chap07-et-instructor-solutions

chap07-et-instructor-solutions - TECHNIQUES OF 7...

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7 TECHNIQUES OF INTEGRATION 7.1 Numerical Integration Preliminary Questions 1. What are T 1 and T 2 for a function on [ 0 , 2 ] such that f ( 0 ) = 3, f ( 1 ) = 4, and f ( 2 ) = 3? SOLUTION Using the given function values T 1 = 1 2 ( 2 )( 3 + 3 ) = 6a n d T 2 = 1 2 ( 1 )( 3 + 8 + 3 ) = 7 . 2. For which graph in Figure 16 will T N overestimate the integral? What about M N ? x y y = f ( x ) x y y = g ( x ) FIGURE 16 T N overestimates the value of the integral when the integrand is concave up; thus, T N will overestimate the integral of y = g ( x ) . On the other hand, M N overestimates the value of the integral when the integrand is concave down; thus, M N will overestimate the integral of y = f ( x ) . 3. How large is the error when the Trapezoidal Rule is applied to a linear function? Explain graphically. The Trapezoidal Rule integrates linear functions exactly, so the error will be zero. 4. Suppose T 4 is used to approximate Z 3 0 f ( x ) dx ,where | f 00 ( x ) |≤ 2forall x . What is the maximum possible error? The maximum possible error in T 4 is max | f 00 ( x ) | ( b a ) 3 12 n 2 = 2 ( 3 0 ) 3 12 ( 4 ) 2 = 9 32 . 5. What are the two graphical interpretations of the Midpoint Rule? The two graphical interpretations of the Midpoint Rule are the sum of the areas of the midpoint rectangles and the sum of the areas of the tangential trapezoids. Exercises In Exercises 1–12, calculate T N and M N for the value of N indicated. 1. Z 2 0 x 2 , N = 4 Let f ( x ) = x 2 .Wedivide [ 0 , 2 ] into 4 subintervals of width 1 x = 2 0 4 = 1 2 with endpoints 0 , 0 . 5 , 1 , 1 . 5 , 2, and midpoints 0 . 25 , 0 . 75 , 1 . 25 , 1 . 75. With this data, we get T 4 = 1 2 · 1 2 ³ 0 2 + 2 ( 0 . 5 ) 2 + 2 ( 1 ) 2 + 2 ( 1 . 5 ) 2 + 2 2 ´ = 2 . 75 ; and M 4 = 1 2 ³ 0 . 25 2 + 0 . 75 2 + 1 . 25 2 + 1 . 75 2 ´ = 2 . 625 . 2. Z 4 0 xdx , N = 4
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SECTION 7.1 Numerical Integration 759 SOLUTION Let f ( x ) = x .Wedivide [ 0 , 4 ] into 4 subintervals of width 1 x = 4 0 4 = 1 with endpoints 0 , 1 , 2 , 3 , 4, and midpoints 0 . 5 , 1 . 5 , 2 . 5 , 3 . 5. With this data, we get T 4 = 1 2 · 1 · ³ 0 + 2 1 + 2 2 + 2 3 + 4 ´ 5 . 14626 ; and M 4 = 1 · ³ 0 . 5 + 1 . 5 + 2 . 5 + 3 . 5 ´ 5 . 38382 . 3. Z 4 1 x 3 dx , N = 6 Let f ( x ) = x 3 [ 1 , 4 ] into 6 subintervals of width 1 x = 4 1 6 = 1 2 with endpoints 1 , 1 . 5 , 2 , 2 . 5 , 3 , 3 . 5 , 4, and midpoints 1 . 25 , 1 . 75 , 2 . 25 , 2 . 75 , 3 . 25 , 3 . 75. With this data, we get T 6 = 1 2 µ 1 2 ³ 1 3 + 2 ( 1 . 5 ) 3 + 2 ( 2 ) 3 + 2 ( 2 . 5 ) 3 + 2 ( 3 ) 3 + 2 ( 3 . 5 ) 3 + 4 3 ´ = 64 . 6875 ; and M 6 = 1 2 ³ 1 . 25 3 + 1 . 75 3 + 2 . 25 3 + 2 . 75 3 + 3 . 25 3 + 3 . 75 3 ´ = 63 . 28125 . 4. Z π / 2 0 sin xdx , N = 8 Let f ( x ) = sin x [ 0 , / 2 ] into 8 subintervals of width 1 x = 2 0 8 = 16 with endpoints 0 , 16 , 2 16 ,... , 8 16 = 2 , and midpoints 32 , 3 32 15 32 . With this data, we get T 8 = 1 2 ³ 16 ´³ p sin ( 0 ) + 2 p sin ( / 16 ) +···+ p sin ( 8 / 16 ) ´ 1 . 18005 ; and M 8 = 16 ³ p sin ( / 32 ) + p sin ( 3 / 32 ) p sin ( 15 / 32 ) ´ 1 . 20344 . 5. Z 4 1 x , N = 6 Let f ( x ) = 1 / x [ 1 , 4 ] into 6 subintervals of width 1 x = 4 1 6 = 1 2 with endpoints 1 , 1 . 5 , 2 , 2 . 5 , 3 , 3 . 5 , 4, and midpoints 1 . 25 , 1 . 75 , 2 . 25 , 2 . 75 , 3 . 25 , 3 . 75. With this data, we get T 6 = 1 2 µ 1 2 ¶µ 1 1 + 2 1 . 5 + 2 2 + 2 2 . 5 + 2 3 + 2 3 . 5 + 1 4 1 . 40536 ; and M 6 = 1 2 µ 1 1 . 25 + 1 1 . 75 + 1 2 . 25 + 1 2 . 75 + 1 3 . 25 + 1 3 . 75 1 . 37693 .
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