Chapter 5 | Integration
515
A ≈ R 6 = ∑ i =1 6
f ( x i )Δ x = f ( x 1 )Δ x + f ( x 2 )Δ x + f ( x 3 )Δ x + f ( x 4 )Δ x + f ( x 5 )Δ x + f ( x 6 )Δ x = f (0.5)0.5+ f (1)0.5+ f (1.5)0.5+ f (2)0.5+ f (2.5)0.5+ f (3)0.5 = (0.125)0.5 + (0.5)0.5 + (1.125)0.5 + (2)0.5 + (3.125)0.5 + (4.5)0.5 = 0.0625 + 0.25 + 0.5625 + 1 + 1.5625 + 2.25 =5.6875.
Example 5.4 Approximating the Area Under a Curve
Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of f ( x ) = x 2 on the interval [0, 2]; use n =4.
Solution First, divide the interval [0, 2] into n equal subintervals. Using n =4, Δ x = (2−0) 4 =0.5. approximation, the heights are f (0) =0, f (0.5) = 0.25, f (1) =1, f (1.5) = 2.25. Then, L 4 = f ( x 0 )Δ x + f ( x 1 )Δ x + f ( x 2 )Δ x + f ( x 3 )Δ x = 0(0.5) + 0.25(0.5) + 1(0.5) + 2.25(0.5) =1.75. each rectangle. The intervals ⎡ ⎣ 0, 0.5 ⎤ ⎦ , ⎡ ⎣ 0.5, 1 ⎤ ⎦ , ⎡ ⎣ 1, 1.5 ⎤ ⎦ , ⎡ ⎣ 1.5, 2 ⎤
This is the width of
⎦ are shown in Figure 5.6 . Using a left-endpoint
Figure 5.6 The graph shows the left-endpoint approximation of the area under f ( x ) = x 2 from 0 to 2.
The right-endpoint approximation is shown in Figure 5.7 . The intervals are the same, Δ x =0.5, but nowuse the right endpoint to calculate the height of the rectangles. We have R 4 = f ( x 1 )Δ x + f ( x 2 )Δ x + f ( x 3 )Δ x + f ( x 4 )Δ x = 0.25(0.5) + 1(0.5) + 2.25(0.5) + 4(0.5) =3.75.
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