3 11 midpoint rule given a definite integral z b a f

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fairly robust numerical methods built in, with loads of options. 3 1.1 Midpoint Rule Given a definite integral Z b a f ( x ) d x, the n th midpoint approximation 4 is: b - a n [ f ( c 1 ) + f ( c 2 ) + · · · + f ( c n )] , where c i = a + (2 i - 1) b - a 2 n . 3 Check out § 17.2.2 of Heikki Ruskeep¨ a’s textbook, Mathematica Navigator (second edition), for more information—yes, the ECC library has a copy. 4 This will be discussed in class. 2
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Example: Using n = 6 compute the midpoint approximation of Z 4 1 x d x = 2 x x 3 4 1 = 14 3 = 4 2 3 = 4 . ¯ 6 and compare your approximate answer to the exact answer. Work: Z 4 1 x d x 0 . 5 h 1 . 25 + 1 . 75 + 2 . 25 + 2 . 75 + 3 . 25 + 3 . 75 i 4 . 66925 Here’s the Mathematica code for this problem. In[11]:= Integrate @ Sqrt @ x D , 8 x,1, 4 <D Out[11]= 14 ÅÅÅÅÅÅÅ 3 In[12]:= N @ %11, 7 D Out[12]= 4.666667 In[13]:= H 4 - 1 L ê 6 * Sum @ Sqrt @ 1 + H 2 i - 1 L H 4 - 1 L ê H 2 * 6 LD , 8 i,1, 6 <D Out[13]= 1 ÅÅÅÅ 2 i k j j 3 ÅÅÅÅ 2 + è!!! 5 ÅÅÅÅÅÅÅÅÅ 2 + è!!! 7 ÅÅÅÅÅÅÅÅÅ 2 + è!!!!!! 11 ÅÅÅÅÅÅÅÅÅÅÅ 2 + è!!!!!! 13 ÅÅÅÅÅÅÅÅÅÅÅ 2 + è!!!!!! 15 ÅÅÅÅÅÅÅÅÅÅÅ 2 y { z z In[16]:= N @ %13, 7 D Out[16]= 4.669245 In[15]:= NIntegrate @ Sqrt @ x D , 8 x,1,4 <D Out[15]= 4.66667 Untitled-1 1 Figure 3: Mathematica Code 1.2 Trapezoid Rule Given a definite integral Z b a f ( x ) d x, the n th trapezoidal approximation 5 is: b - a 2 n [ f ( a ) + 2 f ( c 1 ) + · · · + 2 f ( c n - 1 ) + f ( b )] , where c i = a + ( b - a ) i n . Example: Using n = 6 compute the trapezoidal approximation of Z 4 1 x d x = 2 x x 3 4 1 = 14 3 = 4 2 3 = 4 . ¯ 6 5 This will be discussed in class. 3
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and compare your approximate answer to the exact answer. Work: Z 4 1 x d x 0 . 25 h 1 + 2 1 . 5 + 2 2 + 2 2 . 5 + 2 3 + 2 3 . 5 + 4 i 4 . 66925 Here’s the Mathematica code for this problem. In[11]:= Integrate @ Sqrt @ x D , 8 x,1, 4 <D Out[11]= 14 ÅÅÅÅÅÅÅ 3 In[12]:= N @ %11, 7 D Out[12]= 4.666667 In[13]:= H 4 - 1 L ê 6 * Sum @ Sqrt @ 1 + H 2 i - 1 L H 4 - 1 L ê H 2 * 6 LD , 8 i,1, 6 <D Out[13]= 1 ÅÅÅÅ 2 i k j j 3 ÅÅÅÅ 2 + è!!! 5 ÅÅÅÅÅÅÅÅÅ 2 + è!!! 7 ÅÅÅÅÅÅÅÅÅ 2 + è!!!!!! 11 ÅÅÅÅÅÅÅÅÅÅÅ 2 + è!!!!!! 13 ÅÅÅÅÅÅÅÅÅÅÅ 2 + è!!!!!! 15 ÅÅÅÅÅÅÅÅÅÅÅ 2 y { z z In[16]:= N @ %13, 7 D Out[16]= 4.669245 In[15]:= NIntegrate @ Sqrt @ x D , 8 x,1,4 <D Out[15]= 4.66667 Untitled-1 1 Figure 4: Mathematica Code 1.3 Simpson’s Rule Here, n needs to be even. Given a definite integral Z b a f ( x ) d x, the n th Simpson’s Rule approximation 6 is: b - a 3 n [ f ( a ) + 4 f ( c 1 ) + 2 f ( c 2 ) + · · · + 2 f ( c n - 2 ) + 4 f ( c n - 1 ) + f ( b )] , where c i = a + b - a n i. Example: Using n = 6 compute the Simpson’s Rule approximation of Z 4 1 x d x = 2 x x 3 4 1 = 14 3 = 4 2 3 = 4 . ¯ 6 and compare your approximate answer to the exact answer. Work: Z 4 1 x d x 1 6 h 1 + 4 1 . 5 + 2 2 + 4 2 . 5 + 2 3 + 4 3 . 5 + 2+ i 4 . 66656 6 This will be discussed in class.
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