A Level Mathematics: Integration, questions with answers

Integrate by parts twice: ∫3x²e^(3x)dx = x²e^(3x) − ∫2x e^(3x)dx = x²e^(3x) − (2/3)xe^(3x) + (2/9)e^(3x).
Limits 0 to 1: (1 − 2/3 + 2/9)e³ − 2/9 = (5/9)e³ − 2/9.

Use integration by parts with u = x² (integrating 3e^(3x)), then repeat for the remaining x e^(3x) integral.

Using sin²x = (1 − cos2x)/2: 6sin²x = 3 − 3cos2x.
∫(3 − 3cos2x) dx = 3x − (3/2)sin2x + c.

Use the double-angle identity sin²x = (1−cos2x)/2 to rewrite the integrand before integrating.

(a) Divide 9x³ + 6x² + 12x + k by (3x + 2):
9x³ ÷ 3x = 3x²; 3x²(3x + 2) = 9x³ + 6x²; subtract to leave 12x + k.
12x ÷ 3x = 4; 4(3x + 2) = 12x + 8; subtract to leave k - 8.
Quotient: 3x² + 4, remainder: k - 8, so p(x)/(3x+2) = 3x² + 4 + (k-8)/(3x+2).
(b) Integrate the result of part (a):
∫(3x² + 4)dx = x³ + 4x
∫(k-8)/(3x+2) dx = [(k-8)/3] ln(3x+2)
So ∫p(x)/(3x+2) dx = x³ + 4x + [(k-8)/3] ln(3x+2)
Evaluate from 1 to 6:
At x = 6: 216 + 24 + [(k-8)/3] ln 20 = 240 + [(k-8)/3] ln 20
At x = 1: 1 + 4 + [(k-8)/3] ln 5 = 5 + [(k-8)/3] ln 5
Difference: 235 + [(k-8)/3](ln 20 - ln 5) = 235 + [(k-8)/3] ln 4
Comparing with a + ln 64: since ln 64 = 3 ln 4, take a = 235 and [(k-8)/3] ln 4 = 3 ln 4 …

The full model answer is in the app.

By parts with u = ln 3x, dv = x² dx, so du = (1/x)dx, v = x³/3:
∫ x² ln 3x dx = (x³/3)ln 3x − ∫ (x³/3)(1/x) dx = (x³/3)ln 3x − (1/3)∫x² dx = (x³/3)ln 3x − x³/9
Evaluate between x = 1 and x = 3:
At x = 3: (27/3)ln 9 − 27/9 = 9 ln 9 − 3 = 18 ln 3 − 3
At x = 1: (1/3)ln 3 − 1/9
∫₁³ x² ln 3x dx = (18 ln 3 − 3) − ((1/3)ln 3 − 1/9) = (53/3) ln 3 − 26/9
So a = 53/3, b = 3, c = −26/9.

Integrate by parts, differentiating ln 3x and integrating x², then substitute the limits 1 and 3.

(a) f(x) = A/(2x+1) + B/(x+2) + C/(x+2)².
3−3x² = A(x+2)² + B(2x+1)(x+2) + C(2x+1).
At x=−2: 3−12 = C(−3) → C = 3.
At x=−1/2: 3−3/4 = A(3/2)² → 9/4 = A(9/4) → A = 1.
Comparing x² coefficients: −3 = A + 2B = 1+2B → B = −2.
f(x) = 1/(2x+1) − 2/(x+2) + 3/(x+2)²
(b) ∫f(x)dx = (1/2)ln(2x+1) − 2ln(x+2) − 3/(x+2).
At x=4: (1/2)ln9 − 2ln6 − 3/6 = (1/2)ln9 − 2ln6 − 1/2.
At x=0: (1/2)ln1 − 2ln2 − 3/2 = −2ln2 − 3/2.
Difference: [(1/2)ln9 − 2ln6 − 1/2] − [−2ln2 − 3/2] = (1/2)ln9 − 2ln6 + 2ln2 + 1.
(1/2)ln9 = ln3, and −2ln6+2ln2 = 2ln(2/6) = −2ln3.
Total: ln3 − 2ln3 + 1 = 1 − ln3
∫₀⁴f(x)dx = 1 − ln3

Use the cover-up rule / comparing coefficients to find A, B, C; integrate each partial fraction term and substitute the limits.

(a) ∫ 4/(2x−5) dx = 2 ln(2x−5) + c (since d/dx[ln(2x−5)] = 2/(2x−5)).
Evaluate between limits 4 and 10:
[2 ln(2x−5)] from 4 to 10 = 2 ln(15) − 2 ln(3)
Use the log laws: 2 ln15 − 2 ln3 = 2 ln(15/3) = 2 ln5 = ln(5²) = ln 25
So the integral equals ln 25 (a = 25).
(b) ∫ e^(2x−5) dx = ½ e^(2x−5) + c
Evaluate between limits 4 and 10:
[½ e^(2x−5)] from 4 to 10 = ½e^15 − ½e^3
So the exact value is ½e^15 − ½e^3.

Integrate 4/(2x−5) to 2ln(2x−5) and e^(2x−5) to ½e^(2x−5), then apply the limits and simplify using log laws for part (a).

(a) Dividing: p(x) = (4x - 3)(3x² + 2) + 2. Check: (4x-3)(3x²+2) + 2 = 12x³ + 8x - 9x² - 6 + 2 = 12x³ - 9x² + 8x - 4. So the quotient is 3x² + 2 and the remainder is 2.
(b) p(x)/(4x-3) - 3x² = [3x² + 2 + 2/(4x-3)] - 3x² = 2 + 2/(4x-3). So ∫₂¹² [2 + 2/(4x-3)] dx = [2x + ½ln(4x-3)]₂¹² = (24 + ½ln45) - (4 + ½ln5) = 20 + ½ln(45/5) = 20 + ½ln9 = 20 + ln3.

Perform polynomial division to simplify the integrand, then integrate and apply the limits, using log laws to simplify the final answer.

(a) Using the double angle formulas:
1 − cos 2θ = 2 sin²θ
1 + cos 2θ = 2 cos²θ
So (1 − cos 2θ)/(1 + cos 2θ) = 2sin²θ/2cos²θ = tan²θ ✓
(b) By part (a), the integral is ∫tan²θ dθ = ∫(sec²θ − 1) dθ = tan θ − θ (+ C)
Evaluate from π/6 to π/3:
At θ = π/3: tan(π/3) − π/3 = √3 − π/3
At θ = π/6: tan(π/6) − π/6 = (√3/3) − π/6
Difference = (√3 − π/3) − (√3/3 − π/6) = √3 − √3/3 − π/3 + π/6 = (2√3/3) − π/6

Use the double-angle identities to rewrite the fraction as tan²θ, then integrate tan²θ via the identity tan²θ = sec²θ − 1 and substitute the limits.