Solutions — Area Between Curves and the Trapezoidal Rule

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Find intersections: x + 4 = x² − 2 ⇒ x² − x − 6 = 0 ⇒ (x−3)(x+2) = 0 ⇒ x = −2, 3

Which is on top? At x = 0: y = x + 4 = 4 and y = x² − 2 = −2. So y = x + 4 is on top.

A = ∫₋₂³ [(x+4) − (x²−2)] dx = ∫₋₂³ (x + 6 − x²) dx
= [x²/2 + 6x − x³/3]₋₂³
= (9/2 + 18 − 9) − (4/2 − 12 + 8/3)
= (9/2 + 9) − (2 − 12 + 8/3)
= (9/2 + 9) − (−10 + 8/3)
= 27/2 − (−22/3)
= 27/2 + 22/3 = 81/6 + 44/6 = 125/6 units²
Strip width: w = (3−1)/4 = 0.5
Table of values (f(x) = √x):

x 1 1.5 2 2.5 3

√x 1.0000 1.2247 1.4142 1.5811 1.7321

Estimate = (0.5/2)[1.0000 + 2(1.2247) + 2(1.4142) + 2(1.5811) + 1.7321]
= 0.25[1.0000 + 2.4494 + 2.8284 + 3.1622 + 1.7321]
= 0.25 × 11.1721 = 2.793
(Exact: [2x^3/2/3]₁³ = 2√3·3/3 − 2/3 = 2√27/3 − 2/3 ≈ 2.797. Small overestimate as √x is concave down.)
Find intersections: x² = √x ⇒ x⁴ = x ⇒ x(x³−1) = 0 ⇒ x = 0 or x = 1

Which is on top? At x = 0.25: √0.25 = 0.5 and (0.25)² = 0.0625. So y = √x is on top.

A = ∫₀¹ [√x − x²] dx = [2x^3/2/3 − x³/3]₀¹
= (2/3 − 1/3) − 0 = 1/3 units²
Strip width: w = π/3 x-values: 0, π/3, 2π/3, π
f(0) = 0, f(π/3) = sin(π/3) = √3/2 ≈ 0.8660, f(2π/3) = sin(2π/3) = √3/2 ≈ 0.8660, f(π) = 0

Estimate = (π/3 / 2)[0 + 2(0.8660) + 2(0.8660) + 0]
= (π/6)[3.4641] = π√3/2 ≈ 2.720

Exact value = 2. The rule overestimates by 0.720 (36%). This is a relatively large error because sin x is concave down and the approximation over-estimates with only 3 strips.
On [0,1]: At x = 0.5, y = x = 0.5 > x³ = 0.125. So y = x is on top.
A₁ = ∫₀¹ (x − x³) dx = [x²/2 − x⁴/4]₀¹ = 1/2 − 1/4 = 1/4

On [−1,0]: At x = −0.5, y = x = −0.5 and x³ = −0.125. So y = x³ is on top (less negative).
A₂ = ∫₋₁⁰ (x³ − x) dx = [x⁴/4 − x²/2]₋₁⁰ = 0 − (1/4 − 1/2) = 1/4

Total area = 1/4 + 1/4 = 1/2 units²
On [0,1]: At x = 0.5, e^0.5 ≈ 1.649 > e^−0.5 ≈ 0.607. So y = e^x is on top.

A = ∫₀¹ (e^x − e^−x) dx = [e^x + e^−x]₀¹
= (e + e⁻¹) − (1 + 1)
= e + 1/e − 2
= e + e⁻¹ − 2 units² ≈ 1.086 units²
Strip width: w = 1, n = 6 strips (7 values given)
Estimate = (1/2)[f(0) + 2f(1) + 2f(2) + 2f(3) + 2f(4) + 2f(5) + f(6)]
= (1/2)[2 + 2(5) + 2(9) + 2(11) + 2(10) + 2(7) + 3]
= (1/2)[2 + 10 + 18 + 22 + 20 + 14 + 3]
= (1/2) × 89 = 44.5
Which is on top? At x = π/4, cos(π/4) = sin(π/4) = √2/2. They intersect here.
At x = π/6, cos(π/6) = √3/2 ≈ 0.866 > sin(π/6) = 0.5. So cos x is on top from 0 to π/4.
At x = π/3, sin(π/3) = √3/2 ≈ 0.866 > cos(π/3) = 0.5. So sin x is on top from π/4 to π/2.

A = ∫₀^π/4 (cos x − sin x) dx + ∫_π/4^π/2 (sin x − cos x) dx
= [sin x + cos x]₀^π/4 + [−cos x − sin x]_π/4^π/2
= [(√2/2 + √2/2) − (0 + 1)] + [(−0 − 1) − (−√2/2 − √2/2)]
= [√2 − 1] + [−1 + √2]
= 2√2 − 2 units² ≈ 0.828 units²
Find intersections: x² − 2x = 4 − x² ⇒ 2x² − 2x − 4 = 0 ⇒ x² − x − 2 = 0 ⇒ (x−2)(x+1) = 0 ⇒ x = −1, 2

Which is on top? At x = 0: y = 4 − x² = 4 and y = x² − 2x = 0. So y = 4 − x² is on top.

A = ∫₋₁² [(4 − x²) − (x² − 2x)] dx
= ∫₋₁² (4 + 2x − 2x²) dx
= [4x + x² − 2x³/3]₋₁²
= (8 + 4 − 16/3) − (−4 + 1 + 2/3)
= (12 − 16/3) − (−3 + 2/3)
= (36/3 − 16/3) − (−9/3 + 2/3)
= 20/3 − (−7/3) = 20/3 + 7/3 = 27/3 = 9 units²
Cross-sectional area: strip width w = 1, n = 6 strips, depths: 0, 1.2, 2.5, 3.1, 2.8, 1.9, 0
A = (1/2)[0 + 2(1.2) + 2(2.5) + 2(3.1) + 2(2.8) + 2(1.9) + 0]
= (1/2)[0 + 2.4 + 5.0 + 6.2 + 5.6 + 3.8 + 0]
= (1/2) × 23.0 = 11.5 m²

Flow rate = cross-sectional area × velocity = 11.5 × 2.5 = 28.75 m³/s

x	1	1.5	2	2.5	3
√x	1.0000	1.2247	1.4142	1.5811	1.7321