SalasSV_08_07 - 486 CHAPTER 8 TECHNIQUES OF INTEGRATION 8.7...

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486 ± CHAPTER 8 TECHNIQUES OF INTEGRATION ± 8.7 NUMERICAL INTEGRATION To evaluate a defnite integral by the Formula ± b a f ( x ) dx = F ( b ) F ( a ), we must be able to fnd an antiderivative F and we must to able to evaluate this antiderivative both at a and at b . When this is not possible, the method Fails. The method Fails even For such simple-looking integrals as ± 1 0 x sin xdx or ± 1 0 e x 2 dx . There are no elementary functions with derivative x sin x or e x 2 . Here we take up some simple numerical methods For estimating defnite integrals— methods that you can use whether or not you can fnd an antiderivative. All the methods we describe involve only simple arithmetic and are ideally suited to the computer. We Focus now on ± b a f ( x ) dx . We suppose that f is continuous on [ a , b ] and, For pictorial convenience, assume that f is positive. Take a regular partition P ={ x 0 , x 1 , x 2 , ... , x n 1 , x n } oF [ a , b ], subdividing the interval into n subintervals each oF length ( b a ) / n : [ a , b ] = [ x 0 , x 1 ] ∪···∪ [ x i 1 , x i ] [ x n 1 , x n ] with ± x i = b a n . The region ² i pictured in ±igure 8.7.1, can be approximated in many ways. f i x i – 1 x i b – a n x Figure 8.7.1 f x i – 1 x i x Figure 8.7.2 (1) By the leFt-endpoint rectangle (±igure 8.7.2): area = f ( x i 1 ) ± x i = f ( x i 1 ) ² b a n ³ .
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8.7 NUMERICAL INTEGRATION ± 487 (2) By the right-endpoint rectangle (Figure 8.7.3): area = f ( x i ) ± x i = f ( x i ) ± b a n ² . f x i – 1 x i x Figure 8.7.3 (3) By the midpoint rectangle (Figure 8.7.4): area = f ± x i 1 + x i 2 ² ± x i = f ± x i 1 + x i 2 ²± b a n ² f x i – 1 x i x i – 1 + x i 2 x Figure 8.7.4 (4) By a trapezoid (Figure 8.7.5): area = 1 2 [ f ( x i 1 ) + f ( x i )] ± x i = 1 2 [ f ( x i 1 ) + f ( x i )] ± b a n ² f x i – 1 x i x Figure 8.7.5 x f x i – 1 x i x i – 1 + x i 2 Figure 8.7.6 (5) By a parabolic region (Figure 8.7.6): take the parabola y = Ax 2 + Bx + C that passes through the three points indicated. area = 1 6 ³ f ( x i 1 ) + 4 f ± x i 1 + x i 2 ² + f ( x i ) ´ ± x i = ³ f ( x i 1 ) + 4 f ± x i 1 + x i 2 ² + f ( x i ) ´± b a 6 n ² . You can verify this formula for the area under the parabola by doing Exercises 11 and 12. (If the three points are collinear, the parabola degenerates to a straight line and the parabolic region becomes a trapezoid. The formula then gives the area of the trapezoid.) The approximations to ² i just considered yield the following estimates for µ b a f ( x ) dx . (1) The left-endpoint estimate: L n = b a n [ f ( x 0 ) + f ( x 1 ) +···+ f ( x n 1 )]. (2) The right-endpoint estimate: R n = b a n [ f ( x 1 ) + f ( x 2 ) f ( x n )]. (3) The midpoint estimate: M n = b a n ³ f ± x 0 + x 1 2 ² f ± x n 1 + x n 2 ²´ .
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488 ± CHAPTER 8 TECHNIQUES OF INTEGRATION (4) The trapezoidal estimate ( trapezoidal rule ): T n = b a n ± f ( x 0 ) + f ( x 1 ) 2 + f ( x 1 ) + f ( x 2 ) 2 +···+ f ( x n 1 ) + f ( x n ) 2 ² = b a 2 n [ f ( x 0 ) + 2 f ( x 1 ) 2 f ( x n 1 ) + f ( x n )].
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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SalasSV_08_07 - 486 CHAPTER 8 TECHNIQUES OF INTEGRATION 8.7...

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