Solutions — Area Under a Curve

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1. Show Solution
   x-intercepts: x²−1=0 ⇒ x=±1. On [−1,1]: f(x)≤0.
   A = −∫₋₁¹(x²−1)dx = −[x³/3−x]₋₁¹
   = −[(1/3−1)−(−1/3+1)] = −[−2/3−2/3] = 4/3 units²
2. Show Solution
   3x−x²=0 ⇒ x(3−x)=0 ⇒ x=0 or x=3. On [0,3]: f(x)≥0.
   A = ∫₀³(3x−x²)dx = [3x²/2−x³/3]₀³
   = (27/2−9)−0 = 27/2−18/2 = 9/2 = 4.5 units²
3. Show Solution
   ∫₋₁²(x³−x)dx = [x&sup4;/4−x²/2]₋₁²
   = (4−2)−(1/4−1/2) = 2−(−1/4) = 9/4
   
   For area: roots at x=−1, 0, 1. Check: on [−1,0] f≥0? At x=−0.5: f=(−0.5)³−(−0.5)=−0.125+0.5=0.375>0. On [0,1] f≤0. On [1,2] f≥0.
   A = ∫₋₁⁰f dx + |∫₀¹f dx| + ∫₁²f dx
   = [1/4] + |−1/4| + [9/4−1/4] = 1/4+1/4+2 = 5/2 units²
4. Show Solution
   cos x ≥ 0 on [0, π/2].
   A = ∫₀^π/2 cos x dx = [sin x]₀^π/2 = 1 − 0 = 1 unit²
5. Show Solution
   Roots: x=0, x=2, x=−1. On [−1,0]: f≤0; on [0,2]: f≥0 (check signs).
   ∫₋₁⁰(x³−x²−2x)dx = [x&sup4;/4−x³/3−x²]₋₁⁰ = 0−(1/4+1/3−1) = 5/12
   Wait — expand: x(x−2)(x+1) = x(x²−x−2) = x³−x²−2x
   At x=−1: f=(−1)(−3)(0)=0; at x=−0.5: f=−0.5(−2.5)(0.5)=0.625>0 ⇒ on [−1,0]: f≥0
   At x=1: f=1(−1)(2)=−2<0 ⇒ on [0,2]: f≤0
   A₁ = ∫₋₁⁰ f dx = [x&sup4;/4−x³/3−x²]₋₁⁰ = 0−(1/4+1/3−1) = 5/12
   A₂ = |∫₀² f dx| = |[x&sup4;/4−x³/3−x²]₀²| = |(4−8/3−4)| = 8/3
   Total = 5/12 + 8/3 = 5/12 + 32/12 = 37/12 units²
6. Show Solution
   e^x−2=0 ⇒ x=ln 2. On [0, ln 2]: f(x)=e^x−2≤0.
   A = −∫₀^ln2(e^x−2)dx = −[e^x−2x]₀^ln2
   = −[(2−2ln2)−(1−0)] = −(1−2ln2) = 2ln2−1 units² ≈ 0.386 units²
7. Show Solution
   4−x²=0 ⇒ x=±2. On [−2,2]: f(x)≥0 (downward parabola, vertex at (0,4)).
   A = ∫₋₂²(4−x²)dx = [4x−x³/3]₋₂²
   = (8−8/3)−(−8+8/3) = 16−16/3 = 48/3−16/3 = 32/3 units²
   [Sketch: parabola opening down with vertex (0,4), crossing x-axis at (−2,0) and (2,0). Shaded region between curve and x-axis.]
8. Show Solution
   The student's error: they computed the signed (net) area, not the geometric area.
   sin x ≥ 0 on [0,π] and sin x ≤ 0 on [π, 2π]. The positive and negative regions cancel.
   Correct total area = |∫₀^π sin x dx| + |∫_π^2π sin x dx|
   = [−cos x]₀^π + |[−cos x]_π^2π| = 2 + 2 = 4 units²
9. Show Solution
   Roots: x³−3x=x(x²−3)=0 ⇒ x=0, ±√3
   On [−√3,0]: test x=−1: (−1)(1−3)=2>0 ⇒ f≥0
   On [0,√3]: test x=1: 1(1−3)=−2<0 ⇒ f≤0
   A₁ = ∫_−√3⁰(x³−3x)dx = [x&sup4;/4−3x²/2]_−√3⁰ = 0−(9/4−9/2) = 9/4
   A₂ = |∫₀^√3(x³−3x)dx| = |[x&sup4;/4−3x²/2]₀^√3| = |9/4−9/2| = 9/4
   Total area = 9/4 + 9/4 = 9/2 = 4.5 units²
10. Show Solution
    (a) v(t)=t²−4t+3=(t−1)(t−3)=0 ⇒ t=1 and t=3 (both in [0,4])
    
    (b) On [0,1]: v>0 (test t=0.5: 0.25−2+3=1.25). On [1,3]: v<0. On [3,4]: v>0.
    D = ∫₀¹v dt + |∫₁³v dt| + ∫₃⁴v dt
    = [t³/3−2t²+3t]₀¹ + |[t³/3−2t²+3t]₁³| + [t³/3−2t²+3t]₃⁴
    = (1/3−2+3) + |(9−18+9)−(1/3−2+3)| + (64/3−32+12)−(9−18+9)
    = 4/3 + |0−4/3| + (64/3−20)
    = 4/3 + 4/3 + 4/3 = 4 m total distance
    
    (c) Displacement = ∫₀⁴v dt = [t³/3−2t²+3t]₀⁴ = 64/3−32+12 = 64/3−20 = 4/3 m
    Displacement (4/3 m) is the net change in position; distance (4 m) counts all motion. The particle moved backward in [1,3], which reduced net displacement but added to total distance.