Practice Maths

Solutions — Fundamental Theorem of Calculus and Definite Integrals

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  1. ∫13 (4x − 1) dx = [2x² − x]13
    = (2(9) − 3) − (2(1) − 1)
    = (18 − 3) − (2 − 1)
    = 15 − 1 = 14
  2. ∫02 (x² + 3x) dx = [x³/3 + 3x²/2]02
    = (8/3 + 6) − (0 + 0)
    = 8/3 + 18/3 = 26/3 ≈ 8.67
  3. ∫1e (1/x) dx = [ln|x|]1e
    = ln(e) − ln(1)
    = 1 − 0 = 1
  4. δx = (4 − 0)/4 = 1   Left endpoints: x = 0, 1, 2, 3
    f(0) = 0, f(1) = 1, f(2) = 4, f(3) = 9
    Riemann sum = 1 × (0 + 1 + 4 + 9) = 14

    (Exact value: [x³/3]04 = 64/3 ≈ 21.3 — left endpoints underestimate since f is increasing)
  5. Using the splitting property:
    ∫15 f(x) dx = ∫13 f(x) dx + ∫35 f(x) dx
    7 = ∫13 f(x) dx + 2
    ∫13 f(x) dx = 5
  6. ∫0π/4 2 sin x dx = [−2 cos x]0π/4
    = −2 cos(π/4) − (−2 cos 0)
    = −2 × (√2/2) + 2
    = −√2 + 2 = 2 − √2
  7. ∫02 e3x dx = [e3x/3]02
    = e6/3 − e0/3
    = (e6 − 1)/3
  8. The student's working is correct. Since ∫52 f(x) dx = −10 (given),
    using the reversed-limits property: ∫25 f(x) dx = −∫52 f(x) dx = −(−10) = 10
    The reversal of limits changes the sign, so ∫25 f(x) dx = 10.
  9. δx = (3−1)/4 = 0.5   Right endpoints: x = 1.5, 2, 2.5, 3
    f(1.5) = 1/1.5 ≈ 0.667, f(2) = 0.5, f(2.5) = 0.4, f(3) = 0.333
    Estimate = 0.5(0.667 + 0.5 + 0.4 + 0.333) = 0.5 × 1.9 = 0.950

    Exact: ∫13 (1/x) dx = [ln x]13 = ln 3 ≈ 1.099
    Percentage error = |0.950 − 1.099|/1.099 × 100 ≈ 13.6%
    (Right endpoints underestimate here because 1/x is a decreasing function.)
  10. Since g′(x) = f(x), g is an antiderivative of f.
    By the FTC: ∫14 f(x) dx = [g(x)]14 = g(4) − g(1) = 11 − 3 = 8

    This represents the net change in g(x) as x goes from 1 to 4, which equals the signed area under the graph of f(x) = g′(x) between x = 1 and x = 4.

    Sketch: f(x) curve on [1,4]; the shaded region between f and the x-axis has signed area = 8. The antiderivative g increases by 8 units over this interval.