91影视

To proof$f$is continuous on$[a,b].$

$f$is continuous on $[a,b].$

Therefore, from definition of continuity

$f$is right continuous on $[a,b]$and $f$is left continuous on $[a,b].$

$\underset{h鈫�0^{+}}{\lim }\frac{f(x+h)-f\left(x\right)}{h}=\underset{h鈫�0^{-}}{\lim }\frac{f(x-h)-f\left(x\right)}{h}$ (1)

Now, $F\left(x\right)={鈭�}_a^{x}f\left(t\right)dt$defines the area function for $x鈭�(a,b)$

Now, replace $xbyx+h$

$F(x+h)={鈭�}_a^{x+h}f\left(t\right)dt$

Now, replace $x$by $x-h$

$F(x-h)={鈭�}_0^{x-h}f\left(t\right)dt$

So,

$$\begin{gathered} \underset{h鈫�0^{+}}{\lim }\frac{F(x+h)-F\left(x\right)}{h}=\underset{h鈫�0^{+}}{\lim }\frac{{鈭�}_2^{x+h}f\left(t\right)dt-{鈭�}_a^{x}f\left(t\right)dt}{h} \\ =\underset{h鈫�0^{+}}{\lim }\frac{{鈭�}_2^{x+h}f\left(t\right)dt+{鈭�}_a^{x}f\left(t\right)dt}{h} \\ =\underset{h鈫�0^{+}}{\lim }\frac{{鈭�}_x^{x+h}f\left(t\right)dt}{h} \\ =\underset{h鈫�0^{+}}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \end{gathered}$$ (2)

Again,

$$\begin{gathered} \underset{h鈫�0^{-}}{\lim }\frac{F(x-h)-F\left(x\right)}{-h}=\underset{h鈫�0^{-}}{\lim }\frac{{鈭�}_a^{x-h}f\left(t\right)dt-{鈭�}_a^{x}f\left(t\right)dt}{-h} \\ =\underset{h鈫�0^{-}}{\lim }\frac{{鈭�}_a^{x-h}f\left(t\right)dt+{鈭�}_a^{x}f\left(t\right)dt}{-h} \\ =\underset{h鈫�0^{-}}{\lim }\frac{{鈭�}_x^{x-h}f\left(t\right)dt}{-h} \\ =\underset{h鈫�0^{-}}{\lim }\frac{f(x-h)-f\left(x\right)}{-h} \end{gathered}$$ (3)

So, from (1),(2) and (3)

$\underset{h鈫�0^{+}}{\lim }\frac{F(x+h)-f\left(x\right)}{h}=\underset{h鈫�0^{-}}{\lim }\frac{F(x-h)-F\left(x\right)}{-h}$

Therefore, $F$is right continuous on $[a,b]$and $F$is left continuous on $[a,b].$

Hence, $F\left(x\right)={鈭�}_a^{x}f\left(t\right)dt$is continuous on$[a,b].$