Q. 65 - Calculus

Question

The proof of the latter part of the Second Fundamental Theorem of Calculus in the reading covered only the case $h\to 0^{+}$. Rewrite this proof in your own words, and then write a proof of what happens as $h\to 0^{-}$.

Step-by-Step Solution

Answer

The Second Fundamental Theorem states that if $f$ is continuous on $\left[a,b\right]$ then $F\left(x\right)={\int }_0^{x}f\left(t\right)dt$ for all $x\in \left[a,b\right]$, then

$F$ is continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ and $F$ is an antiderivative of $f$, i.e., $F\text{'}\left(x\right)=f\left(x\right)$.

Step 1. Given information

We have to proof the latter part of the Second Fundamental Theorem of Calculus the case when $h\to 0^{+}$.

Step 2. Proof of the given question

It is required to prove that $F$ is an antiderivative of $f$, i.e. $F\text{'}\left(x\right)=f\left(x\right)$.

Considering the right derivative, $h$ is positive.

So,

$F\left(x\right)={\int }_0^{x}f\left(t\right)dt$

Finding its derivative with respect to $x$,

$$\begin{gathered} F\text{'}\left(x\right)=\underset{h\to 0^{+}}{\lim }\frac{F\left(x+h\right)-F\left(x\right)}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{{\int }_a^{x+h}f\left(t\right)dt-{\int }_a^{x}f\left(t\right)dt}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{{\int }_a^{x+h}f\left(t\right)dt+{\int }_x^{a}f\left(t\right)dt}{h} \\ =\underset{h\to 0^{+}}{\lim }\frac{{\int }_x^{x+h}f\left(t\right)dt}{h} \end{gathered}$$

The Mean Value Theorem is,

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

Applying it to $\underset{h\to 0^{+}}{\lim }\frac{{\int }_x^{x+h}f\left(t\right)dt}{h}$

$$\begin{gathered} \underset{h\to 0^{+}}{\lim }\frac{{\int }_x^{x+h}f\left(t\right)dt}{h} \\ =\underset{h\to 0^{+}}{\lim }f\left(c\right) \end{gathered}$$

This means that there is some $c$ on $\left[x,x+h\right]$.

In the limit as $h$ goes to $0^{+}$, $c$ gets squeezed down to $x$.

Since $f\left(x\right)$ is continuous,

$$\begin{gathered} \underset{h\to \infty }{\lim }f\left(c\right) \\ =\underset{c\to x}{\lim }f\left(c\right) \\ =f\left(x\right) \end{gathered}$$

Therefore, $F\text{'}\left(x\right)=\underset{h\to 0^{+}}{\lim }\frac{F\left(x+h\right)-F\left(x\right)}{h}=f\left(x\right)$ is proved.