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Evaluate the upper and lower sums for f(x) = 1 + x2, −1 ≤ x ≤ 1, with n = 3 and 4.

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Answer:

92/27 and 58/27 for n=3 and 13/4 and 9/4 for n=4

Step-by-step explanation:

n = 3: First let’s find ∆x:

∆x = (b − a)/n = 1 − (−1)3 = 2/3

We will have three intervals: −1 ≤ x ≤ −1/3, −1/3 ≤ x ≤1/3, 1/3 ≤ x ≤ 1.

Upper sum: On the first interval, the highest point occurs at f(−1) = 2. On the second interval, the highest point occurs at f(1/3) = f(−1/3) = 1 + ( 1/3)^2=1 + 1/9 = 10/9

On the third interval, the highest point occurs at f(1) = 2. So A ≈ A upper = 3Σ i=1 f(xi)∆x = [f(−1) + f (1/3)+ f(1)]∆x = (2 +10/9+ 2)*2/3 = 92/27

Lower sum: On the first interval, the lowest point occurs at f(−1/3) = 10/9

On the second interval, the lowest point occurs at f(0) = 1. On the third interval, the lowest point occurs at f(1/3) = 10/9. SoA ≈ A lower =3Σ i=1 f(xi)∆x = f(−1/3) + f(0) + f(1/3) . ∆x = (10/9+ 1 +10/9)· 2/3= 58/27

Apply the same technique for n=4

User Sandy Muspratt
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