Q. 78 - Calculus

Question

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

Step-by-Step Solution

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 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 :

$$\begin{gathered} The Net change theorem for the function f which is differentiable on [a,b] is , \\ {\int }_a^{b}f\text{'}\left(x\right)dx=f\left(b\right)-f\left(a\right) \\ \therefore {\int }_a^{b}f\left(x\right)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(F\left(x_k\right)-F\left(x_{k-1}\right)\right)=\underset{n\to \infty }{\lim }F\left(b\right)-F\left(a\right)=F\left(b\right)-F\left(a\right) \end{gathered}$$