Practice Maths

Topic Review — Further Integration — Solutions

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This review covers all lessons in Further Integration: definite integrals, area under a curve, area between curves, the trapezoidal rule, and applications of integration. Questions are mixed in difficulty.

Review Questions

  1. Evaluate ∫25 (3x² − 4x + 1) dx.
    = [x³ − 2x² + x]25 = (125−50+5) − (8−8+2) = 80 − 2 = 78
  2. Evaluate ∫01 (e2x + 3) dx. Give an exact answer.
    = [e2x/2 + 3x]01 = (e²/2 + 3) − (1/2) = (e²+5)/2
  3. Find the exact area enclosed between the curve y = 4 − x² and the x-axis.
    Roots: x = ±2. Area = ∫−22(4−x²)dx = [4x−x³/3]−22 = (8−8/3)−(−8+8/3) = 16−16/3 = 32/3 units²
  4. Use the trapezoidal rule with n = 4 to estimate ∫02 (x³ + 1) dx. Give your answer to 3 decimal places.
    w = 0.5; values: f(0)=1, f(0.5)=1.125, f(1)=2, f(1.5)=4.375, f(2)=9
    = (0.5/2)[1+2(1.125)+2(2)+2(4.375)+9] = 0.25[1+2.25+4+8.75+9] = 0.25×25 = 6.250
    (Exact: [x4/4+x]02 = 4+2 = 6. Rule slightly overestimates.)
  5. Find the area of the region enclosed between y = 2x − x² and y = 0 (the x-axis).
    Roots: 2x−x²=0 ⇒ x=0,2. Parabola is above x-axis on [0,2].
    A = ∫02(2x−x²)dx = [x²−x³/3]02 = 4−8/3 = 4/3 units²
  6. Find the area enclosed between y = x and y = x² − 2.
    Intersections: x = x²−2 ⇒ x²−x−2=0 ⇒ (x−2)(x+1)=0 ⇒ x=−1,2.
    At x=0: y=x=0 > x²−2=−2, so y=x is on top.
    A = ∫−12[x−(x²−2)]dx = ∫−12(x+2−x²)dx = [x²/2+2x−x³/3]−12 = (2+4−8/3)−(1/2−2+1/3) = 10/3−(−7/6) = 27/6 = 9/2 units²
  7. A particle has velocity v(t) = 6 − 2t m/s. Find the displacement from t = 0 to t = 5. Then find the total distance travelled.
    Displacement = ∫05(6−2t)dt = [6t−t²]05 = 30−25 = 5 m
    v=0 at t=3. D = ∫03(6−2t)dt + |∫35(6−2t)dt| = [6t−t²]03 + |[6t−t²]35| = 9 + |30−25−18+9| = 9+4 = 13 m
  8. The rate of fuel consumption of a vehicle is r(t) = 0.5t + 2 litres per hour. Find the total fuel used over 6 hours, and the average rate of consumption.
    Total = ∫06(0.5t+2)dt = [0.25t²+2t]06 = 9+12 = 21 litres
    Average rate = 21/6 = 3.5 L/hr
  9. Given that ∫04 f(x) dx = 10, ∫02 f(x) dx = 3, find ∫24 f(x) dx. Also find ∫42 f(x) dx.
    ∫24 f(x) dx = ∫04 f(x) dx − ∫02 f(x) dx = 10 − 3 = 7
    ∫42 f(x) dx = −∫24 f(x) dx = −7
  10. Find the average value of f(x) = sin x on [0, π/2]. Determine the value of c in [0, π/2] where f(c) equals this average value.
    Average = (2/π)∫0π/2 sin x dx = (2/π)[−cos x]0π/2 = (2/π)(0−(−1)) = 2/π
    Find c: sin c = 2/π ⇒ c = arcsin(2/π) ≈ 0.690 radians
  11. The table gives water flow rate (megalitres/day) for a week:
    Day01234567
    Rate12151814108911
    Use the trapezoidal rule to estimate total flow over 7 days.
    w=1, n=7 strips, 8 values
    = (1/2)[12+2(15)+2(18)+2(14)+2(10)+2(8)+2(9)+11]
    = 0.5[12+30+36+28+20+16+18+11] = 0.5×171 = 85.5 megalitres
  12. Find the value of k such that ∫0k (2x + 1) dx = 12.
    [x²+x]0k = 12 ⇒ k²+k = 12 ⇒ k²+k−12=0 ⇒ (k+4)(k−3)=0
    k > 0, so k = 3
  13. A particle starts at position x = 2 m and moves with velocity v(t) = 3t² − 12t + 9 m/s. Find the position at t = 4 s, and the total distance travelled from t = 0 to t = 4.
    v = 0: 3t²−12t+9=3(t−1)(t−3)=0 at t=1,3
    Displacement = ∫04(3t²−12t+9)dt = [t³−6t²+9t]04 = 64−96+36 = 4
    Position at t=4: x = 2+4 = 6 m

    Distance: I1=[t³−6t²+9t]01=1−6+9=4; I2=|[...]13|=|27−54+27−4|=|−4|=4; I3=[...]34=64−96+36−(27−54+27)=4−0=4
    Total distance = 4+4+4 = 12 m
  14. A region is bounded above by y = 4 − x² and below by y = 4x − x² − 3. Find the enclosed area, showing all working.
    Find intersections: 4−x² = 4x−x²−3 ⇒ 4 = 4x−3 ⇒ 4x=7 ⇒ x=7/4. Wait, let me recheck:
    4−x² = 4x−x²−3 ⇒ 4 = 4x−3 ⇒ 7 = 4x ⇒ x=7/4... this only gives one intersection.
    Reconsidering: 4−x² − (4x−x²−3) = 7−4x. Zero when x=7/4. This is linear, so only one crossing.
    We need limits from context. Let's find where each curve crosses the x-axis:
    y=4−x²=0: x=±2. y=4x−x²−3=0: x=1,3.
    On [1,2] (where both are above x-axis): test x=1.5: top=4−2.25=1.75, bottom=6−2.25−3=0.75. Top is greater.
    A = ∫12[(4−x²)−(4x−x²−3)]dx = ∫12(7−4x)dx = [7x−2x²]12 = (14−8)−(7−2) = 6−5 = 1 unit²
  15. Evaluate ∫π/6 (2 cos x − 1) dx and interpret its geometric meaning.
    = [2 sin x − x]π/6π/2
    = (2×1 − π/2) − (2×1/2 − π/6)
    = (2 − π/2) − (1 − π/6)
    = 1 − π/2 + π/6 = 1 − π/3 ≈ 1 − 1.047 = −0.047
    The negative result means the function (2cos x − 1) is below the x-axis for part of [π/6, π/2], so the signed area is slightly negative (the area below the x-axis is slightly larger than the area above).