91Ó°ÊÓ

The given data is if f is integrable and concave down on [a,b] and n is any positive integer.

Consider an integral function f that is concave down on [a,b]

Let n be a positive integer.

Consider a partition of [a,b] into sub-intervals of the form $[x_{k-1},x_k]\mathrm{where}1≤k≤n$

Define $∆x=\frac{b-a}{n}\mathrm{and}{x}_k=a+k∆x$

Then n -rectangle midpoint sum and n-rectangle trapezoid sum is determined as follows,

$MID\left(n\right)=\underset{k=1}{\overset{n}{∑}}f\left(\frac{x_{k-1}+x_k}{2}\right)∆xandTRAP\left(n\right)=\underset{k=1}{\overset{n}{∑}}\left(\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}\right)∆x$

Consider a concave down function f with the interval partitioned into 5 trapezoidal sub-intervals.

From the diagram, in each of the sub-intervals, the area of trapezoid is less than or equal to the area under the curve $f\left(x\right)$

Sum up the areas in all the n sub-intervals to obtain the following inequality,

$TRAP\left(n\right)≤{∫}_a^{b}f\left(x\right)dx$

Consider a concave down function f with the interval partitioned into 5 midpoint rectangles.

From the diagram, we observe that the tangent line to the concave down function through the midpoint is above the curve.

That means, in each sub-interval, the midpoint trapezoid has more area than the curve area under the curve$f\left(x\right)$

Since, the midpoint trapezoid have the same area as the midpoint rectangles, sum up the areas in all the n sub-intervals to obtain the following inequality,

${∫}_a^{b}f\left(x\right)dx≤MID\left(n\right)$

Thus, for a concave down function f on [a,b] and a positive integer n,

$TRAP\left(n\right)≤{∫}_a^{b}f\left(x\right)dx≤MID\left(n\right)$