Lecture 24 18.01 Fall 2006 Lecture 24: Numerical Integration Numerical Integration We use numerical integration to ﬁnd the deﬁnite integrals of expressions that look like: � b (a big mess) a We also resort to numerical integration when an integral has no elementary antiderivative. For instance, there is no formula for � x � 3 2 2 cos(t )dt or e−x dx 0 0 Numerical integration yields numbers rather than analytical expressions. We’ll talk about three techniques for numerical integration: Riemann sums, the trapezoidal rule, and Simpson’s rule. 1. Riemann Sum a b Figure 1: Riemann sum with left endpoints: (y0 + y1 + . . . + yn−1 )Δx Here, xi − xi−1 = Δx (or, xi = xi−1 + Δx) a = x0 < x1 < x2 < . . . < xn = b y0 = f (x0 ), y1 = f (x1 ), . . . yn = f (xn ) 1 Lecture 24 18.01 Fall 2006 2. Trapezoidal Rule The trapezoidal rule divides up the area under the function into trapezoids, rather than rectangles. The area of a trapezoid is the height times the average of the parallel bases: � � � � base 1 + base 2 y3 + y4 Area = height = Δx (See Figure 2) 2 2 y4 y3 ∆x Figure 2: Area = a � y3 + y 4 2 � Δx b Figure 3: Trapezoidal rule = sum of areas of trapezoids. � Total Trapezoidal Area = = y0 + y1 y1 + y2 y2 + y3 yn−1 + yn + + + ... + 2 2 2 2 �y yn � 0 Δx + y1 + y2 + ... + yn−1 + 2 2 Δx 2 � Lecture 24 18.01 Fall 2006 Note: The trapezoidal rule gives a more symmetric treatment of the two ends (a and b) than a Riemann sum does — the average of left and right Riemann sums. 3. Simpson’s Rule This approach often yields much more accurate results than the trapezoidal rule does. Here, we match quadratics (i.e. parabolas), instead of straight or slanted lines, to the graph. This approach requires an even number of intervals. y0 y2 y1 x₀ x₁ ∆x x₂ ∆x Figure 4: Area under a parabola. � Area under parabola = (base)(weighted average height) = (2Δx) y0 + 4y1 + y2 6 � Simpson’s rule for n intervals (n must be even!) � � 1 Area = (2Δx) [(y0 + 4y1 + y2 ) + (y2 + 4y3 + y4 ) + (y4 + 4y5 + y6 ) + · · · + (yn−2 + 4yn−1 + yn )] 6 Notice the following pattern in the coeﬃcients: 1 1 4 4 1 1 4 2 4 3 1 1 2 4 4 1 1 Lecture 24 18.01 Fall 2006 1st chunk 0 2nd chunk 1 2 3 4 Figure 5: Area given by Simpson’s rule for four intervals Simpson’s rule: � b f (x) dx ≈ a Δx (y0 + 4y1 + 2y2 + 4y3 + 2y4 + . . . + 4yn−3 + 2yn−2 + 4yn−1 + yn ) 3 The pattern of coeﬃcients in parentheses is: 1 1 4 1 4 2 4 2 4 1 4 2 1 4 1 = = = sum 6 sum 12 sum 18 To double check – plug in f (x) = 1 (n even!). �n� �n �� Δx Δx � (1 + 4 + 2 + 4 + 2 + · · · + 2 + 4 + 1) = 1+1+4 +2 − 1 = nΔx (n even) 3 3 2 2 4 Lecture 24 18.01 Fall 2006 � 1 Example 1. Evaluate 0 integration. 1 dx using two methods (trapezoidal and Simpson’s) of numerical 1 + x2 ∆x ∆x 0 Figure 6: Area under x 0 1 2 1 1 1 (1+x2 ) above [0, 1]. 1/(1 + x2 ) 1 4 5 1 2 By the trapezoidal rule: � Δx 1 1 y0 + y1 + y2 2 2 � 1 = 2 � 1 4 1 (1) + + 2 5 2 � �� 1 1 4 1 1 = + + = 0.775 2 2 2 5 4 By Simpson’s rule: Δx 1/2 (y0 + 4y1 + y2 ) = 3 3 � � 1+4 4 1 + 5 2 �� = 0.78333... Exact answer: � 0 1 �1 1 π π � dx = tan−1 x� = tan−1 1 − tan−1 0 = − 0 = ≈ 0.785 2 1+x 4 4 0 Roughly speaking, the error, | Simpson’s − Exact |, has order of magnitude (Δx)4 . 5