Question

If \(f\) is a positive function and \({f^{\prime \prime }}(x) < 0\)for, show that\({T_n} < \int_a^b f (x)dx < {M_n}\).

Short answer

It is proved that \({T_n} < \int_a^b f (x)dx < {M_n}\)

Step-by-step solution

Definition

Trapezoid Rule

If\((a,b)\)is divided into\(n\)subintervals, each of length\(\Delta x\), with endpoints at\(\left\{ {{x_0},{x_1},{x_2}, \ldots ,{x_n}} \right\}\), then\({T_n} = \frac{{\Delta x}}{2}\left( {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + \cdots + 2f\left( {{x_{n - 2}}} \right) + 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right){\rm{ }}\)

Midpoint Rule

If\((a,b)\)is divided into\(n\)subintervals, each of length\(\Delta x\)and\({m_i}\)is the midpoint of the\(i\)th subinterval, then\({M_n} = \sum\limits_{i = 1}^n f \left( {{m_i}} \right)\Delta x\).

For upper bound

It is given that\({\rm{f}}\)is a positive function and \({f^{\prime \prime }}(x) < 0\)for\(a \le x \le b\).

So\(f(x)\)is concave downward in the interval\((a,b)\).

As the curve is concave downward then tangent line will be above the graph so midpoint approximation will be overestimation of\(\int_a^b f (x)dx\).

So\(\int_a^b f (x)dx < {M_n}\,\,\,\,\,\,\, - - - - (1)\)

For lower bound

Now we look for lower bound using trapezoidal rule.

Since\(f(x)\) is concave downward so the line joining the points \((a,f(a))\) and \((b,f(b))\) will be under the graph

Here\({T_n}\)will be underestimation of\(\int_a^b f (x)dx\)

Then\({T_n} < \int_a^b f (x)dx\,\,\,\,\,\, - - - - - (2)\)

Combining equation\((1)\& \,(2)\)we have:\({T_n} < \int_a^b f (x)dx < {M_n}\)

It is proved that \({T_n} < \int_a^b f (x)dx < {M_n}\)