# sol5 - Chapter 5 1 Section 5.1 1 The velocity v(t in the...

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Chapter 5 1. Section 5.1 1. The velocity v ( t ) in the table below is increasing. Use n = 4 subdivisions to approximate the total distance traveled. Also, ﬁnd t 3 6 9 12 15 f ( t ) 20 21 25 30 40 (a) An upper estimate. (b) A lower estimate. Solution: (a) The thing to notice is that f ( t ) is increasing and getting progressively steeper. This means that an upper estimate can be obtained by using right sums. The right point sum in this case is given by distance 3( f (6) + f (9) + f (12) + f (15)) = 3(116) = 348 (b) For the same reasons given in part (a), a lower estimate for the distance is obtained by using left sums. In this case distance 3( f (3) + f (6) + f (9) + f (12)) = 3(96) = 278 Remark : You can also approximate distance traveled, or equivalently area under curves, by using rectangles of diﬀerent widths. For example, a left handed Riemann sum with nonuniform rectangles would have the following form L ( f ) = n X i =0 f ( a + i Δ x i x i The subscript i with the Δ x indicates that the width of each rectangle varies. 2. Consider the function f ( x ) = 1 1 + x 2 for - 2 x 2 Use Δ x = 1 to ﬁnd the: (a) Left sum (b) Right sum Solution: (a) First note that since we are insisting on Δ x = 1, the number of rectangles we will be using is n = ( b - a ) / x ) = (2 - ( - 2)) / 1 = 4. Furthermore, the points at which we need to evaluate f ( x ) have the form x i = a + i Δ x for i = 0 , .., n - 1 x i = - 2 + i for i = 0 , 1 , 2 , 3 After plugging and chugging , we see that f ( x ) must be evaluated at - 2 , - 1 , 0 , and 1. This generates the left hand sum area Δ x ( f ( - 2) + f ( - 1) + f (0) + f (1)) area f ( - 2) + f ( - 1) + f (0) + f (1) area (1 / 5 + 1 / 2 + 1 + 1 / 2) (b) The reasoning is exactly the same as that given for part (a), the only diﬀerence being that f ( x ) must be evaluated at points of the form x i = a + i Δ x for i = 1 , 2 , ..., n x i = - 2 + i for i = 1 , 2 , 3 , 4 1

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Another plug and chug leads us to plug in - 1 , 0 , 1 , and , 2 in for x . Therefore the right hand sum is area Δ x ( f ( - 1) + f (0) + f (1) + f (2)) area f ( - 1) + f (0) + f (1) + f (2) area (1 / 2 + 1 + 1 / 2 + 1 / 5) 2. Section 5.2 1. Estimate the integral with n = 4 using the following techniques Z 6 2 1 + x 2 dx (a) Left-hand sum Solution: First of all, Δ x for this problem is Δ x = ( a - b ) /n = (6 - 2) / 4 = 1 For the left-hand sum, the points that we plug in for x have the form x i = a + i Δ x for i = 1 , 2 , ..., n x i = 2 + i Δ x for i = 1 , 2 , 3 , 4 (b) Right-hand sum 2. For question 1 above, how far oﬀ can your estimates be from the true value? Solution: The thing to notice (this can be seen from the graph of f ( x ) = 1 + x 2 ) is that the right-hand sum is an over estimate to the integral, while the left-hand sum is an under estimate to the integral. So the integral value lies somewhere in the middle. Letting R ( f ) denote the right hand sum and L denote the left hand sum, we have
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sol5 - Chapter 5 1 Section 5.1 1 The velocity v(t in the...

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