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Chapter 7 · Integrals

Exercise 7.8 · Q21 (MCQ): ∫13dx1+x2\int_1^{\sqrt3}\frac{dx}{1 + x^2}

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Choose the correct answer.

∫13dx1+x2  equals\int_1^{\sqrt3}\frac{dx}{1 + x^2}\ \text{ equals}

(A) π3\frac{\pi}{3}    (B) 2π3\frac{2\pi}{3}    (C) π6\frac{\pi}{6}    (D) π12\frac{\pi}{12}

Exam answer ∫13dx1+x2=[tan⁡−1x]13=π3−π4=π12\begin{aligned} \int_1^{\sqrt3}\frac{dx}{1 + x^2} &= \Big[\tan^{-1}x\Big]_1^{\sqrt3} \\ &= \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} \end{aligned}

Answer: (D) π12\dfrac{\pi}{12}

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

The one idea this problem is built on

11+x2\frac{1}{1 + x^2} is the derivative of tan⁡−1x\tan^{-1}x (standard formula (x), Section 7.2). So by Theorem 2 (Section 7.8.3) the integral is tan⁡−13−tan⁡−11\tan^{-1}\sqrt3 - \tan^{-1}1⁠, two principal values: π3−π4=π12\frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}⁠.

Given, to find, and the plan
Given
The integrand 11+x2\frac{1}{1 + x^2} on [1,3][1, \sqrt3]⁠, and four options.
To find
Which option equals the integral.
Plan
(1) Check continuity. (2) Antiderivative tan⁡−1x\tan^{-1}x⁠, checked by differentiating. (3) Evaluate with principal values. (4) Examine every option, using 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

2Inverse tangent

ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}⁠; tan⁡−1y\tan^{-1}y is the angle in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) whose tangent is yy (Chapter 2). Basics: trigonometric identities for integrals

3Area 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 [1,3][1, \sqrt3]

    1+x2≥1>01 + x^2 \ge 1 > 0 for every real xx⁠, so 11+x2\frac{1}{1 + x^2} is continuous everywhere, in particular on [1,3][1, \sqrt3]⁠.

    Theorem 2 needs continuity on the closed interval; the denominator never vanishes.
    A quotient of polynomials is continuous where the denominator is not 0. Lesson: the fundamental theorem

  2. Antiderivative, and the check

    ∫dx1+x2=tan⁡−1x+C,F(x)=tan⁡−1x.\int\frac{dx}{1 + x^2} = \tan^{-1}x + C, \qquad F(x) = \tan^{-1}x.

    Check: F′(x)=11+x2F'(x) = \frac{1}{1 + x^2} ✓, which is standard formula (x) read backwards.

    This is formula 7.4 (3) with a=1a = 1 as well: 11tan⁡−1x1\frac11\tan^{-1}\frac{x}{1}⁠.
    Standard formula (x), Section 7.2. Basics: integration undoes differentiation

  3. Evaluate F(3)−F(1)F(\sqrt3) - F(1)

    tan⁡π3=3\tan\frac{\pi}{3} = \sqrt3 and π3\frac{\pi}{3} lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, so tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3}⁠. Likewise tan⁡π4=1\tan\frac{\pi}{4} = 1 gives tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠. So

    I=π3−π4=4π−3π12=π12≈0.2617994.I = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \approx 0.2617994.
    1 √3 2 1/4 1/2 1 x y y = 1/(1 + x²) area π/12
    The shaded region under y=11+x2y = \frac{1}{1 + x^2} from 1 to 3\sqrt3 has area π12≈0.262\frac{\pi}{12} \approx 0.262⁠. On this interval the curve falls from 12\frac12 to 14\frac14 (dashed levels).
    Inverse tangent values must be principal values. Another angle with tangent 3\sqrt3⁠, such as 4π3\frac{4\pi}{3}⁠, is not tan⁡−13\tan^{-1}\sqrt3⁠, and using it would give a wrong number.
    Principal range of tan⁡−1\tan^{-1}⁠: (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠, Chapter 2. Basics: trigonometric values

  4. An independent bound

    11+x2\frac{1}{1 + x^2} decreases for x>0x > 0⁠. On [1,3][1, \sqrt3] it goes from 11+1=12\frac{1}{1 + 1} = \frac12 down to 11+3=14\frac{1}{1 + 3} = \frac14⁠. The width is 3−1≈0.7320508\sqrt3 - 1 \approx 0.7320508⁠. So

    14(3−1)≤I≤12(3−1),0.183≤I≤0.366.\begin{aligned}\tfrac14(\sqrt3 - 1) &\le I \le \tfrac12(\sqrt3 - 1), \\ 0.183 &\le I \le 0.366.\end{aligned}
    The area under the curve is trapped between the rectangles of heights 14\frac14 and 12\frac12 on the same base. π12≈0.262\frac{\pi}{12} \approx 0.262 lies inside, which confirms Step 3 without using any antiderivative.
    Smallest height × width ≤ area ≤ largest height × width. Basics: area as a sum of thin strips

  5. Examine every option

    • (A) π3≈1.047\frac{\pi}{3} \approx 1.047⁠. This is tan⁡−13\tan^{-1}\sqrt3 alone: the lower limit was forgotten. It is far above the bound 0.366. Wrong.
    • (B) 2π3≈2.094\frac{2\pi}{3} \approx 2.094⁠. Even larger; it would need an area bigger than the whole rectangle of height 1 on the base. Wrong.
    • (C) π6≈0.524\frac{\pi}{6} \approx 0.524⁠. Above the upper bound 0.366, so impossible. (It is tan⁡−113\tan^{-1}\frac{1}{\sqrt3}⁠, or π3−π6\frac{\pi}{3} - \frac{\pi}{6} with a wrong value for tan⁡−11\tan^{-1}1⁠.) Wrong.
    • (D) π12≈0.262\frac{\pi}{12} \approx 0.262⁠. Equals π3−π4\frac{\pi}{3} - \frac{\pi}{4} and lies inside the bound. Correct.
    Only one option can be equal to the exact value from Step 3, and the bound from Step 4 independently excludes (A), (B) and (C).
Result ∫13dx1+x2=π12≈0.2617994.\int_1^{\sqrt3}\frac{dx}{1 + x^2} = \frac{\pi}{12} \approx 0.2617994.

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

Check it with numbers

A numerical integration of 11+x2\frac{1}{1 + x^2} over [1,3][1, \sqrt3] gives about 0.2617994, and π12≈0.2617994\frac{\pi}{12} \approx 0.2617994⁠. They agree.

tan⁡−13≈1.0471976\tan^{-1}\sqrt3 \approx 1.0471976⁠, tan⁡−11≈0.7853982\tan^{-1}1 \approx 0.7853982⁠; the difference is 0.2617994.

Common mistakes
  1. Choosing (A). That is only F(3)F(\sqrt3)⁠; Theorem 2 subtracts F(1)=π4F(1) = \frac{\pi}{4}⁠.
  2. Mixing up tan⁡−13\tan^{-1}\sqrt3 and tan⁡−113\tan^{-1}\frac{1}{\sqrt3}⁠. tan⁡π3=3\tan\frac{\pi}{3} = \sqrt3 and tan⁡π6=13\tan\frac{\pi}{6} = \frac{1}{\sqrt3}⁠.
  3. Using degrees. 60∘−45∘=15∘60^\circ - 45^\circ = 15^\circ is the right angle, but the integral is the number π12\frac{\pi}{12} (radians).
What to take away
  1. ∫abdx1+x2=tan⁡−1b−tan⁡−1a\int_a^b\frac{dx}{1 + x^2} = \tan^{-1}b - \tan^{-1}a⁠, with principal values.
  2. A quick area bound is a powerful way to eliminate MCQ options.
  3. Remember tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠, tan⁡−13=π3\tan^{-1}\sqrt3 = \frac{\pi}{3}⁠, tan⁡−113=π6\tan^{-1}\frac{1}{\sqrt3} = \frac{\pi}{6}⁠.

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, 3Theorem 2, Section 7.8.3 (Lesson)
∫dx1+x2=tan⁡−1x+C\int\frac{dx}{1 + x^2} = \tan^{-1}x + CStep 2Standard formula (x), Section 7.2
Principal values of tan⁡−1\tan^{-1}Steps 3, 5Chapter 2 (Basics)
Area bounds by rectanglesSteps 4, 5Definite integral as area, Section 7.8.1 (Basics)
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