Question

Prove that the Net Change Theorem (Theorem 4.27) is equivalent to the Fundamental Theorem of Calculus.

Short answer

Ans:${\int }_a^{b}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(F\left(x_k\right)-F\left(x_{k-1}\right)\right)=\underset{n\to \infty }{\lim }F\left(b\right)-F\left(a\right)=F\left(b\right)-F\left(a\right)$

Step-by-step solution

Step 1. Given Information: 

The objective is to prove the Fundamental Theorem of Calculus.

Let, $f$be a continuous function on $[a,b]$and F be any anti-derivative of $f$.

Step 2. Proving with the help of the limit of Reimann sum :

So, the definite integral from $x=a$to $x=b$defined by a limit of Reimann sum is,

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left({x^{°}}_k\right)△x \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^{n}F\text{'}\left({x^{°}}_k\right)△x[F\text{'}=f] \\ \mathrm{Where},\mathrm{for}\mathrm{each}\mathrm{n},△\mathrm{x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}},{\mathrm{x}}_{\mathrm{k}}=\mathrm{a}+\mathrm{k}△\mathrm{x}\mathrm{and}{{\mathrm{x}}^{°}}_{\mathrm{k}}\mathrm{is}\mathrm{apoint}\mathrm{on}\mathrm{the}\mathrm{sub}\mathrm{interval}[{\mathrm{x}}_{\mathrm{k}+1},{\mathrm{x}}_{\mathrm{k}}] \\ \mathrm{NOw}\mathrm{applying}\mathrm{the}\mathrm{Mean}\mathrm{Value}\mathrm{Theorem}\mathrm{to}\mathrm{F}\mathrm{on}[\mathrm{a},\mathrm{b}]=[{\mathrm{x}}_{\mathrm{k}+1},{\mathrm{x}}_{\mathrm{k}}] \\ \mathrm{such}\mathrm{that}\mathrm{there}\mathrm{exists}\mathrm{some}\mathrm{points}{\mathrm{c}}_{\mathrm{k}}\in [{\mathrm{x}}_{\mathrm{k}+1},{\mathrm{x}}_{\mathrm{k}}]\mathrm{such}\mathrm{that}, \\ {\mathrm{F}}^{\text{'}}\left({\mathrm{c}}_{\mathrm{k}}\right)=\frac{\mathrm{F}\left({\mathrm{x}}_{\mathrm{k}}\right)-\mathrm{F}\left({\mathrm{x}}_{\mathrm{k}-1}\right)}{{\mathrm{x}}_{\mathrm{k}}-{\mathrm{x}}_{\mathrm{k}-1}} \end{gathered}$$

Step 3. Substituting the above expression for F'(x°k) in the Reimann sum:

$$\begin{gathered} {\int }_a^b f\left(x\right)dx \\ =\underset{n\to \infty }{\lim } \sum {\left(k=1\right)}^n F^{\text{'}}\left(x_k^{*}\right)\Delta x \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^{{}^n} \left(\frac{\left(F\right(x_k)-F(x_{k-1})}{(x_k-x_{k-1}}\right)\Delta x \\ =\left(F\right(x_1)-F(x_0\left)\right)+\left(F\right(x_2)-F(x_1\left)\right)+\cdots +\left(F\right(x_{(}n-1\left)\right)-F(x_{(}n-2)\left)\right)+\left(F\right(x_n)-F(x_{(}n-1\left)\right)) \\ =-F(x_0)+(F\left(x_1\right)-F\left(x_1\right))+(F\left(x_2\right)-F\left(x_2\right))+(F(x_{(}n-1))-F(x_{(}n-1\left)\right))+F(x_n) \\ =-F(x_0)+0+0+0+F(x_n)=F(b)-F(a) \end{gathered}$$

Step 4. Differentiation of the Net change theorem :

$$\begin{gathered} TheNetchangetheoremforthefunctionfwhichisdifferentiableon[a,b]is, \\ {\int }_a^{b}f\text{'}\left(x\right)dx=f\left(b\right)-f\left(a\right) \\ \therefore {\int }_a^{b}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(F\left(x_k\right)-F\left(x_{k-1}\right)\right)=\underset{n\to \infty }{\lim }F\left(b\right)-F\left(a\right)=F\left(b\right)-F\left(a\right) \end{gathered}$$