Q. 57 - Calculus

Question

Use the definition of the definite integral as a limit of Riemann sums to prove Theorem 4.12(b): For any function f that is integrable on $[a, b]$

${\int }_b^{a}f\left(x\right)dx=-{\int }_a^{b}f\left(x\right)dx$

Step-by-Step Solution

Answer

The theorem 4.12(b)  is proved.

${\int }_b^{a}f\left(x\right)dx=-{\int }_a^{b}f\left(x\right)dx$

Step 1. Given Information

We are given a function f that is integrable. 

Step 2. Proving the theorem

On the interval $[a, b]$, the value of $\Delta x$ is,

$\Delta x=\frac{b-a}{n}$

And on the interval $[b, a]$, the value of $\Delta x$ is,

$$\begin{gathered} \Delta x\text{'}=\frac{a-b}{n} \\ =-\left(\frac{b-a}{n}\right) \\ =-\Delta x \end{gathered}$$

Step 3. Proving the theorem

Proving the theorem, 

$$\begin{gathered} {\int }_b^{a}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{*}\right)\Delta x^{*} \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{*}\right)-\Delta x \\ =-{\int }_a^{b}f\left(x\right)dx \end{gathered}$$

Hence Proved.