let ( f(x) ) be a function which is continuous for all ( x ). let ( l_{100}, r_{100} ) and ( m_{100} ) be…

let ( f(x) ) be a function which is continuous for all ( x ). let ( l_{100}, r_{100} ) and ( m_{100} ) be the riemann sums using 100 subintervals with left, right and middle sample points, respectively, for ( f ) on the interval ( 0,10 ). which of the following statements is false?\n\nif ( f ) is increasing on ( 0,10 ), then ( l_{100} leq r_{100} ).\n\nif ( f ) is decreasing on ( 0,10 ), then ( l_{100} geq r_{100} )\n\nthe definite integral ( int_{0}^{10} f(x) d x ) exists (i.e. ( f ) is riemann integrable on ( 0,10 ) ).\n\nif ( f ) is increasing on ( 0,5) ) and decreasing on ( (5,10 ) then ( l_{100}=r_{100} ).\n\nall three sums ( l_{100}, r_{100}, m_{100} ) exist.

let ( f(x) ) be a function which is continuous for all ( x ). let ( l_{100}, r_{100} ) and ( m_{100} ) be the riemann sums using 100 subintervals with left, right and middle sample points, respectively, for ( f ) on the interval ( 0,10 ). which of the following statements is false?\n\nif ( f ) is increasing on ( 0,10 ), then ( l_{100} leq r_{100} ).\n\nif ( f ) is decreasing on ( 0,10 ), then ( l_{100} geq r_{100} )\n\nthe definite integral ( int_{0}^{10} f(x) d x ) exists (i.e. ( f ) is riemann integrable on ( 0,10 ) ).\n\nif ( f ) is increasing on ( 0,5) ) and decreasing on ( (5,10 ) then ( l_{100}=r_{100} ).\n\nall three sums ( l_{100}, r_{100}, m_{100} ) exist.

Answer

Explanation:

Step1: Analyze the first option

If (f(x)) is increasing on ([0,10]), then (f(x_{i - 1})\leq f(x_{i})) for (i = 1,\cdots,100). The left - Riemann sum (L_{100}=\sum_{i = 1}^{100}f(x_{i-1})\Delta x) and the right - Riemann sum (R_{100}=\sum_{i = 1}^{100}f(x_{i})\Delta x) with (\Delta x=\frac{10 - 0}{100}=0.1). Since (f(x_{i-1})\leq f(x_{i})), we have (L_{100}\leq R_{100}).

Step2: Analyze the second option

If (f(x)) is decreasing on ([0,10]), then (f(x_{i - 1})\geq f(x_{i})) for (i = 1,\cdots,100). Using the formulas (L_{100}=\sum_{i = 1}^{100}f(x_{i-1})\Delta x) and (R_{100}=\sum_{i = 1}^{100}f(x_{i})\Delta x) with (\Delta x = 0.1), we get (L_{100}\geq R_{100}).

Step3: Analyze the third option

Since (f(x)) is continuous on ([0,10]), by the fundamental theorem of calculus, if a function (y = f(x)) is continuous on the closed interval ([a,b]), then (\int_{a}^{b}f(x)dx) exists. Here (a = 0), (b = 10), so (\int_{0}^{10}f(x)dx) exists.

Step4: Analyze the fourth option

Let (f(x)) be increasing on ([0,5)) and decreasing on ((5,10]). The left - Riemann sum (L_{100}=\sum_{i = 1}^{100}f(x_{i-1})\Delta x) and the right - Riemann sum (R_{100}=\sum_{i = 1}^{100}f(x_{i})\Delta x). The sample points for (L_{100}) are (x_{i-1}=0+(i - 1)\times0.1) and for (R_{100}) are (x_{i}=0 + i\times0.1). In general, (L_{100}\neq R_{100}) unless (f(x)) is a constant function (which is not the case here as (f) has different monotonicity on sub - intervals).

Step5: Analyze the fifth option

Since (f(x)) is continuous on ([0,10]), and for a continuous function (y = f(x)) on ([0,10]), and for a partition (P={x_0,x_1,\cdots,x_{100}}) of ([0,10]) with (x_i=0 + i\times0.1), the left - Riemann sum (L_{100}=\sum_{i = 1}^{100}f(x_{i-1})\Delta x), the right - Riemann sum (R_{100}=\sum_{i = 1}^{100}f(x_{i})\Delta x) and the mid - point Riemann sum (M_{100}=\sum_{i = 1}^{100}f(\frac{x_{i-1}+x_{i}}{2})\Delta x) all exist.

Answer:

If (f) is increasing on ([0,5)) and decreasing on ((5,10]) then (L_{100}=R_{100}) (the fourth option).