Exercise 7.1 answers: Integrals

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1. We want a function whose derivative is sin 2x.
2. The derivative of cos 2x is −2 sin 2x, by the chain rule — that is sin 2x, but twice too big and with the wrong sign.
3. Divide by −2 to cancel that: the derivative of −¹⁄₂cos 2x is exactly sin 2x.

Answer−¹⁄₂cos 2x + C

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1. We want a function whose derivative is cos 3x.
2. The derivative of sin 3x is 3 cos 3x — three times too big.
3. Divide by 3: the derivative of ¹⁄₃sin 3x is cos 3x.

Answer¹⁄₃sin 3x + C

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1. We want a function whose derivative is e^2x.
2. The derivative of e^2x is 2e^2x — twice too big.
3. Divide by 2: the derivative of ¹⁄₂e^2x is e^2x.

Answer¹⁄₂e^2x + C

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1. We want a function whose derivative is (ax + b)². Here a and b are constants, with a ≠ 0.
2. By the chain rule, the derivative of (ax + b)³ is 3(ax + b)² × a = 3a(ax + b)² — too big by a factor of 3a.
3. Divide by 3a: the derivative of ¹⁄_3a(ax + b)³ is exactly (ax + b)².

Answer¹⁄_3a(ax + b)³ + C

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1. Split into two pieces and find an anti derivative of each: sin 2x and −4e³ˣ.
2. From sin 2x (as in Q1), an anti derivative is −¹⁄₂cos 2x.
3. The derivative of e³ˣ is 3e³ˣ, so an anti derivative of e³ˣ is ¹⁄₃e³ˣ; multiplying by −4 gives −⁴⁄₃e³ˣ.
4. Add the two pieces.

Answer−¹⁄₂cos 2x − ⁴⁄₃e^3x + C

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1. Split into two pieces: 4e³ˣ and 1.
2. An anti derivative of e³ˣ is ¹⁄₃e³ˣ; multiplying by 4 gives ⁴⁄₃e³ˣ.
3. An anti derivative of 1 is x.
4. Add the two pieces.

Answer⁴⁄₃e^3x + x + C

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1. Multiply out first: x²(1 − 1⁄x²) = x² − 1.
2. An anti derivative of x² is x³⁄3, and of −1 is −x.
3. Add the two pieces.

Answerx³⁄3 − x + C

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1. Integrate term by term; a, b, c are just constants that come along for the ride.
2. An anti derivative of x² is x³⁄3, of x is x²⁄2, and of 1 is x.
3. Multiply each by its constant and add: a·x³⁄3 + b·x²⁄2 + c·x.

Answera·x³⁄3 + b·x²⁄2 + c·x + C

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1. Split into two pieces: 2x² and eˣ.
2. An anti derivative of x² is x³⁄3, so of 2x² it is ²⁄₃x³.
3. An anti derivative of eˣ is eˣ itself.
4. Add the two pieces.

Answer²⁄₃x³ + e^x + C

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1. Expand the square first: (√x)² − 2·√x·¹⁄_√x + (¹⁄_√x)² = x − 2 + ¹⁄_x.
2. An anti derivative of x is ¹⁄₂x², and of −2 is −2x.
3. ¹⁄_x is the one power the power rule cannot handle (it would divide by 0); its anti derivative is log|x|. Because √x appears in the question, x > 0, so log|x| = log x.
4. Add the three pieces.

Answer¹⁄₂x² − 2x + log x + C

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1. Divide every term in the numerator by x²: x³⁄x² + 5x²⁄x² − 4⁄x² = x + 5 − 4⁄x².
2. An anti derivative of x is x²⁄2, and of 5 is 5x.
3. For −4⁄x² = −4x⁻², use the power rule: the anti derivative of x⁻² is x⁻¹⁄(−1) = −1⁄x, so −4 times that is 4⁄x.
4. Add the three pieces.

Answerx²⁄2 + 5x + 4⁄x + C

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1. Divide every term by √x = x^1/2: x³⁄x^1/2 + 3x⁄x^1/2 + 4⁄x^1/2 = x^5/2 + 3x^1/2 + 4x^−1/2.
2. Integrate each power using xⁿ → xⁿ⁺¹⁄(n+1): x^5/2 → ²⁄₇x^7/2.
3. 3x^1/2 → 3 · ²⁄₃x^3/2 = 2x^3/2, and 4x^−1/2 → 4 · 2x^1/2 = 8x^1/2.
4. Add the three pieces.

Answer²⁄₇x^7/2 + 2x^3/2 + 8x^1/2 + C

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1. Factor the numerator by grouping: x³ − x² + x − 1 = x²(x − 1) + 1(x − 1) = (x − 1)(x² + 1).
2. Cancel the common factor (x − 1): the integrand simplifies to x² + 1.
3. An anti derivative of x² is x³⁄3, and of 1 is x.

Answerx³⁄3 + x + C

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1. Multiply out: (1 − x)√x = √x − x√x = x^1/2 − x^3/2.
2. An anti derivative of x^1/2 is ²⁄₃x^3/2.
3. An anti derivative of x^3/2 is ²⁄₅x^5/2.
4. Subtract the second from the first.

Answer²⁄₃x^3/2 − ²⁄₅x^5/2 + C

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1. Multiply out: √x(3x² + 2x + 3) = 3x^5/2 + 2x^3/2 + 3x^1/2.
2. An anti derivative of x^5/2 is ²⁄₇x^7/2, so of 3x^5/2 it is ⁶⁄₇x^7/2.
3. An anti derivative of x^3/2 is ²⁄₅x^5/2, so of 2x^3/2 it is ⁴⁄₅x^5/2.
4. An anti derivative of x^1/2 is ²⁄₃x^3/2, so of 3x^1/2 it is 2x^3/2.
5. Add the three pieces.

Answer⁶⁄₇x^7/2 + ⁴⁄₅x^5/2 + 2x^3/2 + C

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1. Split into three pieces: 2x, −3cos x, and eˣ.
2. An anti derivative of 2x is x²; an anti derivative of cos x is sin x, so of −3cos x it is −3sin x.
3. An anti derivative of eˣ is eˣ.
4. Add the three pieces.

Answerx² − 3sin x + e^x + C

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1. Split into three pieces: 2x², −3sin x, and 5√x.
2. An anti derivative of x² is x³⁄3, so of 2x² it is ²⁄₃x³.
3. An anti derivative of sin x is −cos x, so of −3sin x it is 3cos x.
4. An anti derivative of √x = x^1/2 is ²⁄₃x^3/2, so of 5√x it is ¹⁰⁄₃x^3/2.
5. Add the three pieces.

Answer²⁄₃x³ + 3cos x + ¹⁰⁄₃x^3/2 + C

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1. Open the bracket: sec x(sec x + tan x) = sec²x + sec x tan x.
2. An anti derivative of sec²x is tan x (a standard row).
3. An anti derivative of sec x tan x is sec x (a standard row).
4. Add the two pieces.

Answertan x + sec x + C

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1. sec²x ⁄ cosec²x = (1⁄cos²x) · sin²x = sin²x⁄cos²x = tan²x.
2. Use the identity tan²x = sec²x − 1 to rewrite it as a sum.
3. An anti derivative of sec²x is tan x, and of −1 is −x.
4. Add the two pieces.

Answertan x − x + C

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1. Split the fraction: (2 − 3sin x)⁄cos²x = 2⁄cos²x − 3sin x⁄cos²x = 2sec²x − 3(sin x⁄cos x)(1⁄cos x) = 2sec²x − 3sec x tan x.
2. An anti derivative of sec²x is tan x, so of 2sec²x it is 2tan x.
3. An anti derivative of sec x tan x is sec x, so of −3sec x tan x it is −3sec x.
4. Add the two pieces.

Answer2tan x − 3sec x + C

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1. Write √x + 1⁄√x as powers: x^1/2 + x^−1/2.
2. Integrate each power: x^1/2 → ²⁄₃x^3/2, and x^−1/2 → 2x^1/2.
3. So the anti derivative is ²⁄₃x^3/2 + 2x^1/2 + C.
4. Comparing with the four choices, this matches option (C) exactly.

Answer(C)

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1. Question: If d/dx f(x) = 4x³ − 3/x⁴ such that f(2) = 0. Then f(x).
2. We need the function whose rate of change is 4x³ − 3/x⁴, so integrate term by term. Write 3/x⁴ as 3x⁻⁴.
3. ∫ 4x³ dx = x⁴, and ∫ 3x⁻⁴ dx = 3 · x⁻³/(−3) = −1/x³, so subtracting it gives +1/x³.
4. So f(x) = x⁴ + 1/x³ + C for some constant C.
5. Use f(2) = 0: 2⁴ + 1/2³ + C = 0, that is 16 + 1/8 + C = 0, so C = −129/8.
6. Hence f(x) = x⁴ + 1/x³ − 129/8, which is option (A).
7. Why not the others: (B) and (D) have x³ and 1/x⁴, whose rate of change is 3x² − 4/x⁵, not 4x³ − 3/x⁴. (C) has the right rate of change but the wrong constant: at x = 2 it gives 16 + 1/8 + 129/8 = 129/4, not 0.

Answer(A) x⁴ + 1/x³ − 129/8

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