Suppose S₃is the lower sum of the area enclosed by the graph $( x ) = \frac { 1 } { x }$ , the x-axis, and the lines x = 1 and x = 4 by partitioning [1, 4] into three subintervals [1, 2], [2, 3], and [3, 4]. Then S₃ is A) $\frac { 13 } { 12 }$ B) $\frac { 9 } { 8 }$ C) $\frac { 11 } { 6 }$ D)2 E) $\frac { 13 } { 6 }$

Question

Quizplus · Accepted Answer

The lower sum for the function $f(x) = \frac{1}{x}$ over the interval [1,4] with subintervals [1,2], [2,3], and [3,4] is calculated by taking the minimum value of $f(x)$ on each subinterval and multiplying it by the width of the subinterval. For [1,2], the minimum value is $\frac{1}{2}$, for [2,3], it's $\frac{1}{3}$, and for [3,4], it's $\frac{1}{4}$. Thus, the lower sum is $\frac{1}{2} \times 1 + \frac{1}{3} \times 1 + \frac{1}{4} \times 1 = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6+4+3}{12} = \frac{13}{12}$.

Suppose S3 is the Lower Sum of the Area Enclosed (x)=1x( x ) = \frac { 1 } { x }(x)=x1​

Suppose S₃is the Lower Sum of the Area Enclosed $( x ) = \frac { 1 } { x }$