Q. 16 - Calculus

Question

Explain how we get the inequality

$f\left(m_h\right)h\le {\int }_x^{x+h}f\left(t\right)dt\le f\left(M_h\right)h$

in the proof of the Second Fundamental Theorem of Calculus. Make sure you define $m_h$ and $M_h$ clearly.

Step-by-Step Solution

Answer

Since $m_h<f\left(x\right)<M_h$, $f\left(m_h\right)h\le {\int }_x^{x+h}f\left(t\right)dt\le f\left(M_h\right)h$. This is how we get the required inequality.

Step 1 . Given information

$f\left(m_h\right)h\le {\int }_x^{x+h}f\left(t\right)dt\le f\left(M_h\right)h$.

Step 2 . The objective is to explain how the given inequality is achieved.

The Extreme value Theorem states that if a real-valued function $f$ is continuous in the closed and bounded interval $\left[a,b\right]$, then $f$ must attain a maximum and a minimum, each at least once. That is, there exist numbers $c$ and $d$ in $\left[a,b\right]$ such that:

$f\left(c\right)\ge f\left(x\right)\ge f\left(d\right)$.

So,

$$\begin{gathered} M_h>f\left(x\right)>m_h \\ {m}_k<f\left(x\right)<M_h \end{gathered}$$

where $M_h$ is the maximum value on $\left[a,b\right]$ and $m_h$ is the minimum value on $\left[a,b\right]$.

Hence, on $\left[x,x+h\right]$ the area of the rectangles using left and right sum can be written as, $f\left(m_h\right)h$ and $f\left(M_h\right)h$ respectively.

Since, $m_k<f\left(x\right)<M_h$ so, $f\left(m_h\right)h\le {\int }_x^{x+h}f\left(t\right)dt\le f\left(M_h\right)h$.