INSTRUCTORS Raffi Hovasapian John Zhu

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### Trapezoid Rule

• Understand that
• y acts as (b1+b2) term
• xacts as h term
• Same principle as rectangle rule, just with a different shape

### Trapezoid Rule

Approximate the area under f(x) = x3 from x = 1 to x = 5 using 4 inscribed trapezoids.
• Width of trapezoids = [(5 − 1)/4] = 1
• A ≈ (1) [1/2][f(1) + 2f(2) + 2f(3) + 2f(4) + f(5)]
• A ≈ [1/2][1 + 2(8) + 2(27) + 2(64) + 125]
• A ≈ [1/2][1 + 16 + 54 + 128 + 125]
A ≈ 162
Approximate the area under f(x) = x3 from x = 1 to x = 5 using 8 inscribed trapezoids.
• Width of trapezoids = [(5 − 1)/8] = [1/2]
• A ≈ [1/2] [1/2][f(1) + 2f(1.5) + 2(f(2)) + 2(f(2.5)) + 2(f(3)) + 2(f(3.5)) + 2(f(4)) + 2(f(4.5)) + f(5)]
• A ≈ [1/4][1 + 6.75 + 16 + 31.25 + 54 + 85.75 + 128 + 182.25 + 125]
A ≈ 157.5
Approximate the area under f(x) = x3 from x = −2 to x = 2 using 4 inscribed trapezoids.
• Width of trapezoids = [(2 − (−2))/4] = 1
• A ≈ [1/2] [f(−2) + 2f(−1) + 2f(0) + 2f(1) + f(2)]
• A ≈ [1/2] [−8 − 2 + 0 + 2 + 8]
• A ≈ 0
• This is a property of odd functions. If A1 = Area from −a to 0 of an odd function and A2 = Area from 0 to a of the same odd function, A1 = −A2 or A1 + A2 = 0
A = 0
Approximate the area under f(x) = 3 sinx from x = 0 to x = π using 4 inscribed trapezoids.
• Width of trapezoids = [(π− 0)/4] = [(π)/4]
• A ≈ [(π)/4] [1/2] [f(0) + 2(f([(π)/4])) + 2(f([(π)/2])) + 2(f([(3π)/4])) + f(π)]
• A ≈ [(π)/8][3 sin0 + 6 sin[(π)/4] + 6 sin[(π)/2] + 6 sin[(3π)/4] + 3sinπ]
• A ≈ [(π)/8][0 + [6/(√2)] + 6 + [6/(√2)] + 0]
A ≈ [(π)/8](6 + [12/(√2)])
Approximate the area under f(x) = [1/2]x from x = 0 to x = 3 using 3 inscribed trapezoids.
• Width of trapezoids = [(3 − 0)/3] = 1
• A ≈ [1/2][0 + 2[1/2] + 2(1) + [3/2]]
A ≈ [9/4]
Approximate the area under f(x) = [1/2]x from x = 0 to x = 3 using 6 inscribed trapezoids.
• Width of trapezoids = [(3 − 0)/6] = [1/2]
• A ≈ [1/4][f(0) + 2(f(.5)) + 2(f(1)) + 2(f(1.5)) + 2(f(2)) + 2(f(2.5)) + f(3)]
• A ≈ [1/4][0 + 2[1/4] + 2[1/2] + 2[3/4] + 2 + [5/2] + [3/2]]
• A ≈ [1/4][[1/2] + 1 + [3/2] + 2 + [5/2] + [3/2]]
• It turns out that the trapezoid method is an accurate way to measure the area under a line. In truth, only one trapezoid is needed for a linear function.
A = [9/4]
Approximate the area under f(x) = 3x2 from x = −4 to x = 0 using 4 inscribed trapezoids.
• Width of trapezoids = [(0 − (−4))/4] = 1
• A ≈ [1/2] [f(−4) + 2f(−3) + 2f(−2) + 2f(−1) + f(0)]
• A ≈ [1/2] [48 + 2(27) + 2(12) + 2(3) + 0]
• A ≈ [1/2] [48 + 54 + 24 + 6]
A ≈ 66
Approximate the area under f(x) = 3x2 from x = 0 to x = 4 using 4 inscribed trapezoids.
• Width of trapezoids = [(4 − 0)/4] = 1
• A ≈ [1/2] [f(0) + 2f(1) + 2f(2) + 2f(3) + f(4)]
• A ≈ [1/2] [0 + 6 + 24 + 54 + 48]
• This is a property of even functions. If A1 is the area of an even function from −a to 0, and A2 is the area of the same even function from 0 to a, then A1 = A2
A ≈ 66
Approximate the area under f(x) = 3x2 from x = −4 to x = 4 using 8 inscribed trapezoids.
• If the trapezoid widths were different, the same approximations found in the previous problems won't necessarily hold. But, this problem shares the same width as previous two problems. We can use the approximates previously found.
• A ≈ 66 + 66
A ≈ 132
Approximate the area under f(x) = cosx from x = 0 to x = π using 4 inscribed trapezoids.
• Width of trapezoids = [(π− 0)/4] = [(π)/4]
• A ≈ [(π)/4] [1/2] [f(0) + 2f([(π)/4]) + 2f([(π)/2]) + 2f([(3π)/4]) + f(π)]
• A ≈ [(π)/8] [cos0 + 2(cos[(π)/4]) + 2(cos[(π)/2]) + 2(cos[(3π)/4]) + cosπ]
• A ≈ [(π)/8] [1 + √2 + 0 + (−√2) + (−1)]
A ≈ 0

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.