Lecture13bc - Welcome to Math 122 Lecture 13 Survey of Calculus and its Applications I1 Instructor Quanlei Fang Dept of Math University at Buffalo

Lecture13bc - Welcome to Math 122 Lecture 13 Survey of...

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Instructor: Quanlei Fang Dept. of Math, University at Buffalo, Spring 2009 Welcome to Math 122 Lecture 13 Survey of Calculus and its Applications I1 § 9.4 Approximation of Definite Integrals ! The Midpoint Rule ! The Trapezoidal Rule ! Simpson’s Rule ! Error Analysis
Approximate the following integral by the midpoint rule. We have ! x = ( b a )/ n = . The endpoints of the four subintervals begin at a = 1 and are spaced 1 unit apart. The first midpoint is at a + ! x /2 = . The midpoints are also spaced unit apart. According to the midpoint rule, the integral is approximately equal to
Approximate the following integral by the trapezoidal rule. 2x"3()314#dx;n=3As in the last example, !x= and the endpoints of the subintervals are a0= , a1= , a2= , and a3= . The trapezoidal rule gives
Approximate the following integral by Simpson’s rule. 2 x " 3 ( ) 3 1 4 # dx ; n = 3
Here a= , b= , and f(x) = (2x – 3)3. Differentiating twice, we find that ""f x( )How large could | f""(x)| be if xsatisfies 1 #x# = 4? Since the function 48xis clearly increasing on the interval from 1 to 4 (in fact, it’s increasing everywhere), its greatest value occurs at x= . Therefore, its greatest value is – 72
so we may take A= in the preceding theorem. The error of approximation using the midpoint rule is at most NOTE: We have hitherto determined that the exact value of this integral is 78 and that the midpoint approximation for it (using n= 3) is 72. Therefore, this approximation was in error by 78 – 72 = 6. Our result in this exercise says that our midpoint approximation error should be no greater than 15. Since 6 is no greater than15, this result suggests that our midpoint approximation was done correctly.

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