Exercise 7.10 answers: Integrals

Open NCERT p. 280Matches NCERT’s answer

1. Use the property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx with a = π/2.
2. Replacing x by (π/2 − x) turns cos x into sin x, so I = ∫₀^(π/2) sin² x dx as well.
3. Add the two forms of I: 2I = ∫₀^(π/2) (cos² x + sin² x) dx = ∫₀^(π/2) 1 dx = π/2.
4. So I = π/4.

Answerπ/4

Watch this explained “The move: fold it, then add it”, 6:58 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Let I be the integral. Replace x by (π/2 − x): sin x becomes cos x and cos x becomes sin x.
2. So I also equals ∫₀^(π/2) √(cos x)/(√(cos x) + √(sin x)) dx.
3. Add the original I and this new form: 2I = ∫₀^(π/2) [√(sin x)+√(cos x)]/[√(sin x)+√(cos x)] dx = ∫₀^(π/2) 1 dx = π/2.
4. So I = π/4.

Answerπ/4

Watch this explained “Four integrands, one answer”, 8:13 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Let I be the integral. Replace x by (π/2 − x): the integrand becomes cos^(3/2) x/(cos^(3/2) x + sin^(3/2) x).
2. Add the original I and this new form: 2I = ∫₀^(π/2) 1 dx = π/2.
3. So I = π/4.

Answerπ/4

Watch this explained “Four integrands, one answer”, 8:13 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Evaluate: ∫₀^(π/2) cos⁵ x/(sin⁵ x + cos⁵ x) dx.
2. Let I be the integral. Replace x by (π/2 − x): the integrand becomes sin⁵ x/(cos⁵ x + sin⁵ x).
3. Add the original I and this new form: 2I = ∫₀^(π/2) 1 dx = π/2.
4. So I = π/4.

Answerπ/4

Watch this explained “Four integrands, one answer”, 8:13 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. |x+2| changes sign at x = −2, which lies inside [−5, 5], so cut the integral there.
2. For x < −2, x+2 < 0, so |x+2| = −(x+2); for x > −2, |x+2| = x+2.
3. I = ∫₋₅^(−2) −(x+2) dx + ∫₋₂⁵ (x+2) dx.
4. First piece: −[(x+2)²/2] from −5 to −2 = 0 − (−9/2) = 9/2.
5. Second piece: [(x+2)²/2] from −2 to 5 = 49/2 − 0 = 49/2.
6. Total: I = 9/2 + 49/2 = 29.

Answer29

Watch this explained “Bars are a place to cut”, 14:11 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. |x−5| changes sign at x = 5, which lies inside [2, 8], so cut the integral there.
2. For x < 5, |x−5| = 5−x; for x > 5, |x−5| = x−5.
3. I = ∫₂⁵ (5−x) dx + ∫₅⁸ (x−5) dx.
4. First piece: [5x − x²/2] from 2 to 5 = 12.5 − 8 = 9/2.
5. Second piece: [x²/2 − 5x] from 5 to 8 = (−8) − (−12.5) = 9/2.
6. Total: I = 9/2 + 9/2 = 9.

Answer9

Watch this explained “Bars are a place to cut”, 14:11 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Use ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx with a = 1: replace x by (1−x).
2. I = ∫₀¹ (1−x)(1−(1−x))ⁿ dx = ∫₀¹ (1−x)xⁿ dx = ∫₀¹ (xⁿ − xⁿ⁺¹) dx.
3. Integrate: [xⁿ⁺¹/(n+1) − xⁿ⁺²/(n+2)] from 0 to 1 = 1/(n+1) − 1/(n+2).
4. Combine over a common denominator: I = 1/[(n+1)(n+2)].

Answer1/[(n+1)(n+2)]

Watch this explained “Folding the stretch onto itself”, 4:57 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Let I be the integral. Replace x by (π/4 − x): tan(π/4 − x) = (1 − tan x)/(1 + tan x).
2. So 1 + tan(π/4 − x) = 1 + (1−tan x)/(1+tan x) = 2/(1+tan x).
3. This gives I = ∫₀^(π/4) log[2/(1+tan x)] dx = ∫₀^(π/4) [log 2 − log(1+tan x)] dx = (π/4) log 2 − I.
4. So 2I = (π/4) log 2, giving I = (π/8) log 2.

Answer(π/8) log 2

Watch this explained “The move: fold it, then add it”, 6:58 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Use ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a−x)dx with a = 2: replace x by (2−x), giving I = ∫₀² (2−x)√x dx.
2. Expand: I = ∫₀² (2√x − x^(3/2)) dx = [2·(2/3)x^(3/2) − (2/5)x^(5/2)] from 0 to 2.
3. At x = 2: (4/3)(2√2) − (2/5)(4√2) = 8√2/3 − 8√2/5.
4. Combine: 8√2/3 − 8√2/5 = 16√2/15.

Answer16√2/15

Watch this explained “Folding the stretch onto itself”, 4:57 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Use sin 2x = 2 sin x cos x, so log sin 2x = log 2 + log sin x + log cos x.
2. So 2 log sin x − log sin 2x = log sin x − log cos x − log 2 = log(tan x) − log 2.
3. I = ∫₀^(π/2) log(tan x) dx − (π/2) log 2. Call the first part J.
4. Replace x by (π/2 − x) in J: tan(π/2−x) = cot x = 1/tan x, so log(tan(π/2−x)) = −log(tan x), which makes J = −J, so J = 0.
5. Therefore I = 0 − (π/2) log 2 = −(π/2) log 2.

Answer−(π/2) log 2

Watch this explained “The fold that does not simplify”, 15:25 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. sin²x is an even function: sin²(−x) = sin²x.
2. For an even function, the integral from −a to a is 2 times the integral from 0 to a.
3. So the integral becomes 2 × ∫_0^(π/2) sin²x dx.
4. Use the identity sin²x = (1 − cos 2x)/2.
5. 2 × ∫_0^(π/2) (1−cos2x)/2 dx = ∫_0^(π/2) (1−cos2x) dx.
6. Integrate: [x − (sin 2x)/2] from 0 to π/2 = π/2 − 0 = π/2.

Answerπ/2

Watch this explained “Even and odd on a balanced stretch”, 11:09 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Let I = ∫_0^π x/(1 + sin x) dx.
2. Use ∫_0^a f(x) dx = ∫_0^a f(a − x) dx with a = π. Since sin(π − x) = sin x, I = ∫_0^π (π − x)/(1 + sin x) dx.
3. Add the two forms of I: 2I = ∫_0^π π/(1 + sin x) dx, so I = (π/2) ∫_0^π dx/(1 + sin x).
4. Call K = ∫_0^π dx/(1 + sin x). The integrand is unchanged by x → π − x, so K = 2∫_0^(π/2) dx/(1 + sin x).
5. Replace x by π/2 − x (sin(π/2 − x) = cos x): K = 2∫_0^(π/2) dx/(1 + cos x).
6. 1 + cos x = 2cos²(x/2), so 1/(1 + cos x) = (1/2) sec²(x/2), whose integral is tan(x/2).
7. K = 2[tan(x/2)]_0^(π/2) = 2(tan(π/4) − tan 0) = 2(1 − 0) = 2.
8. So I = (π/2) × 2 = π.

Answerπ

Watch this explained “The rule with the function taken out”, 18:50 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. sin⁷x is an odd function: sin⁷(−x) = −sin⁷x.
2. For an odd function on a symmetric interval −a to a, the integral is 0.
3. So the integral is 0.

Answer0

Watch this explained “Even and odd on a balanced stretch”, 11:09 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Use the doubled-interval property: ∫_0^(2a) f(x)dx = 2∫_0^a f(x)dx when f(2a−x) = f(x).
2. Here a = π and f(x) = cos⁵x. Since cos(2π−x) = cos x, the property applies.
3. So ∫_0^(2π) cos⁵x dx = 2∫_0^π cos⁵x dx.
4. Now fold the new integral using x → π−x: cos(π−x) = −cos x, so cos⁵(π−x) = −cos⁵x.
5. Adding the original and folded forms: 2∫_0^π cos⁵x dx = ∫_0^π [cos⁵x + (−cos⁵x)] dx = 0.
6. So ∫_0^π cos⁵x dx = 0, and therefore ∫_0^(2π) cos⁵x dx = 2 × 0 = 0.

Answer0

Watch this explained “Twenty-one integrals, four tests”, 17:24 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Let f(x) = (sin x − cos x)/(1 + sin x cos x), and let I be the integral from 0 to π/2.
2. Fold using x → (π/2 − x): sin(π/2−x) = cos x and cos(π/2−x) = sin x.
3. f(π/2−x) = (cos x − sin x)/(1+sin x cos x) = −f(x), since the denominator sin x cos x is unchanged.
4. Adding f(x) and its fold: f(x) + f(π/2−x) = 0, so 2I = ∫_0^(π/2) 0 dx = 0.
5. So I = 0.

Answer0

Watch this explained “The move: fold it, then add it”, 6:58 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Let f(x) = log(1+cos x), and I be the integral from 0 to π.
2. Fold using x → (π−x): cos(π−x) = −cos x, so f(π−x) = log(1−cos x).
3. Add: f(x)+f(π−x) = log[(1+cos x)(1−cos x)] = log(1−cos²x) = log(sin²x) = 2 log(sin x), since sin x ≥ 0 on [0,π].
4. So 2I = ∫_0^π 2 log(sin x) dx, which gives I = ∫_0^π log(sin x) dx. Call this J.
5. Since sin(π−x) = sin x, the graph of log(sin x) is symmetric about x = π/2, so J = 2∫_0^(π/2) log(sin x) dx.
6. This chapter's Example 34 shows ∫_0^(π/2) log(sin x) dx = −(π/2) log 2.
7. So J = 2 × [−(π/2) log 2] = −π log 2, and therefore I = −π log 2.

Answer−π log 2

Watch this explained “The fold that does not simplify”, 15:25 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Let f(x) = √x/(√x+√(a−x)), and I be the integral from 0 to a.
2. Fold using x → (a−x): f(a−x) = √(a−x)/(√(a−x)+√x).
3. Add: f(x) + f(a−x) = [√x + √(a−x)]/[√x+√(a−x)] = 1.
4. So 2I = ∫_0^a 1 dx = a.
5. Therefore I = a/2.

Answera/2

Watch this explained “The move: fold it, then add it”, 6:58 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. |x−1| changes sign at x = 1, which lies inside [0,4], so cut the integral there.
2. For 0 ≤ x ≤ 1, x−1 ≤ 0, so |x−1| = 1−x.
3. For 1 ≤ x ≤ 4, x−1 ≥ 0, so |x−1| = x−1.
4. ∫_0^1 (1−x) dx = [x − x²/2]_0^1 = 1 − 1/2 = 1/2.
5. ∫_1^4 (x−1) dx = [(x−1)²/2]_1^4 = 9/2 − 0 = 9/2.
6. Add the two pieces: 1/2 + 9/2 = 5.

Answer5

Watch this explained “Bars are a place to cut”, 14:11 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280One way to think about it

1. Let I = ∫_0^a f(x) g(x) dx.
2. Use the property ∫_0^a h(x)dx = ∫_0^a h(a−x)dx: I = ∫_0^a f(a−x) g(a−x) dx.
3. Since f(x) = f(a−x) is given, replace f(a−x) by f(x): I = ∫_0^a f(x) g(a−x) dx.
4. Add this to the first form of I: 2I = ∫_0^a f(x)[g(x)+g(a−x)] dx.
5. Since g(x)+g(a−x) = 4 is given, this becomes 2I = ∫_0^a f(x) × 4 dx = 4∫_0^a f(x) dx.
6. Dividing by 2: I = 2∫_0^a f(x) dx, which is what had to be shown.

In shortI = 2∫_0^a f(x) dx, as shown above.

Watch this explained “The move: fold it, then add it”, 6:58 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Split the integral into four: ∫x³ dx + ∫x cos x dx + ∫tan⁵x dx + ∫1 dx, each from −π/2 to π/2.
2. x³ is odd, x cos x is odd (odd × even), and tan⁵x is odd: each one changes sign when x is replaced by −x.
3. On an interval balanced about 0, an odd function's two halves cancel, so these three terms contribute 0.
4. A note: tan⁵x grows without limit near ±π/2, so its part is settled only by this cancellation between the two halves — which is the property the question is testing.
5. Only the constant term is left: ∫_(−π/2)^(π/2) 1 dx = π/2 − (−π/2) = π.
6. So the value is π, which is option (C).

Answerπ, option (C)

Watch this explained “Even and odd on a balanced stretch”, 11:09 into Properties of the definite integral, and using symmetry to kill half the work

Open NCERT p. 280Matches NCERT’s answer

1. Question: The value of ∫_0^(π/2) log((4 + 3 sin x)/(4 + 3 cos x)) dx.
2. Let f(x) = log[(4+3sinx)/(4+3cosx)], and I be the integral from 0 to π/2.
3. Fold using x → (π/2−x): sin(π/2−x)=cos x and cos(π/2−x)=sin x, so f(π/2−x) = log[(4+3cosx)/(4+3sinx)].
4. That is the reciprocal ratio inside the log, so f(π/2−x) = −f(x).
5. Adding f(x) and its fold gives 0, so 2I = 0, and I = 0.
6. So the value is 0, option (C).

Answer0, option (C)

Watch this explained “The move: fold it, then add it”, 6:58 into Properties of the definite integral, and using symmetry to kill half the work