Q. 66 - Calculus

Question

Prove that if $f$ is continuous on $[a,b]$ and we define $F\left(x\right)={\int }_a^{x}f\left(t\right)dt,$ then F is continuous on the closed interval $[a,b]$.

Step-by-Step Solution

Answer

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

Step 1. Given information

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

Step 2. Calculation

$f$ is continuous on $[a,b].$

Therefore, from definition of continuity

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

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

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

Now, replace $x by x+h$

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

Now, replace $x$ by $x-h$

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

So,

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

Again,

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

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

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

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

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