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Chapter 7 · Integrals

Exercise 7.8 · Q22 (MCQ): ∫02/3dx4+9x2\int_0^{2/3}\frac{dx}{4 + 9x^2}

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Choose the correct answer.

∫023dx4+9x2  equals\int_0^{\frac23}\frac{dx}{4 + 9x^2}\ \text{ equals}

(A) π6\frac{\pi}{6}    (B) π12\frac{\pi}{12}    (C) π24\frac{\pi}{24}    (D) π4\frac{\pi}{4}

Exam answer ∫023dx4+9x2=19∫023dxx2+(23)2=19⋅32[tan⁡−13x2]023=16(π4−0)=π24\begin{aligned} &\int_0^{\frac23}\frac{dx}{4 + 9x^2} \\ &= \frac19\int_0^{\frac23}\frac{dx}{x^2 + \left(\frac23\right)^2} \\ &= \frac19\cdot\frac32\Big[\tan^{-1}\frac{3x}{2}\Big]_0^{\frac23} \\ &= \frac16\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{24} \end{aligned}

Answer: (C) π24\dfrac{\pi}{24}

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

The one idea this problem is built on

Make the coefficient of x2x^2 equal to 1: 4+9x2=9(x2+(23)2)4 + 9x^2 = 9\left(x^2 + \left(\frac23\right)^2\right)⁠. Then formula 7.4 (3), ∫dxx2+a2=1atan⁡−1xa\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a}⁠, applies with a=23a = \frac23⁠. The upper limit 23\frac23 makes xa=1\frac{x}{a} = 1⁠, and tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠.

Given, to find, and the plan
Given
The integrand 14+9x2\frac{1}{4 + 9x^2} on [0,23]\left[0, \frac23\right]⁠, and four options.
To find
Which option equals the integral.
Plan
(1) Check continuity. (2) Rewrite the denominator and use formula 7.4 (3). (3) Differentiate FF back. (4) Evaluate. (5) Examine every option, with an area bound as an independent test.

Basics you need

1Second fundamental theorem

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) when ff is continuous on [a,b][a, b] and F′=fF' = f⁠. Lesson: definite integrals and the fundamental theorem

2Formula 7.4 (3)

∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a} + C for a>0a > 0⁠. Lesson: integrals of particular functions

3Principal values

tan⁡−10=0\tan^{-1}0 = 0⁠, tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠, both in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠. Basics: trigonometric identities for integrals

4Area bounds

If m≤f(x)≤Mm \le f(x) \le M on [a,b][a, b]⁠, then m(b−a)≤∫abf≤M(b−a)m(b - a) \le \int_a^b f \le M(b - a)⁠. Basics: area as a sum of thin strips

Solution

  1. Check continuity on [0,23]\left[0, \frac23\right]

    4+9x2≥4>04 + 9x^2 \ge 4 > 0 for every real xx⁠, so the integrand is continuous everywhere, in particular on [0,23]\left[0, \frac23\right]⁠.

    Theorem 2 needs continuity on the closed interval; the denominator is never 0.
    9x2≥09x^2 \ge 0⁠. Lesson: the fundamental theorem

  2. Rewrite the denominator and integrate

    Take out the 9: 4+9x2=9(x2+49)=9(x2+(23)2)4 + 9x^2 = 9\left(x^2 + \frac49\right) = 9\left(x^2 + \left(\frac23\right)^2\right)⁠. So with a=23a = \frac23⁠, 1a=32\frac1a = \frac32⁠:

    ∫dx4+9x2=19∫dxx2+(23)2=19⋅32tan⁡−13x2=16tan⁡−13x2+C.\begin{aligned}\int\frac{dx}{4 + 9x^2} &= \frac19\int\frac{dx}{x^2 + \left(\frac23\right)^2} \\ &= \frac19\cdot\frac32\tan^{-1}\frac{3x}{2} \\ &= \frac16\tan^{-1}\frac{3x}{2} + C.\end{aligned}

    Take F(x)=16tan⁡−13x2F(x) = \frac16\tan^{-1}\frac{3x}{2}⁠.

    Formula 7.4 (3) is stated for x2+a2x^2 + a^2 with coefficient 1 on x2x^2⁠, so the 9 must come out first. xa=x⋅32=3x2\frac{x}{a} = x\cdot\frac32 = \frac{3x}{2}⁠.
    Formula (3), Section 7.4. Lesson: integrals of particular functions

  3. Check: differentiate FF back

    F′(x)=16⋅11+9x24⋅32=14⋅44+9x2=14+9x2. ✓\begin{aligned}F'(x) &= \frac16\cdot\frac{1}{1 + \frac{9x^2}{4}}\cdot\frac32 = \frac14\cdot\frac{4}{4 + 9x^2} \\ &= \frac{1}{4 + 9x^2}.\ \checkmark\end{aligned}
    Chain rule with inner function u=3x2u = \frac{3x}{2}⁠, u2=9x24u^2 = \frac{9x^2}{4}⁠, dudx=32\frac{du}{dx} = \frac32⁠; and 16⋅32=14\frac16\cdot\frac32 = \frac14⁠.
    Standard formula (x), Section 7.2. Basics: integration undoes differentiation

  4. Evaluate F(23)−F(0)F\left(\frac23\right) - F(0)

    At x=23x = \frac23⁠: 3x2=1\frac{3x}{2} = 1⁠. At x=0x = 0⁠: 3x2=0\frac{3x}{2} = 0⁠. So

    I=16tan⁡−11−16tan⁡−10=16⋅π4−0=π24.I = \frac16\tan^{-1}1 - \frac16\tan^{-1}0 = \frac16\cdot\frac{\pi}{4} - 0 = \frac{\pi}{24}.

    In decimals, π24≈0.1308997\frac{\pi}{24} \approx 0.1308997⁠.

    1/3 2/3 1 1/8 1/4 x y y = 1/(4 + 9x²) area π/24
    The shaded region under y=14+9x2y = \frac{1}{4 + 9x^2} from 0 to 23\frac23 has area π24≈0.131\frac{\pi}{24} \approx 0.131⁠. The curve falls from 14\frac14 to 18\frac18 on this interval.
    The limit 23\frac23 was chosen so that the inverse tangent is at the standard value 1. Principal values: tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠, tan⁡−10=0\tan^{-1}0 = 0⁠.
    tan⁡π4=1\tan\frac{\pi}{4} = 1⁠. Basics: trigonometric values

  5. An independent bound

    The integrand decreases for x>0x > 0⁠, from 14\frac14 at x=0x = 0 to 14+4=18\frac{1}{4 + 4} = \frac18 at x=23x = \frac23⁠. The width is 23\frac23⁠. So

    18⋅23≤I≤14⋅23,i.e.112≤I≤16.\frac18\cdot\frac23 \le I \le \frac14\cdot\frac23, \quad\text{i.e.}\quad \frac{1}{12} \le I \le \frac16.

    In decimals, 0.0833≤I≤0.16670.0833 \le I \le 0.1667⁠, and π24≈0.1309\frac{\pi}{24} \approx 0.1309 lies inside.

    The area under the curve is trapped between two rectangles on the same base. This test uses no antiderivative, so it independently confirms Step 4.
    Smallest height × width ≤ area ≤ largest height × width. Basics: area as a sum of thin strips

  6. Examine every option

    • (A) π6≈0.5236\frac{\pi}{6} \approx 0.5236⁠. Four times the correct value, and above the upper bound 16≈0.1667\frac16 \approx 0.1667⁠. Wrong.
    • (B) π12≈0.2618\frac{\pi}{12} \approx 0.2618⁠. Twice the correct value, and also above the upper bound. Wrong.
    • (C) π24≈0.1309\frac{\pi}{24} \approx 0.1309⁠. Equals 16⋅π4\frac16\cdot\frac{\pi}{4} and lies inside the bound. Correct.
    • (D) π4≈0.7854\frac{\pi}{4} \approx 0.7854⁠. This is tan⁡−11\tan^{-1}1 alone, without the factor 16\frac16⁠. Far above the bound. Wrong.
    The exact value from Step 4 matches only (C), and the bound from Step 5 rules out (A), (B) and (D) without any further algebra.
Result ∫023dx4+9x2=π24≈0.1308997.\int_0^{\frac23}\frac{dx}{4 + 9x^2} = \frac{\pi}{24} \approx 0.1308997.

The correct option is (C). (Textbook answer: C.)

Check it with numbers

A numerical integration of 14+9x2\frac{1}{4 + 9x^2} over [0,23]\left[0, \frac23\right] gives about 0.1308997, and π24≈0.1308997\frac{\pi}{24} \approx 0.1308997⁠. They agree.

Common mistakes
  1. Using a=2a = 2 or a=3a = 3⁠. After taking out 9, a2=49a^2 = \frac49⁠, so a=23a = \frac23⁠.
  2. Forgetting the 19\frac19 that comes out of the denominator. The constant factor is 19⋅32=16\frac19\cdot\frac32 = \frac16⁠, not 32\frac32⁠.
  3. Choosing (D). π4\frac{\pi}{4} is only tan⁡−11\tan^{-1}1⁠; the factor 16\frac16 is still needed.
What to take away
  1. ∫dxp2+q2x2=1pqtan⁡−1qxp+C\int\frac{dx}{p^2 + q^2x^2} = \frac{1}{pq}\tan^{-1}\frac{qx}{p} + C for p,q>0p, q > 0⁠: here p=2p = 2⁠, q=3q = 3⁠, so 1pq=16\frac{1}{pq} = \frac16⁠.
  2. Always make the x2x^2 coefficient 1 before reading off aa⁠.
  3. Rectangle bounds quickly eliminate options that are too big or too small.

Every fact used, and where it comes from

FactUsed inSource
∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a) for continuous ffSteps 1, 4Theorem 2, Section 7.8.3 (Lesson)
∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a} + CStep 2Formula (3), Section 7.4 (Lesson)
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}Step 3Standard formula (x), Section 7.2
tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠, tan⁡−10=0\tan^{-1}0 = 0Step 4Principal values, Chapter 2 (Basics)
Area bounds by rectanglesSteps 5, 6Definite integral as area, Section 7.8.1 (Basics)
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