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1 Example 2 Evaluate the lower Riemann sum L(P,f ) and the upper Riemann sum U(P,f ) where P is the regular partition of [1,2] into five subintervals, and f(x)=1/x. Solution In Example 3.2C (2) we showed that P is the partition P = {1, 6/5, 7/5, 8/5, 9/5, 2}. To locate the minimum of f on the subintervals of [1,2] and compute L(P,f ), sketch the graph of f(x)=1/x: Then t 1 *=6/5, the point where f has its minimum value on the subinterval [1,6/5], t 2 *=7/5, the point where f has its minimum value on the subinterval [6/5,7/5], t 3 *=8/5, the point where f has its minimum value on the subinterval [7/5,8/5], t 4 *=9/5, the point where f has its minimum value on the subinterval [8/5,9/5] and t 5 *=2, the point where f has its minimum value on the subinterval [9/5,2]. Then L(P,f)=R(P,T*,f )= f(t 1 *)(x 1 -x 0 )+ f(t 2 *)(x 2 -x 1 )+ f(t 3 *)(x 3 -x 2 )+ f(t 4 *)(x 4 -x 3 )+ f(t 5 *)(x 5 -x 4 ) where x 0 =1, x 1 =6/5, x 2 =7/5, x 3 =8/5, x 4 =9/5, x 5 =2. Observe that x k - x k-1 =1/5 for every k. Hence L(P,f) = f(6/5)[1/5] + f(7/5)[1/5] + f(8/5)[1/5] + f(9/5)[1/5] + f(2)[1/5] = 1/5[5/6+ 5/7 + 5/8 + 5/9 + 1/2] = 1/6+ 1/7 + 1/8 + 1/9 + 1/10 0.646
![1 Example 2 Evaluate the lower Riemann sum L(P,f ) and the upper Riemann sum U(P,f ) where P is the regular partition of [1,2] into five subintervals, and f(x)=1/x.](http://images.slideplayer.com./26/8401000/slides/slide_1.jpg "Solution In Example 3.2C (2) we showed that P is the partition P = {1, 6/5, 7/5, 8/5, 9/5, 2}. To locate the minimum of f on the subintervals of [1,2] and compute L(P,f ), sketch the graph of f(x)=1/x: Then t 1 *=6/5, the point where f has its minimum value on the subinterval [1,6/5], t 2 *=7/5, the point where f has its minimum value on the subinterval [6/5,7/5], t 3 *=8/5, the point where f has its minimum value on the subinterval [7/5,8/5], t 4 *=9/5, the point where f has its minimum value on the subinterval [8/5,9/5] and t 5 *=2, the point where f has its minimum value on the subinterval [9/5,2]. Then L(P,f)=R(P,T*,f )= f(t 1 *)(x 1 -x 0 )+ f(t 2 *)(x 2 -x 1 )+ f(t 3 *)(x 3 -x 2 )+ f(t 4 *)(x 4 -x 3 )+ f(t 5 *)(x 5 -x 4 ) where x 0 =1, x 1 =6/5, x 2 =7/5, x 3 =8/5, x 4 =9/5, x 5 =2. Observe that x k - x k-1 =1/5 for every k. Hence L(P,f) = f(6/5)[1/5] + f(7/5)[1/5] + f(8/5)[1/5] + f(9/5)[1/5] + f(2)[1/5] = 1/5[5/6+ 5/7 + 5/8 + 5/9 + 1/2] = 1/6+ 1/7 + 1/8 + 1/9 + 1/10 ")
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2 To locate the maximum of f on the subintervals of [1,2] and compute U(P,f ), sketch the graph of f(x)=1/x: Then t 1 **=1, the point where f has its maximum value on the subinterval [1,6/5], t 2 **=6/5, the point where f has its maximum value on the subinterval [6/5,7/5], t 3 **=7/5, the point where f has its maximum value on the subinterval [7/5,8/5], t 4 **=8/5, the point where f has its maximum value on the subinterval [8/5,9/5] and t 5 **=9/5, the point where f has its maximum value on the subinterval [9/5,2]. Then U(P,f)=R(P,T**,f )= f(t 1 **)(x 1 -x 0 )+f(t 2 **)(x 2 -x 1 )+f(t 3 **)(x 3 -x 2 )+f(t 4 **)(x 4 -x 3 )+f(t 5 **)(x 5 -x 4 ) where x k - x k-1 =1/5 for every k. Thus U(P,f) = f(1)[1/5] + f(6/5)[1/5] + f(7/5)[1/5] + f(8/5)[1/5] + f(9/5)[1/5] = 1/5[1 + 5/6+ 5/7 + 5/8 + 5/9 ] = 1/5 + 1/6+ 1/7 + 1/8 + 1/9 0.746
![2 To locate the maximum of f on the subintervals of [1,2] and compute U(P,f ), sketch the graph of f(x)=1/x: Then t 1 **=1, the point where f has its maximum value on the subinterval [1,6/5], t 2 **=6/5, the point where f has its maximum value on the subinterval [6/5,7/5], t 3 **=7/5, the point where f has its maximum value on the subinterval [7/5,8/5], t 4 **=8/5, the point where f has its maximum value on the subinterval [8/5,9/5] and t 5 **=9/5, the point where f has its maximum value on the subinterval [9/5,2].](http://images.slideplayer.com./26/8401000/slides/slide_2.jpg "Then U(P,f)=R(P,T**,f )= f(t 1 **)(x 1 -x 0 )+f(t 2 **)(x 2 -x 1 )+f(t 3 **)(x 3 -x 2 )+f(t 4 **)(x 4 -x 3 )+f(t 5 **)(x 5 -x 4 ) where x k - x k-1 =1/5 for every k. Thus U(P,f) = f(1)[1/5] + f(6/5)[1/5] + f(7/5)[1/5] + f(8/5)[1/5] + f(9/5)[1/5] = 1/5[1 + 5/6+ 5/7 + 5/8 + 5/9 ] = 1/5 + 1/6+ 1/7 + 1/8 + 1/9 ")
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3 Let A be the area bounded by the graph of y=f(x) = 1/x, the x-axis and the lines x=1, x=2. Proposition 3.2.32 says: A ½ [L(P,f)+U(P,f)] = ½ [0.646 + 0.746] = 0.696 with error less than ½ [U(P,f)-L(P,f)] = ½ [0.746 – 0.646] = 0.500. Using the Riemann sum of Example 3.2C (2), the inequality L(P,f) R(P,T,f) U(P,f) is illustrated by: 0.646 0.692 0.746 L(P,f) 0.646 U(P,f) 0.746
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