dx}\sin x = \cos x$ ⁠. Formula (xi) holds because $e^x$ is its own derivative. The three separate constants add up to one arbitrary constant $C$ ⁠.

Formulae (i), (ii) and (xi), Section 7.2. Basics: trigonometric facts for integrals
Check: differentiate the answer back
$\begin{aligned} &\frac{d}{dx}\left(x^2 - 3\sin x + e^x + C\right) \\ &\quad = 2x - 3\cos x + e^x + 0. \end{aligned}$
This is the integrand, for every real $x$ ⁠, so the answer is correct.

$\frac{d}{dx}x^2 = 2x$ ⁠, $\frac{d}{dx}\sin x = \cos x$ (so $-3\sin x$ gives $-3\cos x$ ⁠), $\frac{d}{dx}e^x = e^x$ ⁠, and a constant has derivative 0. All three functions are defined and differentiable for every real $x$ ⁠, so no domain restriction is needed.

Derivatives column of the list in Section 7.2; Property I, Section 7.2.1. Basics: integration undoes differentiation

Result

For all real $x$ ⁠,

\begin{aligned} &\int (2x - 3\cos x + e^x)\,dx \\ &\quad = x^2 - 3\sin x + e^x + C. \end{aligned}

(Textbook answer: $x^2 - 3\sin x + e^x + C$ ⁠.)

Check it with numbers

At $x = 1$ (radians) the integrand is $2 - 3\cos 1 + e$ $\approx 2 - 1.6209069 + 2.7182818$ $= 3.0973749$ ⁠. With $F(x) = x^2 - 3\sin x + e^x$ ⁠: $F(1.001) \approx 1.19696987$ and $F(0.999) \approx 1.19077512$ ⁠, so the slope of $F$ at 1 is about

\frac{1.19696987 - 1.19077512}{0.002} \approx 3.0974,

which matches the integrand value 3.0973749 to four decimal places.

(Each value of $F$ is rounded to 8 decimal places, so the slope is reliable to about 4 decimal places.)

Common mistakes

Sign of the cosine integral. $\int \cos x\,dx = +\sin x$ ⁠. (It is $\int \sin x\,dx$ that equals $-\cos x$ ⁠.)
$\int 2x\,dx = 2$ or $x^2/2$ ⁠. The correct answer is $x^2$ ⁠: raise the power, divide by 2, and the 2 in front cancels.
$\int e^x\,dx = \frac{e^{x+1}}{x+1}$ ⁠. The power formula is for $x^n$ with a constant exponent, not for $e^x$ ⁠, whose variable is in the exponent.

What to take away

Linearity (Property V) reduces a sum to a list of standard integrals.
Know formulae (i), (ii), (iii), (xi) both ways: $(\sin x)' = \cos x$ means $\int\cos x\,dx = \sin x + C$ ⁠.
Differentiate back to check the signs of trigonometric terms.

Every fact used, and where it comes from

Fact	Used in	Source
Integrate term by term, constants in front	Step 1	Property V, Section 7.2.1 (Lesson: integration as anti-differentiation)
$\int x^n dx = \frac{x^{n+1}}{n+1} + C$	Step 2	Formula (i), Section 7.2
$\int \cos x\,dx = \sin x + C$	Step 2	Formula (ii), Section 7.2
$\int e^x\,dx = e^x + C$	Step 2	Formula (xi), Section 7.2
$(x^2)' = 2x$ ⁠, $(\sin x)' = \cos x$ ⁠, $(e^x)' = e^x$	Step 3	Derivatives column, list in Section 7.2; Property I, Section 7.2.1

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Exercise 7.1 · Q16: ∫(2x−3cos⁡x+ex) dx\int (2x - 3\cos x + e^x)\,dx∫(2x−3cosx+ex)dx

Basics you need

Solution

Split the integral (Property V)

Use the standard formulae

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.1 · Q16: $\int (2x - 3\cos x + e^x)\,dx$