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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 t1*=6/5, the point where f has its minimum value on the subinterval [1,6/5], t2*=7/5, the point where f has its minimum value on the subinterval [6/5,7/5], t3*=8/5, the point where f has its minimum value on the subinterval [7/5,8/5] , t4*=9/5, the point where f has its minimum value on the subinterval [8/5,9/5] and t5*=2, the point where f has its minimum value on the subinterval [9/5,2]. Then

L(P,f)=R(P,T*,f )= f(t1*)(x1-x0)+ f(t2*)(x2-x1)+ f(t3*)(x3-x2)+ f(t4*)(x4-x3)+ f(t5*)(x5-x4)

where x0=1, x1=6/5, x2=7/5, x3=8/5, x4=9/5, x5=2. Observe that xk - xk-1=1/5 for every k. HenceL(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

)(x

xf

y

x6/5 7/5 8/5 9/5 21

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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 t1**=1, the point where f has its maximum value on the subinterval [1,6/5], t2**=6/5, the point where f has its maximum value on the subinterval [6/5,7/5], t3**=7/5, the point where f has its maximum value on the subinterval [7/5,8/5] , t4**=8/5, the point where f has its maximum value on the subinterval [8/5,9/5] and t5**=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(t1**)(x1-x0)+f(t2**)(x2-x1)+f(t3**)(x3-x2)+f(t4**)(x4-x3)+f(t5**)(x5-x4)

where xk - xk-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

y

x6/5 7/5 8/5 9/5 21

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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