Exercise 7.8 answers: Integrals

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1. The antiderivative of x + 1 is x²/2 + x.
2. Put x = 1: 1/2 + 1 = 3/2.
3. Put x = −1: 1/2 − 1 = −1/2.
4. Subtract: 3/2 − (−1/2) = 2.

Answer2

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1. The antiderivative of 1/x is log|x| (the power rule cannot reach this one, so a log rescues it).
2. Put x = 3: log 3.
3. Put x = 2: log 2.
4. Subtract: log 3 − log 2 = log(3/2).

Answerlog(3/2)

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1. Split the sum into four separate terms and integrate each with the power rule.
2. Antiderivative: x⁴ − (5/3)x³ + 3x² + 9x.
3. Put x = 2: 16 − 40/3 + 12 + 18 = 98/3.
4. Put x = 1: 1 − 5/3 + 3 + 9 = 34/3.
5. Subtract: 98/3 − 34/3 = 64/3.

Answer64/3

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1. Guess −cos 2x, then check: its rate of change is 2 sin 2x, which is twice too big.
2. So the correct antiderivative is −(1/2) cos 2x.
3. Put x = π/4: −(1/2) cos(π/2) = −(1/2)(0) = 0.
4. Put x = 0: −(1/2) cos 0 = −1/2.
5. Subtract: 0 − (−1/2) = 1/2.

Answer1/2

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1. Guess sin 2x, then check: its rate of change is 2 cos 2x, twice too big.
2. So the correct antiderivative is (1/2) sin 2x.
3. Put x = π/2: (1/2) sin π = 0.
4. Put x = 0: (1/2) sin 0 = 0.
5. Subtract: 0 − 0 = 0.

Answer0

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1. The antiderivative of e^x is e^x itself.
2. Put x = 5: e⁵.
3. Put x = 4: e⁴.
4. Subtract: e⁵ − e⁴.

Answere⁵ − e⁴

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1. Write tan x as sin x / cos x. Its antiderivative is −log|cos x|.
2. Put x = π/4: −log(1/√2) = (1/2) log 2.
3. Put x = 0: −log 1 = 0.
4. Subtract: (1/2) log 2 − 0 = (1/2) log 2.

Answer(1/2) log 2

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1. The antiderivative of cosec x is log|cosec x − cot x| (found by multiplying top and bottom by cosec x + cot x).
2. Put x = π/4: cosec(π/4) − cot(π/4) = √2 − 1, so log(√2 − 1).
3. Put x = π/6: cosec(π/6) − cot(π/6) = 2 − √3, so log(2 − √3).
4. Subtract: log(√2 − 1) − log(2 − √3).

Answerlog(√2 − 1) − log(2 − √3)

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1. This is a standard form: the antiderivative of 1/√(1 − x²) is sin^-1x.
2. Put x = 1: sin^-1(1) = π/2.
3. Put x = 0: sin^-1(0) = 0.
4. Subtract: π/2 − 0 = π/2.

Answerπ/2

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1. This is a standard form: the antiderivative of 1/(1 + x²) is tan^-1x.
2. Put x = 1: tan^-1(1) = π/4.
3. Put x = 0: tan^-1(0) = 0.
4. Subtract: π/4 − 0 = π/4.

Answerπ/4

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1. For 1/(x² − 1), the standard antiderivative is (1/2) log|(x − 1)/(x + 1)|.
2. Put x = 3: (1/2) log(2/4) = (1/2) log(1/2).
3. Put x = 2: (1/2) log(1/3).
4. Subtract: (1/2)[log(1/2) − log(1/3)] = (1/2) log(3/2).

Answer(1/2) log(3/2)

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1. Use the identity cos²x = (1 + cos 2x)/2 to rewrite the integrand.
2. Antiderivative: x/2 + (sin 2x)/4.
3. Put x = π/2: π/4 + 0 = π/4.
4. Put x = 0: 0 + 0 = 0.
5. Subtract: π/4 − 0 = π/4.

Answerπ/4

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1. The numerator x is half the rate of change of the denominator x² + 1.
2. So the antiderivative is (1/2) log(x² + 1).
3. Put x = 3: (1/2) log 10.
4. Put x = 2: (1/2) log 5.
5. Subtract: (1/2)(log 10 − log 5) = (1/2) log 2.

Answer(1/2) log 2

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1. Split the numerator: (2x + 3)/(5x² + 1) = 2x/(5x² + 1) + 3/(5x² + 1).
2. The first piece integrates to (1/5) log(5x² + 1), since 2x is a fifth of the denominator's rate of change (10x).
3. The second piece is a standard inverse-tangent form: 3/(5x² + 1) integrates to (3/√5) tan^-1(√5 x).
4. At x = 1: (1/5) log 6 + (3/√5) tan^-1(√5).
5. At x = 0: both terms are 0.
6. Answer: (1/5) log 6 + (3√5/5) tan^-1(√5).

Answer(1/5) log 6 + (3√5/5) tan^-1(√5)

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1. Evaluate: ∫₀¹ x eˣ² dx.
2. Substitute u = x², so du = 2x dx, i.e. x dx = du/2.
3. The integral becomes (1/2) ∫ e^u du = (1/2) e^u.
4. Go back to x: (1/2) e^x².
5. Put x = 1: (1/2) e.
6. Put x = 0: (1/2)(1) = 1/2.
7. Subtract: (1/2)e − 1/2 = (e − 1)/2.

Answer(e − 1)/2

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1. The top and bottom have the same degree (both squared), so this fraction is improper — divide first.
2. 5x²/(x² + 4x + 3) = 5 − (20x + 15)/(x² + 4x + 3).
3. Factor the denominator: x² + 4x + 3 = (x + 1)(x + 3), and split (20x+15)/[(x+1)(x+3)] into partial fractions: −5/2 over (x+1) plus 45/2 over (x+3).
4. So the integrand is 5 + (5/2)/(x + 1) − (45/2)/(x + 3).
5. Antiderivative: 5x + (5/2) log|x + 1| − (45/2) log|x + 3|.
6. Put x = 2: 10 + (5/2) log 3 − (45/2) log 5.
7. Put x = 1: 5 + (5/2) log 2 − (45/2) log 4.
8. Subtract: 5 + (5/2) log(3/2) + (45/2) log(4/5).

Answer5 + (5/2) log(3/2) + (45/2) log(4/5)

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1. Split into three terms and integrate each: 2sec²x → 2 tan x, x³ → x⁴/4, and 2 → 2x.
2. Antiderivative: 2 tan x + x⁴/4 + 2x.
3. Put x = π/4: 2(1) + (π/4)⁴/4 + 2(π/4) = 2 + π⁴/1024 + π/2.
4. Put x = 0: 0.
5. Subtract: 2 + π/2 + π⁴/1024.

Answer2 + π/2 + π⁴/1024

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1. Use the identity cos x = cos²(x/2) − sin²(x/2), so sin²(x/2) − cos²(x/2) = −cos x.
2. The antiderivative of −cos x is −sin x.
3. Put x = π: −sin π = 0.
4. Put x = 0: −sin 0 = 0.
5. Subtract: 0 − 0 = 0.

Answer0

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1. Split the numerator: (6x + 3)/(x² + 4) = 6x/(x² + 4) + 3/(x² + 4).
2. The first piece integrates to 3 log(x² + 4), since 6x is the denominator's own rate of change.
3. The second piece is a standard inverse-tangent form: 3/(x² + 4) integrates to (3/2) tan^-1(x/2).
4. At x = 2: 3 log 8 + (3/2) tan^-1(1) = 3 log 8 + 3π/8.
5. At x = 0: 3 log 4 + 0.
6. Subtract: 3 log 8 − 3 log 4 + 3π/8 = 3 log 2 + 3π/8.

Answer3 log 2 + 3π/8

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1. Evaluate: (x eˣ + sin(πx/4)) dx.
2. Split into two integrals: ∫ x e^x dx and ∫ sin(πx/4) dx.
3. For x e^x, integrate by parts with x differentiating and e^x integrating: the antiderivative is (x − 1)e^x.
4. For sin(πx/4), guess −cos(πx/4) and adjust by the constant from the chain rule: the antiderivative is −(4/π) cos(πx/4).
5. Put x = 1: (1 − 1)e + (−(4/π) cos(π/4)) = 0 − (4/π)(√2/2) = −2√2/π.
6. Put x = 0: (0 − 1)(1) + (−(4/π)(1)) = −1 − 4/π.
7. Subtract: [−2√2/π] − [−1 − 4/π] = 1 + 4/π − 2√2/π.

Answer1 + 4/π − 2√2/π

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1. The antiderivative of 1/(1 + x²) is tan^-1x.
2. Put x = √3: tan^-1(√3) = π/3.
3. Put x = 1: tan^-1(1) = π/4.
4. Subtract: π/3 − π/4 = π/12.
5. That matches option (D).

Answer(D) π/12

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1. Write 4 + 9x² as b² + a²x² with a = 3, b = 2.
2. Use the standard result ∫ dx/(b² + a²x²) = (1/(ab)) tan⁻¹(ax/b).
3. So the antiderivative here is (1/6) tan⁻¹(3x/2).
4. Put x = 2/3: (1/6) tan⁻¹(1) = (1/6)(π/4) = π/24.
5. Put x = 0: (1/6) tan⁻¹(0) = 0.
6. Subtract: the value is π/24 − 0 = π/24, which is option (C).

Answerπ/24, option (C)

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