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Chapter 7 · Integrals

Exercise 7.7 · Q5: ∫1−4x−x2 dx\int\sqrt{1 - 4x - x^2}\,dx

Exercise 7.7Integrals of some more types · Section 7.6.2
The question

Integrate the function

1−4x−x2.\sqrt{1 - 4x - x^2}.
Exam answer

Let I=∫1−4x−x2 dxI = \displaystyle\int\sqrt{1 - 4x - x^2}\,dx⁠.

1−4x−x2=1−[(x+2)2−4]=(5)2−(x+2)2\begin{aligned} 1 - 4x - x^2 &= 1 - \left[(x + 2)^2 - 4\right] \\ &= \left(\sqrt5\right)^2 - (x + 2)^2 \end{aligned}

Put x+2=yx + 2 = y⁠. Then dx=dydx = dy⁠.

I=∫(5)2−y2 dyI = \int\sqrt{\left(\sqrt5\right)^2 - y^2}\,dy

Using ∫a2−y2 dy=y2a2−y2\displaystyle\int\sqrt{a^2 - y^2}\,dy = \frac{y}{2}\sqrt{a^2 - y^2} +a22sin⁡−1ya+C,+ \frac{a^2}{2}\sin^{-1}\frac{y}{a} + C,

I=y25−y2+52sin⁡−1y5+C\begin{aligned} I &= \frac{y}{2}\sqrt{5 - y^2} \\ &\quad + \frac52\sin^{-1}\frac{y}{\sqrt5} + C \end{aligned}

∴ I=x+221−4x−x2\therefore\ I = \dfrac{x + 2}{2}\sqrt{1 - 4x - x^2} +52sin⁡−1(x+25)+C+ \dfrac52\sin^{-1}\left(\dfrac{x + 2}{\sqrt5}\right) + C⁠.

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

The one idea this problem is built on

When x2x^2 comes with a minus sign, take the minus sign out before completing the square:

1−4x−x2=5−(x+2)2.1 - 4x - x^2 = 5 - (x + 2)^2.

That is "a2a^2 minus a square" with a=5a = \sqrt5⁠, so after y=x+2y = x + 2 formula 7.6.2 (iii) applies and the answer contains sin⁡−1\sin^{-1}⁠, just as in Q1 and in the textbook's Example 24.

Given, to find, and the plan
Given
The integrand 1−4x−x2\sqrt{1 - 4x - x^2}⁠.
To find
∫1−4x−x2 dx\int\sqrt{1 - 4x - x^2}\,dx⁠, and where the answer is valid.
Plan
(1) Complete the square with the minus sign taken out. (2) Find the domain. (3) Put y=x+2y = x + 2⁠. (4) Apply 7.6.2 (iii) with a=5a = \sqrt5 and return to xx⁠. (5) Differentiate back.

Basics you need

1Completing the square with −x2-x^2

Write −x2−4x=−(x2+4x)-x^2 - 4x = -(x^2 + 4x)⁠, complete the square inside the bracket, then multiply the minus sign back in. Basics: completing the square

2Formula 7.6.2 (iii)

∫a2−y2 dy=y2a2−y2+a22sin⁡−1ya+C\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>0a > 0⁠; derived by parts in Exercise 7.7 Q1. Lesson: integration by parts

3Derivative of sin⁡−1\sin^{-1}

ddxsin⁡−1u=u′1−u2\frac{d}{dx}\sin^{-1}u = \frac{u'}{\sqrt{1 - u^2}} for ∣u∣<1|u| < 1 (standard formula (viii), Section 7.2, with the chain rule). Basics: substitution and the chain rule

Solution

  1. Complete the square

    1−4x−x2=1−(x2+4x)=1−[(x+2)2−4]=5−(x+2)2.\begin{aligned} 1 - 4x - x^2 &= 1 - (x^2 + 4x) \\ &= 1 - \left[(x + 2)^2 - 4\right] \\ &= 5 - (x + 2)^2. \end{aligned}
    Inside the bracket, half the coefficient of xx is 2, so x2+4x=(x+2)2−4x^2 + 4x = (x + 2)^2 - 4⁠. The minus sign in front of the bracket turns −4-4 into +4+4⁠, and 1+4=51 + 4 = 5⁠. Check by expanding: 5−(x2+4x+4)=1−4x−x25 - (x^2 + 4x + 4) = 1 - 4x - x^2⁠. The square now has the minus sign, so the shape is a2−y2a^2 - y^2 with a2=5a^2 = 5⁠.
    Completing the square, as in Example 24. Basics: completing the square

  2. The domain

    We need 5−(x+2)2≥05 - (x + 2)^2 \ge 0⁠, that is ∣x+2∣≤5|x + 2| \le \sqrt5⁠:

    −2−5≤x≤−2+5,-2 - \sqrt5 \le x \le -2 + \sqrt5,

    roughly −4.236≤x≤0.236-4.236 \le x \le 0.236⁠.

    (x+2)2≤5(x + 2)^2 \le 5 means x+2x + 2 lies between −5-\sqrt5 and 5\sqrt5⁠. Outside that interval the integrand is not real.

  3. Put y=x+2y = x + 2

    Then dy=dxdy = dx and

    ∫1−4x−x2 dx=∫(5)2−y2 dy.\int\sqrt{1 - 4x - x^2}\,dx = \int\sqrt{\left(\sqrt5\right)^2 - y^2}\,dy.
    The shift has derivative 1, so dx=dydx = dy⁠. The new variable runs over −5≤y≤5-\sqrt5 \le y \le \sqrt5⁠.
    Integration by substitution, Section 7.3.1. Basics: substitution

  4. Apply 7.6.2 (iii) with a=5a = \sqrt5 and return to xx

    ∫5−y2 dy=y25−y2+52sin⁡−1y5+C.\begin{aligned} \int\sqrt{5 - y^2}\,dy &= \frac{y}{2}\sqrt{5 - y^2} \\ &\quad + \frac52\sin^{-1}\frac{y}{\sqrt5} + C. \end{aligned}

    With y=x+2y = x + 2 and 5−y2=1−4x−x25 - y^2 = 1 - 4x - x^2⁠:

    ∫1−4x−x2 dx=x+221−4x−x2+52sin⁡−1(x+25)+C.\begin{aligned} \int\sqrt{1 - 4x - x^2}\,dx &= \frac{x + 2}{2}\sqrt{1 - 4x - x^2} \\ &\quad + \frac52\sin^{-1}\left(\frac{x + 2}{\sqrt5}\right) + C. \end{aligned}
    a2=5a^2 = 5 gives the coefficient a22=52\frac{a^2}{2} = \frac52⁠, and a=5a = \sqrt5 goes inside sin⁡−1\sin^{-1}⁠. On the domain, ∣x+25∣≤1\left|\frac{x + 2}{\sqrt5}\right| \le 1⁠, so sin⁡−1\sin^{-1} is defined everywhere it is needed.
    Formula 7.6.2 (iii). Lesson: integration by parts

  5. Check: differentiate the answer

    Write y=x+2y = x + 2⁠, r=5−y2r = \sqrt{5 - y^2}⁠, so drdx=−yr\frac{dr}{dx} = -\frac{y}{r} for r>0r > 0⁠. Then

    ddx[y2 r]=r2−y22r,ddx[52sin⁡−1y5]=52⋅1/51−y2/5=52r.\begin{aligned} \frac{d}{dx}\left[\frac{y}{2}\,r\right] &= \frac{r}{2} - \frac{y^2}{2r}, \\ \frac{d}{dx}\left[\frac52\sin^{-1}\frac{y}{\sqrt5}\right] &= \frac52\cdot\frac{1/\sqrt5}{\sqrt{1 - y^2/5}} \\ &= \frac{5}{2r}. \end{aligned}

    Adding, over 2r2r⁠:

    r2−y2+52r=2(5−y2)2r=r=1−4x−x2. ✓\begin{aligned} \frac{r^2 - y^2 + 5}{2r} &= \frac{2(5 - y^2)}{2r} \\ &= r = \sqrt{1 - 4x - x^2}.\ \checkmark \end{aligned}
    5⋅1−y25=5−y2=r\sqrt5\cdot\sqrt{1 - \frac{y^2}{5}} = \sqrt{5 - y^2} = r⁠, which turns the sin⁡−1\sin^{-1} derivative into 52r\frac{5}{2r}⁠. With r2=5−y2r^2 = 5 - y^2 the numerator is 2r22r^2⁠. This holds strictly inside the domain; at the two 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 ∫1−4x−x2 dx=x+221−4x−x2+52sin⁡−1(x+25)+C,\begin{aligned} \int\sqrt{1 - 4x - x^2}\,dx &= \frac{x + 2}{2}\sqrt{1 - 4x - x^2} \\ &\quad + \frac52\sin^{-1}\left(\frac{x + 2}{\sqrt5}\right) + C, \end{aligned}

valid for −2−5≤x≤−2+5-2 - \sqrt5 \le x \le -2 + \sqrt5⁠. (Textbook answer: 52sin⁡−1(x+25)\frac52\sin^{-1}\left(\frac{x + 2}{\sqrt5}\right) +x+221−4x−x2+C+ \frac{x + 2}{2}\sqrt{1 - 4x - x^2} + C⁠, the same two terms in the other order.)

Check it with numbers

By Section 7.8, ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) for our answer FF⁠, 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.

IntervalSum of thin stripsF(b)−F(a)F(b) - F(a)
[−3,0][-3, 0]5.92699085.92699085.92699085.9269908
[−4.2,−3][-4.2, -3]1.75823801.75823801.75823801.7582380
Common mistakes
  1. Sign slip when completing the square. 1−(x2+4x)=1−(x+2)2+41 - (x^2 + 4x) = 1 - (x + 2)^2 + 4⁠, so the constant is 55⁠, not −3-3⁠.
  2. sin⁡−1x+25\sin^{-1}\frac{x + 2}{5} instead of sin⁡−1x+25\sin^{-1}\frac{x + 2}{\sqrt5}⁠. Inside sin⁡−1\sin^{-1} we divide by a=5a = \sqrt5⁠, not by a2a^2⁠.
  3. Using a log formula. The square carries the minus sign here, so the answer has sin⁡−1\sin^{-1}⁠, not a logarithm.
What to take away
  1. With −x2-x^2⁠, take out the minus sign first: c−bx−x2=(c+b24)−(x+b2)2c - bx - x^2 = \left(c + \frac{b^2}{4}\right) - \left(x + \frac b2\right)^2⁠.
  2. "Number minus square" under the root always leads to 7.6.2 (iii) and a sin⁡−1\sin^{-1} term.

Every fact used, and where it comes from

FactUsed inSource
Completing the squareSteps 1, 2Section 7.4 (7); Example 24 (Basics: completing the square)
Substitution y=x+2y = x + 2Step 3Section 7.3.1
∫a2−y2 dy\int\sqrt{a^2 - y^2}\,dy formulaStep 4Formula 7.6.2 (iii), Section 7.6.2
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}Step 5Standard formula (viii), Section 7.2
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)Numerical checkSection 7.8 (Lesson: definite integrals)
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