Integration

Find ∫ (6x² + 4x) dx.

• A2x³ + 4x² + c
• B6x³ + 2x² + c
• C12x + 4 + c
• D2x³ + 2x² + c✓

Answer: Integrating term by term: ∫6x² dx = 2x³ and ∫4x dx = 2x², giving 2x³ + 2x² + c.

Evaluate the definite integral ∫₁³ 2x dx.

• A6
• B4
• C9
• D8✓

Answer: ∫2x dx = x², so evaluating from 1 to 3 gives 3² − 1² = 9 − 1 = 8.

Find the area between the curve y = 3x² and the x-axis from x = 0 to x = 2.

• A6
• B12
• C24
• D8✓

Answer: Area = ∫₀² 3x² dx = [x³]₀² = 8 − 0 = 8.

Evaluate ∫₀⁴ (2x + 1) dx.

• A24
• B18
• C20✓
• D16

Answer: ∫(2x + 1) dx = x² + x, so from 0 to 4 the value is (16 + 4) − 0 = 20.

Find ∫ (8x³ − 6x + 5) dx.

• A24x² − 6 + c
• B2x⁴ − 6x² + 5x + c
• C2x⁴ − 3x² + 5x
• D2x⁴ − 3x² + 5x + c✓

Answer: Integrate term by term: ∫8x³ dx = 2x⁴, ∫−6x dx = −3x², ∫5 dx = 5x, plus the constant. So the answer is 2x⁴ − 3x² + 5x + c. The 24x² − 6 + c distractor differentiates instead; 2x⁴ − 6x² + 5x + c mishandles the −6x term; 2x⁴ − 3x² + 5x omits +c.

Find ∫ (4/x² + 3√x) dx. (Write 4/x² as 4x^(−2) and √x as x^(1/2).)

• A−8x^(−3) + (3/2)x^(−1/2) + c
• B4x^(−1) + 2x^(3/2) + c
• C−4x^(−1) + (3/2)x^(3/2) + c
• D−4x^(−1) + 2x^(3/2) + c✓

Answer: ∫4x^(−2) dx = 4x^(−1)/(−1) = −4x^(−1) = −4/x. ∫3x^(1/2) dx = 3x^(3/2)/(3/2) = 2x^(3/2). So the answer is −4/x + 2x^(3/2) + c. The −8x^(−3) + (3/2)x^(−1/2) + c distractor differentiates; 4x^(−1) + 2x^(3/2) + c has a sign error; −4x^(−1) + (3/2)x^(3/2) + c mishandles the coefficient of x^(3/2).

Evaluate the definite integral ∫₁² (3x² + 2) dx.

• A9✓
• B7
• C11
• D8

Answer: ∫(3x² + 2) dx = x³ + 2x. Evaluating: at x = 2, 8 + 4 = 12; at x = 1, 1 + 2 = 3. So the integral is 12 − 3 = 9. Integrating only the 3x² term gives 8 − 1 = 7, which drops the contribution of the + 2; the values 11 and 8 come from arithmetic slips in the substitution.

Find the area enclosed between the curve y = x² and the line y = 2x.

• A8/3
• B4/3✓
• C2/3
• D16/3

Answer: They meet where x² = 2x, i.e. x = 0 and x = 2. Between these, 2x ≥ x², so area = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 12/3 − 8/3 = 4/3. Integrating x² on its own gives 8/3, the area under the curve rather than the area between the two graphs, while 2/3 and 16/3 come from using the wrong upper limit (x = 1 and x = 4 respectively).

Use the substitution u = x² + 1 to find ∫ 2x(x² + 1)³ dx.

• A(x² + 1)⁴ + c
• B(1/4)(x² + 1)⁴ + c✓
• C(1/2)(x² + 1)⁴ + c
• D8x²(x² + 1)² + c

Answer: With u = x² + 1, du = 2x dx, so the integral becomes ∫u³ du = u⁴/4 + c = (1/4)(x² + 1)⁴ + c. Common errors are forgetting to divide by 4 altogether, or dividing by 2 instead of 4.

Use integration by parts to find ∫ x e^x dx.

• Ax e^x + e^x + c
• B(x²/2) e^x + c
• Cx e^x − e^x + c✓
• Dx e^x + c

Answer: Take u = x (du = dx) and dv = e^x dx (v = e^x). Then ∫x e^x dx = x e^x − ∫e^x dx = x e^x − e^x + c. The x e^x + e^x + c distractor has the wrong sign on the second term; (x²/2) e^x + c integrates x as if e^x were constant; x e^x + c omits the second term entirely.

A curve passes through the point (2, 5) and has gradient function dy/dx = 3x² − 4x. Find the equation of the curve.

• Ay = x³ − 2x² + 1
• By = x³ − 2x² + 5✓
• Cy = 6x − 4 + 5
• Dy = x³ − 2x² + 9

Answer: Integrate: y = x³ − 2x² + c. Substitute (2, 5): 5 = 8 − 8 + c, so c = 5. Thus y = x³ − 2x² + 5. The c = 1 version uses the wrong constant; the c = 9 version uses the wrong constant; y = 6x − 4 + 5 differentiates the gradient function instead of integrating.

Find ∫ (2x + 3)⁴ dx using the reverse chain rule.

• A(1/10)(2x + 3)⁵ + c✓
• B(1/5)(2x + 3)⁵ + c
• C(2x + 3)⁵ + c
• D8(2x + 3)³ + c

Answer: Raise the power and divide by the new power and the inner coefficient: ∫(2x + 3)⁴ dx = (2x + 3)⁵/(5 × 2) + c = (1/10)(2x + 3)⁵ + c. Dividing only by the new power gives (1/5)(2x + 3)⁵ + c; dividing by neither leaves (2x + 3)⁵ + c; and 8(2x + 3)³ + c is the derivative of (2x + 3)⁴, not its integral.

Evaluate ∫₀^(π/2) cos x dx.

• A0
• Bπ/2
• C1✓
• D−1

Answer: ∫cos x dx = sin x. Evaluating from 0 to π/2 gives sin(π/2) − sin(0) = 1 − 0 = 1. The −1 distractor results from differentiating cos to −sin; 0 and π/2 are arithmetic/limit confusions.

The gradient of a curve is dy/dx = 6/x² = 6x^(−2), and the curve passes through (3, 1). Find y in terms of x.

• Ay = −6x^(−1) + 3✓
• By = 6x^(−1) − 1
• Cy = −6/x + 1
• Dy = −12x^(−3) + 1

Answer: Integrate: y = ∫6x^(−2) dx = 6x^(−1)/(−1) + c = −6/x + c. Substitute (3, 1): 1 = −6/3 + c = −2 + c, so c = 3. Thus y = −6/x + 3. Leaving the constant as 1 without solving for it gives −6/x + 1; dropping the minus sign from the integration gives 6/x − 1; and −12x^(−3) is the derivative of 6x^(−2), not its integral.

Find ∫ 6x² dx.

• A12x + c
• B2x³ + c✓
• C6x² + c
• Dx³ + c

Answer: Raise the power by one and divide by the new power: ∫6x² dx = 2x³ + c. The constant of integration is required for an indefinite integral — losing it costs a mark.

Find ∫ 4x³ dx.

• Ax⁴/2 + c
• B12x² + c
• C4x³ + c
• Dx⁴ + c✓

Answer: Raise the power by one and divide by the new power: ∫4x³ dx = x⁴ + c. The constant of integration is required for an indefinite integral — losing it costs a mark.

Find ∫ 10x dx.

• A5x² + c✓
• B10 + c
• C10x + c
• D5x²/2 + c

Answer: Raise the power by one and divide by the new power: ∫10x dx = 5x² + c. The constant of integration is required for an indefinite integral — losing it costs a mark.

Evaluate ∫ from 1 to 2 of 3x² dx.

• A7✓
• B8
• C7/2
• D12

Answer: Integrate to get x³, then evaluate at the limits: [8] − [1] = 7. Forgetting to subtract the value at the lower limit is the usual error here.

Evaluate ∫ from 1 to 3 of 2x dx.

• A2
• B9
• C4
• D8✓

Answer: Integrate to get x², then evaluate at the limits: [9] − [1] = 8. Forgetting to subtract the value at the lower limit is the usual error here.

Evaluate ∫ from 1 to 2 of 4x³ dx.

• A48
• B16
• C15/2
• D15✓

Answer: Integrate to get x⁴, then evaluate at the limits: [16] − [1] = 15. Forgetting to subtract the value at the lower limit is the usual error here.

Find the area enclosed by y = x², the x-axis and the lines x = 0 and x = 2.

• A8/3✓
• B4
• C8
• D2

Answer: Area = ∫₀² x² dx = [x³/3]₀² = 2³/3 = 8/3. The area under a curve is the integral, not the value of the function at the endpoint.

Find the area enclosed by y = x², the x-axis and the lines x = 0 and x = 3.

• A3
• B27
• C9/2
• D9✓

Answer: Area = ∫₀³ x² dx = [x³/3]₀³ = 3³/3 = 9. The area under a curve is the integral, not the value of the function at the endpoint.

Which substitution is most useful for ∫ 2x(x² + 1)⁵ dx?

• Au = x⁵, matching the power
• Bu = 2x, because it is the outside factor
• Cu = x² + 1, because du = 2x dx appears in the integral✓
• Du = (x² + 1)⁵, the whole bracket

Answer: Choose u so that du matches the rest of the integrand. With u = x² + 1, du/dx = 2x, so 2x dx = du and the integral becomes ∫u⁵ du — immediate. Substitution works when the derivative of your chosen inner function is already sitting there as a factor.

Why does an indefinite integral require '+ c'?

• ABecause the limits of integration have not yet been substituted into the result
• BBecause the integral sign itself carries an implied constant of proportionality
• CDifferentiating any constant gives zero✓
• DBecause integration increases the power

Answer: Infinitely many functions share the same derivative, differing only by a constant, so integration recovers a whole family and + c stands for the one you cannot determine. A DEFINITE integral has limits, and the constants cancel when you subtract.

When is integration by parts the appropriate method?

• AWhenever the integrand happens to contain a fraction with a polynomial denominator
• BFor a product of two unrelated functions✓
• CWhenever the derivative of the inner function already appears as a factor
• DWhenever the integrand contains a trig function

Answer: Parts handles products such as x·eˣ or x·ln x, where neither factor is the derivative of the other. When the inner derivative IS present as a factor, substitution is the faster route — that is the case to watch for.

Use integration by parts to find ∫ x cos 2x dx.

• A(x/2) sin 2x + (1/4) cos 2x + c✓
• B(x/2) sin 2x − (1/4) cos 2x + c
• C(x/2) sin 2x + (1/2) cos 2x + c
• D−(x/2) cos 2x + (1/4) sin 2x + c

Answer: Take u = x and dv/dx = cos 2x, so v = (1/2) sin 2x. Then ∫ x cos 2x dx = (x/2) sin 2x − ∫ (1/2) sin 2x dx = (x/2) sin 2x + (1/4) cos 2x + c. The version with −(1/4) cos 2x gets the sign of ∫ sin 2x dx wrong; differentiating it does not return x cos 2x.

Find the exact value of ∫₀^π x² sin x dx.

• Aπ² − 4✓
• Bπ² + 4
• Cπ² − 2
• D2π² − 4

Answer: By parts twice: ∫ x² sin x dx = −x² cos x + ∫ 2x cos x dx = −x² cos x + 2x sin x + 2 cos x. At π: π² + 0 − 2; at 0: 0 + 0 + 2. The integral is (π² − 2) − 2 = π² − 4. Forgetting to subtract the lower-limit value 2 leaves π² − 2.

Find the exact value of ∫₁^(e²) ln x dx.

• Ae² − 1
• Be² + 1✓
• C2e²
• D3e² − 1

Answer: By parts with u = ln x, dv/dx = 1: ∫ ln x dx = x ln x − x. Evaluate: (e² × 2 − e²) − (1 × 0 − 1) = e² + 1. Subtracting the lower limit as +1 rather than −1 gives e² − 1; 2e² keeps only the x ln x term.

Use a suitable substitution to find the exact value of ∫₀² x²/√(1 + x³) dx.

• A2/3
• B8/3
• C4/9
• D4/3✓

Answer: Let u = 1 + x³, du = 3x² dx, so the integral is (1/3)∫₁⁹ u^(−1/2) du = (1/3)[2√u]₁⁹ = (2/3)(3 − 1) = 4/3. Leaving out the factor 2 from integrating u^(−1/2) gives 2/3.

Find the exact value of ∫₀^(ln 3) e^(2x)/(eˣ + 1) dx.

• A2 + ln 2
• B2 − ln 4
• C2 − ln 2✓
• D4 − ln 2

Answer: Let u = eˣ + 1, so du = eˣ dx and eˣ = u − 1; the limits become 2 and 4. The integral is ∫₂⁴ (u − 1)/u du = [u − ln u]₂⁴ = (4 − ln 4) − (2 − ln 2) = 2 − ln 2. The value 2 − ln 4 forgets to subtract ln 2 at the lower limit.

Given that (x + 4)/(x(x + 2)²) = 1/x − 1/(x + 2) − 1/(x + 2)², find the exact value of ∫₁² (x + 4)/(x(x + 2)²) dx.

• Aln(3/2) + 1/12
• Bln(3/2) − 1/12✓
• Cln(2/3) − 1/12
• Dln 6 − 1/12

Answer: Integrate term by term: ln x − ln(x + 2) + 1/(x + 2), because ∫ −(x + 2)^(−2) dx = +(x + 2)^(−1). At 2: ln 2 − ln 4 + 1/4; at 1: 0 − ln 3 + 1/3. Difference: ln(2 × 3/4) + 1/4 − 1/3 = ln(3/2) − 1/12. Integrating the squared term with the wrong sign gives ln(3/2) + 1/12.

Use partial fractions to find the exact value of ∫₁³ 3/(x(x + 3)) dx.

• A3 ln 2
• Bln 2✓
• C(1/3) ln 2
• Dln(3/2)

Answer: 3/(x(x + 3)) = 1/x − 1/(x + 3). The integral is [ln(x/(x + 3))]₁³ = ln(3/6) − ln(1/4) = ln(1/2) + ln 4 = ln 2. Getting A = 1/3 instead of 1 (the cover-up value for 1/(x(x + 3))) and then forgetting the factor 3 gives (1/3) ln 2.

Solve the differential equation dy/dx = 2xy, given that y = 3 when x = 0.

• Ay = e^(x²) + 2
• By = 3e^(2x)
• Cy = 3e^(x²/2)
• Dy = 3e^(x²)✓

Answer: Separate: ∫ (1/y) dy = ∫ 2x dx, so ln y = x² + c and y = Ae^(x²). y = 3 at x = 0 gives A = 3. The form e^(x²) + 2 adds the constant after exponentiating; its derivative is 2xe^(x²), which is not 2xy.

Water drains from a tank so that the depth h cm after t minutes satisfies dh/dt = −0.2√h. The initial depth is 16 cm. Find the time taken for the tank to empty.

• A20 minutes
• B80 minutes
• C10 minutes
• D40 minutes✓

Answer: Separate: ∫ h^(−1/2) dh = ∫ −0.2 dt, so 2√h = −0.2t + c. At t = 0, h = 16, so c = 8 and √h = 4 − 0.1t. The tank is empty when √h = 0, i.e. t = 40 minutes. Forgetting the factor 2 from integrating h^(−1/2) gives √h = 4 − 0.2t and 20 minutes.

A population N of rabbits after t months is modelled by dN/dt = 0.1N(1 − N/1000), with N = 100 when t = 0. The solution is N = 1000/(1 + 9e^(−t/10)). Find the time at which the population reaches 500, to 3 significant figures.

• A15.0 months
• B6.93 months
• C22.0 months✓
• D0.220 months

Answer: Set 1000/(1 + 9e^(−t/10)) = 500, so 1 + 9e^(−t/10) = 2 and e^(−t/10) = 1/9. Then t = 10 ln 9 = 21.97 ≈ 22.0 months. Writing 9e^(−t/10) = 2, forgetting to subtract the 1, gives t = 10 ln 4.5 = 15.0; 0.220 divides ln 9 by 10 instead of multiplying.

The mass m kg of a melting block of ice decreases at a rate proportional to the square root of its mass. Which differential equation models this, where k is a positive constant and t is time?

• Adm/dt = −k√t
• Bdm/dt = −k/√m
• Cdm/dt = −k√m✓
• Ddm/dt = −km²

Answer: The rate of change of mass is dm/dt; 'proportional to the square root of its mass' means k√m, and 'decreases' gives the minus sign. −k√t makes the rate depend on time rather than mass, and −k/√m is inversely proportional.

Find the exact value of ∫₀^(π/6) (cos 3x + sec²x) dx.

• A(1 + √3)/3✓
• B1 + √3/3
• C3 + √3/3
• D(1 + 3√3)/3

Answer: ∫ cos 3x dx = (1/3) sin 3x and ∫ sec²x dx = tan x. Evaluate: (1/3) sin(π/2) + tan(π/6) − 0 = 1/3 + √3/3 = (1 + √3)/3. The value 1 + √3/3 integrates cos 3x as sin 3x, missing the factor 1/3.

A curve has parametric equations x = t², y = t³, for t ≥ 0. Find the area of the region bounded by the curve, the x-axis and the line x = 4.

• A4
• B32/5
• C16
• D64/5✓

Answer: x = 4 when t = 2. Area = ∫ y (dx/dt) dt = ∫₀² t³ × 2t dt = [2t⁵/5]₀² = 64/5. Integrating y with respect to t, without the factor dx/dt = 2t, gives ∫₀² t³ dt = 4.