Chapter 7 · Integrals

Exercise 7.7 · Q7: $\int\sqrt{1 + 3x - x^2}\,dx$

Exercise 7.7Integrals of some more types · Section 7.6.2

The question

Integrate the function

\sqrt{1 + 3x - x^2}.

Exam answer

Let $I = \displaystyle\int\sqrt{1 + 3x - x^2}\,dx$ ⁠.

\begin{aligned} &1 + 3x - x^2 \\ &= 1 - \left[\left(x - \frac32\right)^2 - \frac94\right] \\ &= \left(\frac{\sqrt{13}}{2}\right)^2 - \left(x - \frac32\right)^2 \end{aligned}

Put $x - \frac32 = y$ ⁠. Then $dx = dy$ ⁠.

I = \int\sqrt{\left(\frac{\sqrt{13}}{2}\right)^2 - y^2}\,dy

Using $\displaystyle\int\sqrt{a^2 - y^2}\,dy = \frac{y}{2}\sqrt{a^2 - y^2}$ $+ \frac{a^2}{2}\sin^{-1}\frac{y}{a} + C,$

\begin{aligned} I &= \frac{y}{2}\sqrt{\frac{13}{4} - y^2} \\ &\quad + \frac{13}{8}\sin^{-1}\frac{2y}{\sqrt{13}} + C \end{aligned}

$\therefore\ I = \dfrac{2x - 3}{4}\sqrt{1 + 3x - x^2}$ $+ \dfrac{13}{8}\sin^{-1}\left(\dfrac{2x - 3}{\sqrt{13}}\right)$ $+ C$ ⁠.

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

The one idea this problem is built on

The same method as Q5, now with fractions. Taking the minus sign out and completing the square gives

1 + 3x - x^2 = \frac{13}{4} - \left(x - \frac32\right)^2,

" $a^2$ minus a square" with $a = \frac{\sqrt{13}}{2}$ ⁠. Formula 7.6.2 (iii) then gives a $\sin^{-1}$ answer. The work is in keeping the fractions tidy.

Given, to find, and the plan

Given: The integrand $\sqrt{1 + 3x - x^2}$ ⁠.
To find: $\int\sqrt{1 + 3x - x^2}\,dx$ ⁠, and where the answer is valid.
Plan: (1) Complete the square. (2) Find the domain. (3) Put $y = x - \frac32$ ⁠. (4) Apply 7.6.2 (iii) with $a = \frac{\sqrt{13}}{2}$ ⁠. (5) Return to $x$ and simplify. (6) Differentiate back.

Basics you need

1Completing the square with $-x^2$

Take the minus sign out of the $x$ terms, complete the square inside, and multiply the minus sign back in. Half of $-3$ is $-\frac32$ ⁠, and $\left(-\frac32\right)^2 = \frac94$ ⁠. Basics: completing the square

2Formula 7.6.2 (iii)

$\int\sqrt{a^2 - y^2}\,dy = \frac{y}{2}\sqrt{a^2 - y^2} + \frac{a^2}{2}\sin^{-1}\frac{y}{a} + C$ ⁠, $a > 0$ ⁠; derived in Exercise 7.7 Q1. Lesson: integration by parts

3Derivative of $\sin^{-1}$

$\frac{d}{dx}\sin^{-1}u = \frac{u'}{\sqrt{1 - u^2}}$ for $|u| < 1$ (standard formula (viii), Section 7.2, with the chain rule). Basics: substitution and the chain rule

Solution

Complete the square
$\begin{aligned} 1 + 3x - x^2 &= 1 - (x^2 - 3x) \\ &= 1 - \left[\left(x - \tfrac32\right)^2 - \tfrac94\right] \\ &= \tfrac{13}{4} - \left(x - \tfrac32\right)^2. \end{aligned}$
$\left(x - \frac32\right)^2 = x^2 - 3x + \frac94$ ⁠, so $x^2 - 3x = \left(x - \frac32\right)^2 - \frac94$ ⁠. The minus sign in front turns $-\frac94$ into $+\frac94$ ⁠, and $1 + \frac94 = \frac{13}{4}$ ⁠. So $a^2 = \frac{13}{4}$ and $a = \frac{\sqrt{13}}{2}$ ⁠.

Completing the square, as in Example 24. Basics: completing the square
The domain

We need $\left(x - \frac32\right)^2 \le \frac{13}{4}$ ⁠, that is $\left|x - \frac32\right| \le \frac{\sqrt{13}}{2}$ ⁠:
$\frac{3 - \sqrt{13}}{2} \le x \le \frac{3 + \sqrt{13}}{2},$
roughly $-0.303 \le x \le 3.303$ ⁠.

These end points are the roots of $x^2 - 3x - 1 = 0$ ⁠, namely $\frac{3 \pm \sqrt{9 + 4}}{2}$ ⁠. Between them $1 + 3x - x^2 \ge 0$ ⁠.
Put $y = x - \frac32$

Then $dy = dx$ and
$\int\sqrt{1 + 3x - x^2}\,dx = \int\sqrt{\left(\tfrac{\sqrt{13}}{2}\right)^2 - y^2}\,dy.$
A shift has derivative 1, so $dx = dy$ ⁠. The root now has the shape $\sqrt{a^2 - y^2}$ ⁠.

Integration by substitution, Section 7.3.1. Basics: substitution
Apply 7.6.2 (iii) with $a = \frac{\sqrt{13}}{2}$

Here $\frac{a^2}{2} = \frac{13}{8}$ and $\frac{y}{a} = \frac{2y}{\sqrt{13}}$ ⁠:
$\begin{aligned} \int\sqrt{\tfrac{13}{4} - y^2}\,dy &= \frac{y}{2}\sqrt{\tfrac{13}{4} - y^2} \\ &\quad + \frac{13}{8}\sin^{-1}\frac{2y}{\sqrt{13}} + C. \end{aligned}$
$\frac{a^2}{2} = \frac{13/4}{2} = \frac{13}{8}$ ⁠. Dividing by $a = \frac{\sqrt{13}}{2}$ is the same as multiplying by $\frac{2}{\sqrt{13}}$ ⁠.

Formula 7.6.2 (iii). Lesson: integration by parts
Return to $x$ and simplify

With $y = x - \frac32$ ⁠: $\frac{y}{2} = \frac{2x - 3}{4}$ ⁠, $\frac{2y}{\sqrt{13}} = \frac{2x - 3}{\sqrt{13}}$ ⁠, and $\frac{13}{4} - y^2 = 1 + 3x - x^2$ ⁠. So
$\begin{aligned} \int\sqrt{1 + 3x - x^2}\,dx &= \frac{2x - 3}{4}\sqrt{1 + 3x - x^2} \\ &\quad + \frac{13}{8}\sin^{-1}\left(\frac{2x - 3}{\sqrt{13}}\right) + C. \end{aligned}$
$\frac{x - 3/2}{2} = \frac{2x - 3}{4}$ (multiply top and bottom by 2), and $2\left(x - \frac32\right) = 2x - 3$ ⁠. Writing the answer without fractions inside fractions matches the textbook's form.
Check: differentiate the answer

Write $y = x - \frac32$ ⁠, $r = \sqrt{\frac{13}{4} - y^2} = \sqrt{1 + 3x - x^2}$ ⁠, so $\frac{dr}{dx} = -\frac{y}{r}$ for $r > 0$ ⁠. Let $S = \frac{13}{8}\sin^{-1}\frac{2y}{\sqrt{13}}$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\left[\frac{y}{2}\,r\right] &= \frac{r}{2} - \frac{y^2}{2r}, \\ \frac{dS}{dx} &= \frac{13}{8}\cdot\frac{2/\sqrt{13}}{\sqrt{1 - 4y^2/13}} \\ &= \frac{13}{8}\cdot\frac{2}{2r} = \frac{13/4}{2r}. \end{aligned}$
Adding, over $2r$ ⁠:
$\frac{r^2 - y^2 + \frac{13}{4}}{2r} = \frac{2\left(\frac{13}{4} - y^2\right)}{2r} = r.\ \checkmark$
$\sqrt{13}\cdot\sqrt{1 - \frac{4y^2}{13}} = \sqrt{13 - 4y^2}$ and $\sqrt{13 - 4y^2} = 2\sqrt{\frac{13}{4} - y^2} = 2r$ ⁠. With $r^2 = \frac{13}{4} - y^2$ the numerator is $2r^2$ ⁠. Valid strictly inside the domain; at the end points the integrand is 0 and the answer's one-sided slope is 0 too.

Product rule; chain rule; standard formula (viii), Section 7.2. Basics: integration undoes differentiation

Result

\begin{aligned} \int\sqrt{1 + 3x - x^2}\,dx &= \frac{2x - 3}{4}\sqrt{1 + 3x - x^2} \\ &\quad + \frac{13}{8}\sin^{-1}\left(\frac{2x - 3}{\sqrt{13}}\right) + C, \end{aligned}

valid for $\frac{3 - \sqrt{13}}{2} \le x \le \frac{3 + \sqrt{13}}{2}$ ⁠. (Textbook answer: $\frac{(2x - 3)}{4}\sqrt{1 + 3x - x^2}$ $+ \frac{13}{8}\sin^{-1}\left(\frac{2x - 3}{\sqrt{13}}\right) + C$ ⁠, the same.)

Check it with numbers

By Section 7.8, $\int_a^b f(x)\,dx = F(b) - F(a)$ for our answer $F$ ⁠, on any interval where the integrand is continuous. The middle column is a computer's sum of very many thin strips, found without our answer.

Interval	Sum of thin strips	$F(b) - F(a)$
$[0, 3]$	$4.6940796$	$4.6940796$
$[-0.2, 1]$	$1.6214991$	$1.6214991$

Common mistakes

$a^2 = \frac{13}{4}$ but $a = \frac{13}{2}$ ⁠. The square root of $\frac{13}{4}$ is $\frac{\sqrt{13}}{2}$ ⁠.
Coefficient $\frac{13}{4}$ in front of $\sin^{-1}$ ⁠. It is $\frac{a^2}{2} = \frac{13}{8}$ ⁠.
Sign of the shift. $x^2 - 3x$ completes to $\left(x - \frac32\right)^2$ ⁠, so $y = x - \frac32$ ⁠, not $x + \frac32$ ⁠.

What to take away

Fractions do not change the method: take the minus out, complete the square, read off $a^2$ ⁠.
Clear nested fractions at the end: $\frac{x - 3/2}{2} = \frac{2x - 3}{4}$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Completing the square	Steps 1, 2	Section 7.4 (7); Example 24 (Basics: completing the square)
Substitution $y = x - \frac32$	Step 3	Section 7.3.1
$\int\sqrt{a^2 - y^2}\,dy$ formula	Step 4	Formula 7.6.2 (iii), Section 7.6.2
$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}$	Step 6	Standard formula (viii), Section 7.2
$\int_a^b f(x)\,dx = F(b) - F(a)$	Numerical check	Section 7.8 (Lesson: definite integrals)

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Exercise 7.7 · Q7: ∫1+3x−x2 dx\int\sqrt{1 + 3x - x^2}\,dx∫1+3x−x2​dx

Basics you need

Solution

Complete the square

The domain

Put y=x−32y = x - \frac32y=x−23​

Apply 7.6.2 (iii) with a=132a = \frac{\sqrt{13}}{2}a=213​​

Return to xxx and simplify

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.7 · Q7: $\int\sqrt{1 + 3x - x^2}\,dx$

Put $y = x - \frac32$

Apply 7.6.2 (iii) with $a = \frac{\sqrt{13}}{2}$

Return to $x$ and simplify